problem solving linear equations

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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Word Problems Linear Equations

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\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

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• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
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Types of Linear Programming Problems

Linear programming is a mathematical technique for optimizing operations under a given set of constraints. The basic goal of linear programming is to maximize or minimize the total numerical value. It is regarded as one of the most essential strategies for determining optimum resource utilization. Linear programming challenges include a variety of problems involving cost minimization and profit maximization, among others. They will be briefly discussed in this article.

The purpose of this article is to provide students with a comprehensive understanding of the different types of linear programming problems and their solutions.

What is Linear Programming?

Linear programming (LP) is a mathematical optimization technique used to maximize or minimize a linear objective function, subject to a set of linear equality and inequality constraints. It is widely used in various fields such as economics, engineering, operations research, and management science to find the best possible outcome given limited resources.

Components of Linear Programming

Components of linear programming include:

• Objective Function: This is a linear function that needs to be optimized (maximized or minimized). It represents the quantity to be maximized or minimized, such as profit, cost, time, etc.
• Decision Variables: These are the variables that represent the choices or decisions to be made. They are the unknown quantities that the optimization process seeks to determine. Decision variables must be continuous and can take any real value within a specified range.
• Constraints: These are restrictions or limitations on the decision variables that must be satisfied. Constraints can be expressed as linear equations or inequalities. They represent the limitations imposed by available resources, capacity constraints, demand requirements, etc.
• Feasible Region: The feasible region is the set of all possible combinations of decision variables that satisfy all constraints. It is defined by the intersection of the constraint boundaries.
• Optimal Solution: This is the best possible solution that maximizes or minimizes the objective function while satisfying all constraints. In graphical terms, it is the point within the feasible region that maximizes or minimizes the objective function.

Linear programming provides a systematic and efficient approach to decision-making in situations where resources are limited and objectives need to be optimized.

Different Types of Linear Programming Problems

The following are the types of linear programming problems:

• Manufacturing problems
• Diet problems
• Transportation problems
• Optimal alignment problem

Let’s discuss more about each of them.

Manufacturing Problems

In these problems, we evaluate the number of units of various items that should be produced and sold by a company when each product requires a given number of workforce, machine hours, labour hours per unit of product, warehouse space per unit of output, and so on, to maximize profit.

Manufacturing problems involve maximizing the production rate or net profits of manufactured products, which might measure the available workspace, the number of workers, machine hours, packing materials used, raw materials required, the product’s market value, and other factors. These are commonly used in the industrial sector to anticipate a company’s future capital increase over time.

Diet Problems

In these challenges, we assess how many components or nutrients a diet should contain in order to lower the cost of the desired diet while guaranteeing the minimal amount of each vitamin.

As the name suggests, diet-related problems can be resolved by eating more particular foods that are rich in essential nutrients and can support the adoption of a particular diet plan. Finding a set of meals that will satisfy a set of daily nutritional demands for the least amount of money is the aim of a diet problem.

Transportation Problems

In these problems , we create a transportation schedule to discover the most cost-effective method of carrying a product from various plants/factories to various markets.

The study of transportation routes or how items from diverse production sources are transported to various markets to minimize the total transportation cost is linked to transportation difficulties. Analyzing such challenges is crucial for large firms with several production units and a broad customer base.

Optimal Assignment Problems

This problem addresses a company’s completion of a given task/assignment by selecting a specific number of employees to complete the assignment within the required timeframe, assuming that each person works on only one job. Event planning and management in major organizations, for example, are examples of such problems.

Constraints and Objective Function of Each Linear Programming Problem

Steps for solving linear programming problems.

Step 1: Identify the decision variables : The first step is to determine which choice factors control the behaviour of the objective function. A function that needs to be optimised is an objective function. Determine the decision variables and designate them with X, Y, and Z symbols.

Step 2: Form an objective function : Using the decision variables, write out an algebraic expression that displays the quantity we aim to maximize.

Step 3: Identify the constraints : Choose and write the given linear inequalities from the data.

Step 4: Draw the graph for the given data : Construct the graph by using constraints for solving the linear programming problem.

