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The equations section of QuickMath allows you to solve and plot virtually any equation or system of equations. In most cases, you can find exact solutions to your equations. Even when this is not possible, QuickMath may be able to give you approximate solutions to almost any level of accuracy you require. It also contains a number of special commands for dealing with quadratic equations.

The Solve command can be uses to solve either a single equation for a single unknown from the basic solve page or to simultaneously solve a system of many equations in many unknowns from the advanced solve page . The advanced command allows you to specify whether you want approximate numerical answers as well as exact ones, and how many digits of accuracy (up to 16) you require. It also allows you to eliminate certain variables from the equations.

Go to the Solve page

The Plot command, from the Graphs section, will plot any function of two variables. In order to plot a single function of x, go to the basic equation plotting page , where you can enter the equation and specify the upper and lower limits on x that you want the graph to be plotted for. The advanced plotting page allows you to plot up to 6 equations on the one graph, each with their own color. It also gives you control over such things as whether or not to show the axes, where the axes should be located, what the aspect ratio of the plot should be and what the range of the dependent variable should be. All equations can be given in the explicit y = f(x) form or the implicit g(x,y) = c form.

Go to the Equation Plotting page

The Quadratics page contains 13 separate commands for dealing with the most common questions concerning quadratics. It allows you to : factor a quadratic function (by two different methods); solve a quadratic equation by factoring the quadratic, using the quadratic formula or by completing the square; rewrite a quadratic function in a different form by completing the square; calculate the concavity, x-intercepts, y-intercept, axis of symmetry and vertex of a parabola; plot a parabola; calculate the discriminant of a quadratic equation and use the discriminant to find the number of roots of a quadratic equation. Each command generates a complete and detailed custom-made explanation of all the steps needed to solve the problem.

Go to the Quadratics page

Introduction to Equations

By an equation we mean a mathematical sentence that states that two algebraic expressions are equal. For example, a (b + c) =ab + ac, ab = ba, and x 2 -1 = (x-1)(x+1) are all equations that we have been using. We recall that we defined a variable as a letter that may be replaced by numbers out of a given set, during a given discussion. This specified set of numbers is sometimes called the replacement set. In this chapter we will deal with equations involving variables where the replacement set, unless otherwise specified, is the set of all real numbers for which all the expressions in the equation are defined.

If an equation is true after the variable has been replaced by a specific number, then the number is called a solution of the equation and is said to satisfy it. Obviously, every solution is a member of the replacement set. The real number 3 is a solution of the equation 2x-1 = x+2, since 2*3-1=3+2. while 1 is a solution of the equation (x-1)(x+2) = 0. The set of all solutions of an equation is called the solution set of the equation.

In the first equation above {3} is the solution set, while in the second example {-2,1} is the solution set. We can verify by substitution that each of these numbers is a solution of its respective equation, and we will see later that these are the only solutions.

A conditional equation is an equation that is satisfied by some numbers from its replacement set and not satisfied by others. An identity is an equation that is satisfied by all numbers from its replacement set.

Example 1 Consider the equation 2x-1 = x+2

The replacement set here is the set of all real numbers. The equation is conditional since, for example, 1 is a member of the replacement set but not of the solution set.

Example 2 Consider the equation (x-1)(x+1) =x 2 -1 The replacement set is the set of all real numbers. From our laws of real numbers if a is any real number, then (a-1)(a+1) = a 2 -1 Therefore, every member of the replacement set is also a member of the solution set. Consequently this equation is an identity.  

The replacement set for this equation is the set of real numbers except 0, since 1/x and (1- x)/x are not defined for x = 0. If a is any real number in the replacement set, then

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About solving equations

A value c c is said to be a root of a polynomial p(x) p x if p(c)=0 p c = 0 ..

. This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

How Wolfram|Alpha solves equations

For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and reliability. Other operations rely on theorems and algorithms from number theory, abstract algebra and other advanced fields to compute results. These methods are carefully designed and chosen to enable Wolfram|Alpha to solve the greatest variety of problems while also minimizing computation time.

Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

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Solving Equations

Solving two step equations.

When we solve equations we are finding an unknown number

We can use inverse operations to solve an equation: The opposite of adding is subtracting The opposite of multiplying is dividing.

Example 1 Solve: 3a + 4 = 19

This equation says 3 times a plus 4 is equal to 19

We need to find out what number a is, to do that we need to get a by itself

The first step is to get rid of the 4 The opposite of adding 4 is subtracting 4

To keep both sides of the equation equal we need to subtract 4 from both sides of the equation

We now have: 3a + 4 - 4 = 19 - 4

This simplifies to: 3a = 15

We now have 3 times a is equal to 15

a is multiplied by 3. The opposite of multiplication is division.

We need to divide both sides by 3

This gives us: 3a ⁄ 3 = 15 ⁄ 3

This can be simplified to: a = 5

Example 2 Solve: 43 = 5b - 7

This time we need to get b by itself

The first step is to get rid of the 7 The opposite of subtracting 7 is adding 7

We need to add 7 to both sides of the equation

We now have: 43 + 7 = 5b - 7 + 7

This simplifies to: 50 = 5b

To get b by itself we need to get rid of the 5 At the moment b is multiplied by 5, the opposite of multiplying by 5 is dividing by 5

We need to divide both sides by 5

This gives us: 50 ⁄ 5 = 5b ⁄ 5

This can be simplified to: 10 = b

We can rewrite this with b first: b = 10

Example 3 Solve: c ⁄ 4 + 2 = 8

This time we need to get c by itself

We start by getting rid of the 2 The opposite of adding 2 is subtracting 2

We need to subtract 2 from both sides of the equation

We now have: c ⁄ 4 + 2 - 2 = 8 - 2

This simplifies to: c ⁄ 4 = 6

To get c by itself we need to get rid of the 4 At the moment c is divided by 4, the opposite of dividing by 4 is multiplying by 4

We need to multiply both sides by 4

This gives us: 4c ⁄ 4 = 6 × 4

This can be simplified to: c = 24

Example 4 Solve: d - 9 ⁄ 2 = 7

We need to get d by itself

This time the whole left side is divided by 2, the first step is to multiply both sides by 2

This gives us: 2(d - 9) ⁄ 2 = 7 × 2

Which simplifies to: d - 9 = 14

To get d by itself we need to add 9 to both sides

This gives: d - 9 + 9 = 14 + 9

Which simplifies to: d = 23

Sometimes the answer to an equation is not a whole number. In these cases we can leave our answer as a fraction.

Example 5 Solve: 5z + 3 = 17

the first step is to subtract 3 from both sides

This gives us: 5z + 3 - 3 = 17 - 3

Which simplifies to: 5z = 14

The final step to get z by itself is to divide both sides by 5

This gives: 5z ⁄ 5 = 14 ⁄ 5

We can leave our answer as a fraction: z = 14 ⁄ 5

Solving Equations with an Unknown on Both Sides

When the unknown (what we are working out) appears on both sides of the equation, the first step is to get them on the same side

It is easiest to do this when we get rid of the smallest unknown, the one with the smallest number in front of it (coefficient)

Example 6 Solve: 7a - 3 = 4a + 9

In this questions we need to find out what a is. There is an a term on both sides of the equation.

On the left side of the equation we have 7a On the right side we have 4a 4a is smaller than 7a so we will get rid of 4a

To get rid of 4a will will subtract 4a from both sides of the equation

This gives us: 7a - 3 - 4a = 4a + 9 - 4a

Which simplifies to: 3a - 3 = 9

We now continue to solve the equation: the next step is to add 3 to both sides

3a - 3 + 3 = 9 + 3

This simplifies to: 3a = 12

We now divide both sides by 3 to get a by itself: 3a ⁄ 3 = 12 ⁄ 3

This can be simplified to: a = 4

Example 7 Solve: 2b - 9 = 4 - 6b

This time we have a b term on both sides of the equation

We have 2b on the left and negative 6b on the right side. The smaller b term is negative 6b.

This time we will start by adding 6b to both sides of the equation

2b - 9 + 6b = 4 - 6b + 6b

This simplifies to: 8b - 9 = 4

We now add 9 to both sides: 8b - 9 + 9 = 4 + 9

This simplifies to: 8b = 13

We finally divide both sides by 8: 8b ⁄ 8 = 13 ⁄ 8

We leave our answer as a fraction: b = 13 ⁄ 8

Example 8 Solve: c - 9 = 3c - 2

This time we have a c term on both sides of the equation. Where we have a 'c' term this means we have 1c.