Step 5: Draw the feasible region : Every constraint on the problem is satisfied by this portion of the graph. Anywhere in the feasible zone is a viable solution for the objective function.

Step 6: Choosing the optimal point : Choose the point for which the given function has maximum or minimum values.

Solved Problems of Linear Programming Problems

Question 1. A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

We define our unknowns:

Let the number of regular gadgets manufactured each day = x

and the number of premium gadgets manufactured each day = y

The objective function is

P = 20x + 30y

We now write the constraints. The fourth sentence states that the company can make at most 7 gadgets a day. This translates as

Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get

x + 2y ≤ 12

Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.

2x + y ≤ 12

The fact that x and y can never be negative is represented by the following two constraints:

x ≥ 0, and y ≥ 0.

We have formulated the problem as follows :

Maximize P=20x + 30y Subject to : x + y ≤ 7, x + 2y ≤ 122, x + y ≤ 12, x ≥ 0, y ≥ 0

In order to solve the problem, we next graph the constraints and feasible region.

Again, we have shaded the feasible region, where all constraints are satisfied.

Since the extreme value of the objective function always takes place at the vertices of the feasible region, we identify all the critical points. They are listed as (0, 0), (0, 6), (2, 5), (5, 2), and (6, 0). To maximize profit, we will substitute these points in the objective function to see which point gives us the maximum profit each day. The results are listed below.

FAQ on Linear programming

How many methods are there in lpp.

There are different methods to solve a linear programming problem. Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods.

What are the four 4 special cases in linear programming?

Four special cases and difficulties arise at times when using the graphical approach to solving LP problems: (1) infeasibility, (2) unboundedness, (3) redundancy, and (4) alternate optimal solutions.

What are the 3 components of linear programming?

The basic components of the LP are as follows: Decision Variables. Constraints. Objective Functions.

What are the applications of LPP?

LPP applications may include production scheduling, inventory policies, investment portfolio, allocation of advertising budget, construction of warehouses, etc.

What are the limitations of LPP?

Constraints (limitations) should be expressed in mathematical form. Relationships between two or more variables should be linear. The values of the variables should always be non-negative or zero. There should always be finite and infinite inputs and output numbers.

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Unit 3: Linear equations and inequalities

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1.1: Solving Linear Equations and Inequalities

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Learning Objectives

• Verify linear solutions.
• Use the properties of equality to solve basic linear equations.
• Clear fractions from equations.
• Identify linear inequalities and check solutions.
• Solve linear inequalities and express the solutions graphically on a number line and in interval notation.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Simplify \(2-6(4-7)^2\) without using a calculator.

If you missed this problem, review here . (Note that this will open a different textbook in a new window.)

2. Evaluate \(6x−4\) when \(x=−2\).

3. Evaluate \(-5x^2−x+9\) when \(x=-3\).

4. Simplify \(7x−1−4x+5\).

\(3x+4\) 

Solving Basic Linear Equations

An equation 129 is a statement indicating that two algebraic expressions are equal. A linear equation with one variable 130 , \(x\), is an equation that can be written in the standard form \(ax + b = 0\) where \(a\) and \(b\) are real numbers and \(a ≠ 0\). For example

\(3 x - 12 = 0\)

A solution 131 to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation \(3x − 12 = 0\) is \(x\) and the solution is \(x = 4\). To verify this, substitute the value \(4\) in for \(x\) and check that you obtain a true statement.

\(\begin{aligned} 3 x - 12 & = 0 \\ 3 ( \color{Cerulean}{4}\color{Black}{ )} - 12 & = 0 \\ 12 - 12 & = 0 \\ 0 & = 0 \:\: \color{Cerulean}{✓} \end{aligned}\)

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example \(\PageIndex{1}\):

Is \(a=2\) a solution to \(−10a+5=−25\)?

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

\(\begin{align*} - 10 a + 5 &= -25 \\ - 10 ( \color{Cerulean}{2} \color{Black}{ ) +} 5 & = -25 \\ -20 + 5 & = -25 \\ -15 &\neq 25\:\: \color{red}{✗}\end{align*}\)  

No, \(a=2\) does not satisfy the equation and is therefore not a solution.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations 132 as equations with the same solution set.