We have 1c on the left and 3c on the right side. The smaller c term is 1c.

We will start by subtracting c from both sides of the equation.

c - 9 - c = 3c - 2 - c

This simplifies to: -9 = 2c - 2

We now add 2 to both sides: -9 + 2 = 2c - 2 + 2

This simplifies to: -7 = 2c

We finally divide both sides by 2: -7 ⁄ 2 = 2c ⁄ 2

We leave our answer as a fraction: c = -7 ⁄ 2

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Solving Equations

Solving equations involves finding the value of the unknown variables in the given equation. The condition that the two expressions are equal is satisfied by the value of the variable. Solving a linear equation in one variable results in a unique solution, solving a linear equation involving two variables gives two results. Solving a quadratic equation gives two roots. There are many methods and procedures followed in solving an equation. Let us discuss the techniques in solving an equation one by one, in detail.

What is the Meaning of Solving Equations?

Solving equations is computing the value of the unknown variable still balancing the equation on both sides. An equation is a condition on a variable such that two expressions in the variable have equal value. The value of the variable for which the equation is satisfied is said to be the solution of the equation. An equation remains the same if the LHS and the RHS are interchanged. The variable for which the value is to be found is isolated and the solution is obtained. Solving an equation depends on what type of equation that we are dealing with. The equations can be linear equations, quadratic equations, rational equations, or radical equations.

Steps in Solving an Equation

The aim of solving an equation is to find the value of the variable that satisfies the condition of the equation true. To isolate the variable, the following operations are performed still balancing the equation on both sides. By doing so LHS remains equal to RHS, and eventually, the balance remains undisturbed throughout.

• Addition property of equality : Add the same number to both the sides. If a = b, then a + c = b + c
• Subtraction property of equality : Subtract the same number from both sides. If a = b, then a - c = b - c
• Multiplication property of equality: Multiply the same number on both sides. If a = b, then ac = bc
• Division property of equality : Divide by the same number on both sides. If a = b, then a/c = b/c (where c ≠ 0)

After performing this systematic balancing method of solving an equation by a series of identical arithmetical operations on both sides of the equation, we separate the variable on one of the sides and the ultimate step is the solution of the equation.

Solving Equations of One Variable

A linear equation of one variable is of the form ax + b = 0, where a, b, c are real numbers. The following steps are followed while solving an equation that is linear.

• Remove the parenthesis and use the distributive property if required.
• Simplify both sides of the equation by combining like terms.
• If there are fractions , multiply both sides of the equation by the LCD (Least common denominator) of all the fractions.
• If there are decimals , multiply both sides of the equation by the lowest power of 10 to convert them into whole numbers.
• Bring the variable terms to one side of the equation and the constant terms to the other side using the addition and subtraction properties of equality.
• Make the coefficient of the variable as 1, using the multiplication or division properties of equality.
• isolate the variable and get the solution.

Consider this example: 3(x + 4) = 24 + x

We simplify the LHS using the distributive property.

3x + 12 = 24 + x

Group the like terms together using the transposing method. This becomes 3x - x = 24-12

Simplify further ⇒ 2x = 12

Use the division property of equality, 2x/2 = 12/2

isolate the variable x. x = 6 is the solution of the equation.

Use any one of the following techniques to simplify the linear equation and solve for the unknown variable. The trial and error method, balancing method and the transposing method are used to isolate the variable.

Solving an Equation by Trial And Error Method

Consider 12x = 60. To find x, we intuitively try to find that 12 times what number is 60. We find that 5 is the required number. Solving equations by trial and error method is not always easy.

Solving an Equation by Balancing Method

We need to isolate the variable x for solving an equation. Let us use the separation of variables method or the balancing method to solve it. Consider an equation 2x + 3 = 17.

We first eliminate 3 in the first step. To keep the balance while solving the equation, we subtract 3 from either side of the equation.

Thus 2x + 3 - 3 = 17 - 3

We have 2x = 14

Now to isolate x, we divide by 2 on both sides. (Division property of equality)

2x/2 = 14/2

Thus, we isolate the variable using the properties of equality while solving an equation in the balancing method.