\(\left. \begin{aligned} 3 x - 5 & = 16 \\ 3 x & = 21 \\ x & = 7 \end{aligned} \right\} \quad \color{Cerulean}{Equivalent \:equations}\)

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, \(\{7\}\). To obtain equivalent equations, use the following properties of equality 133 . Given algebraic expressions \(A\) and \(B\), where \(c\) is a nonzero number:

Table 1.1.1

Multiplying or dividing both sides of an equation by \(0\) is carefully avoided. Dividing by \(0\) is undefined and multiplying both sides by \(0\) results in the equation \(0 = 0\).

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form \(ax + b = c\), then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Example \(\PageIndex{2}\):

Solve: \(7x − 2 = 19\).

\(\begin{aligned} 7 x - 2 & = 19 \\ 7 x - 2 \color{Cerulean}{+ 2} & = 19 \color{Cerulean}{+ 2} & & \color{Cerulean}{Add\: 2\: to\: both\: sides.} \\ 7 x & = 21 \\ \frac { 7 x } { \color{Cerulean}{7} } & = \frac { 21 } { \color{Cerulean}{7} } & & \color{Cerulean}{Divide\: both\: sides\: by\: 7.} \\ x & = 3 \end{aligned}\)

The solution is \(3\).

Example \(\PageIndex{3}\):

Solve: \(56 = 8 + 12y\).

When no sign precedes the term, it is understood to be positive. In other words, think of this as \(56 = +8 + 12y\). Therefore, we begin by subtracting \(8\) on both sides of the equal sign.

\(\begin{aligned} 56 \color{Cerulean}{- 8} & = 8 + 12 y \color{Cerulean}{- 8} \\ 48 & = 12 y \\ \frac { 48 } { \color{Cerulean}{12} } & = \frac { 12 y } { \color{Cerulean}{12} } \\ 4 & = y \end{aligned}\)

It does not matter on which side we choose to isolate the variable because the symmetric property 134 states that \(4 = y\) is equivalent to \(y = 4\).

The solution is \(4\).

Example \(\PageIndex{4}\):

Solve: \(\frac { 5 } { 3 } x + 2 = - 8\).

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient \(\frac{5}{3}\) .

\begin{aligned} \frac { 5 } { 3 } x + 2 & = - 8 \\ \frac { 5 } { 3 } x + 2 \color{Cerulean}{- 2} & = - 8 \color{Cerulean}{- 2}\quad \color{Cerulean}{Subtract\: 2\: on\: both\: sides.} \\ \frac { 5 } { 3 } x & = - 10 \\ \color{Cerulean}{\frac { 3 } { 5 }} \color{Black}{ \cdot} \frac { 5 } { 3 } x & = \color{Cerulean}{\frac { 3 } { \cancel{5} }} \color{Black}{\cdot} ( \overset{-2}{\cancel{-10}} )\quad \color{Cerulean}{Multiply \:both \:sides\: by\: \frac{3}{5}.} \\ 1x & = 3 \cdot ( - 2 ) \\ x & = - 6 \end{aligned}

The solution is \(−6\).

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Exercise \(\PageIndex{1}\)

Solve: \(\frac { 2 } { 3 } x + \frac { 1 } { 2 } = - \frac { 5 } { 6 }\).

Video Solution: www.youtube.com/v/cQwqXs9AD6M

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

• Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
• Step 2a: Add or subtract as needed to isolate the variable.
• Step 2b: Divide or multiply as needed to isolate the variable.
• Step 3: Check to see if the answer solves the original equation.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Example \(\PageIndex{5}\):

Solve: \(- 4 a + 2 - a = 1\).

First combine the like terms on the left side of the equal sign.

\(\begin{aligned} - 4 a + 2 - a = 1 & \quad \color{Cerulean}{ Combine\: same-side\: like\: terms.} \\ - 5 a + 2 = 1 & \quad\color{Cerulean} { Subtract\: 2\: on\: both\: sides.} \\ - 5 a = - 1 & \quad\color{Cerulean} { Divide\: both\: sides\: by\: - 5.} \\ a = \frac { - 1 } { - 5 } = \frac { 1 } { 5 } \end{aligned}\)

Always use the original equation to check to see if the solution is correct.

\(\begin{aligned} - 4 a + 2 - a & = - 4 \left( \color{OliveGreen}{\frac { 1 } { 5 }} \right) + 2 - \color{OliveGreen}{\frac { 1 } { 5 }} \\ & = - \frac { 4 } { 5 } + \frac { 2 } { 1 } \cdot \color{Cerulean}{\frac { 5 } { 5 }}\color{Black}{ -} \frac { 1 } { 5 } \\ & = \frac { - 4 + 10 + 1 } { 5 } \\ & = \frac { 5 } { 5 } = 1 \:\:\color{Cerulean}{✓} \end{aligned}\)

The solution is \(\frac{1}{5}\) .

Given a linear equation in the form \(ax + b = cx + d\), we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Example \(\PageIndex{6}\):

Solve: \(−2y − 3 = 5y + 11\).

Subtract \(5y\) on both sides so that we can combine the terms involving y on the left side.

\(\begin{array} { c } { - 2 y - 3 \color{Cerulean}{- 5 y}\color{Black}{ =} 5 y + 11 \color{Cerulean}{- 5 y} } \\ { - 7 y - 3 = 11 } \end{array}\)

From here, solve using the techniques developed previously.

\(\begin{aligned} - 7 y - 3 & = 11 \quad\color{Cerulean}{Add\: 3\: to\: both\: sides.} \\ - 7 y & = 14 \\ y & = \frac { 14 } { - 7 } \quad\color{Cerulean}{Divide\: both\: sides\: by\: -7.} \\ y & = - 2 \end{aligned}\)

The solution is \(−2\).

Solving will often require the application of the distributive property.

Example \(\PageIndex{7}\):

Solve: \(- \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x )\).

Simplify the linear expressions on either side of the equal sign first.

\(\begin{aligned} - \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x ) & \quad\color{Cerulean} { Distribute } \\ - 5 x + 1 + 3 = 7 - 14 x & \quad\color{Cerulean} { Combine\: same-side\: like\: terms. } \\ - 5 x + 4 = 7 - 14 x & \quad\color{Cerulean} { Combine\: opposite-side\: like\: terms. } \\ 9 x = 3 & \quad\color{Cerulean} { Solve. } \\ x = \frac { 3 } { 9 } = \frac { 1 } { 3 } \end{aligned}\)

The solution is \(\frac{1}{3}\) .

Example \(\PageIndex{8}\):

Solve: \(5(3−a)−2(5−2a)=3\).

Begin by applying the distributive property.

\(\begin{aligned} 5 ( 3 - a ) - 2 ( 5 - 2 a ) & = 3 \\ 15 - 5 a - 10 + 4 a & = 3 \\ 5 - a & = 3 \\ - a & = - 2 \end{aligned}\)

Here we point out that \(−a\) is equivalent to \(−1a\); therefore, we choose to divide both sides of the equation by \(−1\).

\(\begin{array} { c } { - a = - 2 } \\ { \frac { - 1 a } { \color{Cerulean}{- 1} }\color{Black}{ =} \frac { - 2 } { \color{Cerulean}{- 1} } } \\ { a = 2 } \end{array}\)

Alternatively, we can multiply both sides of \(−a=−2\) by negative one and achieve the same result.

\(\begin{aligned} - a & = - 2 \\ \color{Cerulean}{( - 1 )}\color{Black}{ (} - a ) & = \color{Cerulean}{( - 1 )}\color{Black}{ (} - 2 ) \\ a & = 2 \end{aligned}\)

The solution is \(2\).

Exercise \(\PageIndex{2}\)

Solve: \(6 - 3 ( 4 x - 1 ) = 4 x - 7\).

Video Solution: www.youtube.com/v/NAIAZrFjU-o

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example \(\PageIndex{9}\):

Solve: \(\frac { 1 } { 3 } x + \frac { 1 } { 5 } = \frac { 1 } { 5 } x - 1\).

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the \(LCD (3, 5) = 15\).