Solving an Equation by Transposing Method

While solving an equation, we change the sides of the numbers. This process is called transposing. While transposing a number, we change its sign or reverse the operation. Consider 5y + 2 = 22.

We need to find y, so isolate it. Hence we transpose the number 2 to the other side. The equation becomes,

Now taking 5 to the other side, we reverse the operation of multiplication to division. y = 20/5 = 4

Solving an Equation That is Quadratic

There are equations that yield more than one solution. Quadratic polynomials are of degree two and the zeroes of a quadratic polynomial represent the quadratic equation.

Consider (x+3) (x+2)= 0. This is quadratic in nature. We just equate each of the expressions in the LHS to 0.

Either x+3 = 0 or x+2 =0.

We arrive at x = -3 and x = -2.

A quadratic equation is of the form ax 2 + bx + c = 0. Solving an equation that is quadratic, results in two roots : α and β.

Steps involved in solving a quadratic equation are:

By Completing The Squares Method

By factorization method, by formula method.

Solving an equation of quadratic type by completing the squares method is quite easy as we apply our knowledge of algebraic identity: (a+b) 2

• Write the equation in the standard form ax 2 + bx + c = 0.
• Divide both the sides of the equation by a.
• Move the constant term to the other side
• Add the square of one-half of the coefficient of x on both sides.
• Complete the left-hand side as a square and simplify the right-hand side.
• Take the square root on both sides and solve for x.

For more information about solving equations (quadratic) by completing the squares, click here .

Solving an equation of quadratic type using the factorization method , follow the steps discussed here. Write the given equation in the standard form and by splitting the middle terms, factorize the equation. Rewrite the equation obtained as a product of two linear factors. Equate each linear factor to zero and solve for x. Consider 2x 2 + 19x + 30 =0. This is of the standard form ax 2 + bx + c = 0.

Split the middle term in such a way that the product of the terms should equal the product of the coefficient of x 2 and c and the sum of the terms should be b. Here the product of the terms should be 60 and the sum should be 19. Thus, split 19x as 4x and 15x (as the sum of 4 and 15 is 19 and their product is 60).

2x 2 + 4x + 15x + 30 = 0

Take the common factor out of the first two terms and the common factors out of the last two terms.

2x(x + 2) + 15(x + 2) = 0

Factoring (x+2) again, we get

(x + 2)(2x + 15) = 0

x = -2 and x = -15/2

Solving an equation that is quadratic involves such steps while splitting the middle terms on factorization.

Solving an equation of quadratic type using the formula

x = [-b ± √[(b 2 -4ac)]/2a helps us find the roots of the quadratic equation ax 2 + bx + c = 0. Plugging in the values of a, b, and c in the formula, we arrive at the solution.

Consider the example: 9x 2 -12 x + 4 = 0

a= 9, b = -12 and c = 4

x = [-b ± √[(b 2 -4ac)]/2a

= [12 ± √[((-12) 2 -4×9×4)] / (2 × 9)

= [12 ± √(144 - 144)] / 18

= (12 ± 0)/18

x = 12/18 = 2/3

Solving an Equation That is Rational

An equation with at least one polynomial expression in its denominator is known as the rational equation. Solving an equation that is rational involves the following steps. Reduce the fractions to a common denominator and then solve the equation of the numerators .

Consider x/(x-1) = 5/3

On cross-multiplication, we get

3x = 5(x-1)

3x = 5x - 5

3x - 5x = - 5

Solving an Equation That is Radical

An equation in which the variable is under a radical is termed the radical equation. Solving an equation that is a radical involves a few steps. Express the given radical equation in terms of the index of the radical and balance the equation. Solve for the variable.

Consider √(x+1) = 4

Now square both the sides to balance it. [ √(x+1)] 2 = 4 2

Thus x = 16-1 =15

Important Notes on Solving Equations:

• Solving an equation is finding the value of the variable in the equation.
• The solution of an equation satisfies the condition of the given equation.
• Solving an equation of linear type can be also done  graphically .
• If the right side part of an equation is zero, then for solving equation, just graph the left side of the equation and the x-intercept (s) of the graph would be the solution(s).