\(\begin{aligned} \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 3 } x + \frac { 1 } { 5 } \right) & = \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 5 } x - 1 \right) \quad \color{Cerulean}{Multiply\: both\: sides\: by\: 15.} \\ \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 3 } x + \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } & = \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } x - \color{Cerulean}{15}\color{Black}{ \cdot} 1\quad\color{Cerulean}{Simplify.} \\ 5 x + 3 & = 3 x - 15\quad\quad\quad\color{Cerulean}{Solve.} \\ 2 x & = - 18 \\ x & = \frac { - 18 } { 2 } = - 9 \end{aligned}\)

The solution is \(−9\).

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

Table 1.1.2

We simplify expressions and solve equations. If you multiply an expression by \(6\), you will change the problem. However, if you multiply both sides of an equation by \(6\), you obtain an equivalent equation.

Table 1.1.3

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Table 1.1.4

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “ let x represent… ” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

Example \(\PageIndex{10}\):

When \(6\) is subtracted from twice the sum of a number and \(8\) the result is \(5\). Represent this as an algebraic equation and find the number.

Let n represent the unknown number.

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

Table 1.1.5

The key was to focus on the phrase “ twice the sum ,” this prompted us to group the sum within parentheses and then multiply by \(2\). After translating the sentence into a mathematical statement we then solve.

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 5 \\ 2 n + 16 - 6 & = 5 \\ 2 n + 10 & = 5 \\ 2 n & = - 5 \\ n & = \frac { - 5 } { 2 } \end{aligned}\)

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 2 \left( \color{Cerulean}{- \frac { 5 } { 2 }}\color{Black}{ +} 8 \right) - 6 \\ & = 2 \left( \frac { 11 } { 2 } \right) - 6 \\ & = 11 - 6 \\ & = 5 \quad\color{Cerulean}{✓}\end{aligned}\)       

The number is \(−\frac{5}{2}\).

General guidelines for setting up and solving word problems follow.

• Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
• Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
• Step 3: Translate and set up an algebraic equation that models the problem.
• Step 4: Solve the resulting algebraic equation.
• Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Linear Inequalities

A linear inequality 138 is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities, all of which are solved in this section:

A solution to a linear inequality 139 is a real number that will produce a true statement when substituted for the variable.

Example \(\PageIndex{11}\):

Are \(x=−4\) and \(x=6\) solutions to \(5x+7<22\)?

Substitute the values in for \(x\), simplify, and check to see if we obtain a true statement.

Table 1.1.6

\(x=−4\) is a solution and \(x=6\) is not

Linear inequalities have either infinitely many solutions or no solution. If there are infinitely many solutions, we graph the solution set on a number line and/or express the solution using interval notation.

Expressing Solutions to Linear Inequalities

What number would make the inequality \(x>3\) true? Are you thinking, "\(x\) could be four"? That’s correct, but \(x\) could be 6, too, or 137, or even 3.0001. Any number greater than three is a solution to the inequality \(x>3\). We show all the solutions to the inequality \(x>3\) on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.

We can also represent inequalities using interval notation . There is no upper end to the solution to this inequality. In interval notation, we express \(x>3\) as \((3,\infty)\). The symbol \(\infty\) is read as “ infinity .” It is not an actual number. Figure \(\PageIndex{2}\) shows both the number line and the interval notation.

We use the left parenthesis symbol, (, to show that the endpoint of the inequality is not included. The left bracket symbol, [, would show that the endpoint is included.

The inequality \(x\leq 1\) means all numbers less than or equal to one. Here we need to show that one is a solution, too. We do that by putting a bracket at \(x=1\). We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions (Figure \(\PageIndex{3}\)). There is no lower end to those numbers. We write \(x\leq 1x\leq 1 \)in interval notation as \((−\infty,1]\). The symbol \(−\infty\) is read as “negative infinity.”

Figure \(\PageIndex{4}\) shows both the number line and interval notation.

INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION

The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals. Notice that \(\infty\) and \(-\infty\) always use parentheses in interval notation, never brackets.

Example \(\PageIndex{12}\)

Graph each inequality on the number line and write in interval notation.

• \(x\geq −3\)
• \(x<2.5\)
• \(x\leq −\frac{3}{5}\)

Exercise \(\PageIndex{3}\)

Graph each inequality on the number line and write in interval notation:

• \(x\leq −4\)
• \(x\geq 0.5\)
• \(x<−\frac{2}{3}\).

Solving Linear Inequalities

All but one of the techniques learned for solving linear equations apply to solving linear inequalities. You may add or subtract any real number to both sides of an inequality, and you may multiply or divide both sides by any positive real number to create equivalent inequalities. For example:

\[\begin{align*} 10 &> - 5 \\[4pt] 10{\color{Cerulean}{-7}}\,&{\color{Black}{>}} -5{\color{Cerulean}{-7}} & & {\color{Cerulean}{Subtract\: 7\: on\: both\: sides.}}\\[4pt]  3 &> - 12 & & \color{Cerulean}{✓}\quad\color{Cerulean}{True.} \\[20pt] 10 &>-5\\[4pt] \frac{10}{\color{Cerulean}{5}}\,&\color{Black}{>}\frac{-5}{\color{Cerulean}{5}} & & \color{Cerulean}{Divide\: both\: sides\: by\: 5.}\\[4pt] 2 &>-1 & & \color{Cerulean}{✓\:\:True} \end{align*} \]

Subtracting \(7\) from each side and dividing each side by positive \(5\) results in an inequality that is true.

Example \(\PageIndex{13}\):

Solve and graph the solution set: \(5x+7<22\).

\(\begin{array} { c } { 5 x + 7 < 22 } \\ { 5 x + 7 \color{Cerulean}{- 7}\color{Black}{ < 22}\color{Cerulean}{ - 7} } \\ { 5 x < 15 } \\ { \frac { 5 x } {\color{Cerulean}{ 5} } < \frac { 15 } { \color{Cerulean}{5} } } \\ { x < 3 } \end{array}\)

It is helpful to take a minute and choose a few values in and out of the solution set, substitute them into the original inequality, and then verify the results. As indicated, you should expect \(x=0\) to solve the original inequality and that \(x=5\) should not.

Table 1.1.7

Checking in this manner gives us a good indication that we have solved the inequality correctly.

We can express this solution in two ways: using set notation and interval notation.

\(\begin{array} { r } { \{ x | x < 3 \} } &\color{Cerulean}{Set\: notation} \\ { ( - \infty , 3 ) } &\color{Cerulean}{Interval\: notation} \end{array}\)

In this text we will choose to present answers using interval notation.

\((−∞,  3) \)

When working with linear inequalities, a different rule applies when multiplying or dividing by a negative number. To illustrate the problem, consider the true statement \(10 > −5\) and divide both sides by \(−5\).

\(\begin{array} { l } { 10 > - 5 } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{>} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Divide\: both\: sides\: by\: -5.} \\ { - 2 \color{red}{>}\color{Black}{ 1} \quad \color{red}{✗} \color{Cerulean}{ False } } \end{array}\)

Dividing by \(−5\) results in a false statement. To retain a true statement, the inequality must be reversed.

\(\begin{array} { l } { 10 \color{OliveGreen}{>}\color{Black}{ - 5} } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{<} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Reverse\: the\: inequality.} \\ { - 2 \color{OliveGreen}{<}\color{Black}{ 1} \quad \color{Cerulean}{✓} \color{Cerulean}{ True } } \end{array}\)

The same problem occurs when multiplying by a negative number. This leads to the following new rule: when multiplying or dividing by a negative number, reverse the inequality . It is easy to forget to do this so take special care to watch for negative coefficients. In general, given algebraic expressions \(A\) and \(B\), where \(c\) is a positive nonzero real number, we have the following properties of inequalities 140 :

Table 1.1.8

We use these properties to obtain an equivalent inequality 141 , one with the same solution set, where the variable is isolated. The process is similar to solving linear equations.

Example \(\PageIndex{14}\):

Solve and graph the solution set: \(−2(x+8)+6≥20\).