☛ Related Articles:

• Solving Equations Calculator
• Simultaneous linear equations
• One variable linear equations and inequations
• Simple equations and their applications

Examples of Solving an Equation

Example 1. Use the balancing method of solving equations: (x-2) / 5 - (x-4) / 2 = 2

The given equation is (x-2) / 5 - (x-4) / 2 = 2.

Solving an equation that is rational involves the following steps.

Simplify the LHS. Take the LCD of the denominators. LCD is 10.

[2(x-2) -5(x-4)]/10 = 2

Use distributive property and simplify the numerator.

We get [2x- 4 -5x+20]/10 = 2

Use the multiplication property of equality to get rid of the denominator.

10 × [2x- 4 -5x+20]/10 = 10 × 2

Simplifying we get - 3x + 16 = 20

isolate the term with the variable using the addition property of equality

-3x + 16 - 16 = 20 - 16

isolate the variable using the division property of equality

-3x/3 = 4/3

Answer: Thus solving (x-2)/ 5 - (x-4)/2 = 2, x = -4/3

Example 2. Use the transposing method of solving an equation 0.4(a+10)= 2 - 0.6a

Solving equations that are linear involving decimals involves the following steps.

Given 0.4(a+10)= 2 - 0.6a

0.4 a + 0.4 × 10 = 2- 0.6a

0.4 a + 4 = 2- 0.6a

0.4 a + 0.6a = 2-4

Answer: The solution is a = -2

Example 3. What is the value of p on solving an equation: 4 (p - 3) - (p - 5) = 4?

Given: 4 (p - 3) - (p - 5) = 4

Let us use the transposing method in solving equations.

4p - 12 - p + 5 = 4 (distributive property)

Answer: The value of p = 11/3

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Practice Questions on Solving an Equation

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FAQs on Solving Equations

What is solving an equation.

Solving an equation is finding the value of the unknown variables in the given equation. The process of solving an equation depends on the type of the equation .

What are The Steps in Solving Equations?

Identify the type of equation: linear, quadratic, logarithmic, exponential, radical or rational.

• Remove the brackets, if any in the given equation. Apply the distributive property.
• Add the same number to both the sides
• Subtract the same number from both the sides
• Multiply the same number on both the sides
• Divide by the same number on both sides.

What are The Golden Rule in Solving an Equation?

The type of the equation is identified. If it is a linear equation, separating the variables method or transposing method is used. If it is a quadratic equation, completing the squares, splitting the middle terms using factorization is used or by formula method.

How Do You use 3 Steps in Solving an Equation?

The 3 steps in solving an equation are to

• remove the brackets, if any using the distributive property,
• simplify the equation by adding or subtracting the like terms ,
• isolating the variable and solving it.

How Do You Solve Linear Equations?

While solving an equation that is linear, we isolate the variable whose value is to be found. We either use transposing method or the balancing method.

How Do You Solve Quadratic Equations?

While solving an equation that is quadratic, we write the equation in the standard form ax 2 + bx + c = 0, and then solve using the formula method or factorization method or completing the squares method.

How Do You Solve Radical Equations?

While solving an equation that is radical , we remove the radical sign, by raising both the sides of the equation to the index of the radical, isolate the variable and solve for x.

How Do You Solve Rational Equations?

While solving an equation that is rational, we simplify the expression on each side of the equation, cross multiply , combine the like terms and then isolate the variable to solve for x.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

Every week, we teach lessons on solving equations to students in schools and districts across the US as part of our online one-on-one math tutoring programs. On this page we’ve broken down everything we’ve learnt about teaching this topic effectively.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

[FREE] Solving Equations Worksheet (Grade 8)

Use this worksheet to check your 8th grade students’ understanding of solving equations. 15 questions with answers to identify areas of strength and support!

Common Core State Standards

How does this relate to 8 th grade and high school math?

• Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

• Combine like terms .
• Simplify the equation by using the opposite operation to both sides.
• Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

• Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
• Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
• Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
• Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

• Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
• Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
• Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
• Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
• Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solve equations with fractions
• Linear equations
• Writing linear equations
• Substitution
• Identity math
• One step equation

Practice solving equations questions

1. Solve 4x-2=14.

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

• Inequalities
• Types of graph
• Math formulas
• Coordinate plane

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• \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
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• How do you solve word problems?
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• High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,...