\(\begin{aligned} - 2 ( x + 8 ) + 6 & \geq 20 \quad\color{Cerulean}{Distribute.} \\ - 2 x - 16 + 6 & \geq 20 \quad\color{Cerulean}{Combine\: like\: terms.} \\ - 2 x - 10 & \geq 20 \quad\color{Cerulean}{Solve\: for\: x.} \\ - 2 x & \geq 30 \quad\color{Cerulean}{Divide\: both\: sides\: by\: -2.} \\ \frac { - 2 x } { \color{Cerulean}{- 2} } & \color{OliveGreen}{\leq} \frac { \color{Black}{30} } { \color{Cerulean}{- 2} } \quad\color{Cerulean}{Reverse\: the\: inequality.} \\ x & \leq - 15 \end{aligned}\)

Interval notation \((−∞,  −15] \)

Example \(\PageIndex{15}\):

Solve and graph the solution set: \(−2(4x−5)<9−2(x−2)\).

\(\begin{array} { c } { - 2 ( 4 x - 5 ) < 9 - 2 ( x - 2 ) } \\ { - 8 x + 10 < 9 - 2 x + 4 } \\ { - 8 x + 10 < 13 - 2 x } \\ { - 6 x < 3 } \\ { \frac { - 6 x } { \color{Cerulean}{- 6} } \color{OliveGreen}{>} \frac { \color{Black}{3} } { \color{Cerulean}{- 6} } }\color{Cerulean}{Reverse\:the\:inequality.} \\ { x > - \frac { 1 } { 2 } } \end{array}\)

Interval notation \((−\frac{1}{2}, ∞)\)

Example \(\PageIndex{16}\):

Solve and graph the solution set: \(\frac{1}{2}x−2≥\frac{1}{2}(\frac{7}{4}x−9)+1\).

\(\begin{array} { c } { \frac { 1 } { 2 } x - 2 \geq \frac { 1 } { 2 } \left( \frac { 7 } { 4 } x - 9 \right) + 1 } \\ { \frac { 1 } { 2 } x - 2 \geq \frac { 7 } { 8 } x - \frac { 9 } { 2 } + 1 } \\ { \frac { 1 } { 2 } x - \frac { 7 } { 8 } x \geq - \frac { 7 } { 2 } + 2 } \\ { - \frac { 3 } { 8 } x \geq - \frac { 3 } { 2 } } \\ { \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left(\color{Black}{ - \frac { 3 } { 8 } x} \right) \leq \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left( \color{Black}{-} \frac { 3 } { 2 } \right) \quad \color{Cerulean} { Reverse\: the\: inequality. } } \\ { x \leq 4 } \end{array}\)

Interval notation: \((−∞,  4]\)

Exercise \(\PageIndex{4}\)

Solve and graph the solution set: \(10 - 5 ( 2 x + 3 ) \leq 25\)

\([ - 3 , \infty )\);

Video Solution: www.youtube.com/v/COLLNtwYFm8

Translation of Linear Inequalities

Some of the key words and phrases that indicate inequalities are summarized below:

Table 1.1.9

Key Takeaways

• Solving general linear equations involves isolating the variable, with coefficient \(1\), on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
• If solving a linear equation leads to a true statement like \(0 = 0\), then the equation is an identity and the solution set consists of all real numbers, \(ℝ\).
• If solving a linear equation leads to a false statement like \(0 = 5\), then the equation is a contradiction and there is no solution, \(Ø\).
• Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
• Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.
• Inequalities typically have infinitely many solutions. The solutions are presented graphically on a number line or using interval notation or both.
• All but one of the rules for solving linear inequalities are the same as solving linear equations. If you divide or multiply an inequality by a negative number, reverse the inequality to obtain an equivalent inequality.

129 Statement indicating that two algebraic expressions are equal.

130 An equation that can be written in the standard form \(ax + b = 0\), where \(a\) and \(b\) are real numbers and \(a ≠ 0\).

131 Any value that can replace the variable in an equation to produce a true statement.

132 Equations with the same solution set.

133 Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers.

134 Allows you to solve for the variable on either side of the equal sign, because \(x = 5\) is equivalent to \(5 = x\).

135 Equations that are true for particular values.

136 An equation that is true for all possible values.

137 An equation that is never true and has no solution.