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Solve Systems of Equations Using Matrices

A system of linear equations is a set of equations where each equation is linear in nature, meaning it forms a straight line when graphed. Solving such systems is crucial in various fields, including engineering, physics, economics, and computer science. Matrices provide an efficient way to handle and solve these systems, especially when dealing with large sets of equations.

What Are Systems of Linear Equations?

A system of linear equations consists of two or more linear equations involving the same set of variables.

The solution to a system of linear equations is the set of values for the variables that satisfy all equations simultaneously.

Example of Systems of Linear Equations

Two Equations with Two Variables

Three Equations with Three Variables

x + y + z = 6

• 2x − y + 3z = 14
• 4x + 3y + 2z = 24

What Are Matrices?

Matrices are rectangular arrays of numbers, symbols, or expressions arranged in rows and columns. In the context of solving linear equations, matrices are used to represent the coefficients of the equations and manipulate them to find the solutions.

Example of Systems of Linear Equations as Matrices

Matrix Representation of a System of Equations for the system

The matrix form is [Tex]\begin{pmatrix} 1 & 3 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}[/Tex]

Augmented Matrix, The augmented matrix combines the coefficient matrix and the constants i.e.,

[Tex]\left[ \begin{array}{cc|c} 1 & 3 & 5 \\ 2 & -1 & 4 \end{array} \right] [/Tex]

How to Solve Problems Step-by-Step Solution

To solve a system of linear equations using matrices, follow these steps

Step 1. Form the Augmented Matrix: Write the system of equations as an augmented matrix. Step 2. Perform Row Operations: Use row operations to simplify the matrix to row echelon form or reduced row echelon form. Step 3. Back-Substitution: If using row echelon form, solve for the variables starting from the last row and moving upwards. Step 4. Check Your Solution: Substitute the values of the variables back into the original equations to verify the solution.

Example: Solve the system using Gaussian elimination

Step 1: Form the augmented matrix.

[Tex]\left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 3 & 14 \\ 4 & 3 & 2 & 24 \end{array} \right] [/Tex]

Step 2: Perform row operations to simplify.

• R 2 = [2, -1, 3, 14] – 2 × [1, 1, 1, 6] = [2 – 2, -1 – 2, 3 – 2, 14 – 12] = [0, -3, 1, 2]
• R 3 = [4, 3, 2, 24] – 4 × [1, 1, 1, 6] = [4 – 4, 3 – 4, 2 – 4, 24 – 24] = [0, -1, -2, 0]
• Further simplify to row echelon form.

Step 3: Back-substitute to solve for x, y, and z.

From row 3:

-1y – 2z = 0

y + 2z = 0 (multiply by -1 to simplify)

Now, Solve for y in terms of z

From row 2:

-3y + z = 2

Substitute y = -2z into this equation:

-3(-2z) + z = 2

Now, Solve for y

Now that we have z = 2/7, substitute this into y = -2z:

y = -2 × 2/7 = -4/7

Now, Solve for x

From row 1:

Substitute y = – 4/7and z = 2/7 into this equation:

x – 4/7 + 2/7 = 6

x – 2/7 = 6

7x – 2 = 42

Step 4: Verify the solution by substituting back into the original equations.

Thus, the solution to the system is:

x = 44/7, y = -4/7, z = 2/7

Solved Problems on Systems of Equations Using Matrices

Problem 1: Solve the system of equations using matrices.

• 2x − 3y = -4

Form the Augmented Matrix: [Tex]\left[ \begin{array}{cc|c} 1 & 1 & 5 \\ 2 & -3 & 4 \end{array} \right] [/Tex] Row Operations: [Tex]Subtract 2×R 1 ​  from R 2 ​ : \left[ \begin{array}{cc|c} 1 & 1 & 5 \\ 0 & -5 & -14 \end{array} \right][/Tex] Back-Substitute: From R 2 : [Tex]? = 14/ 5[/Tex] Substitute into R 1 : [Tex]x+5/14=5[/Tex] , solve for x. Solution: x = 5/11 ,y = 5/14

Problem 2: Solve using the inverse matrix method.

• 3x + 2y = 6

Matrix Form: [Tex]A = \begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} B = \begin{bmatrix} x \\ y \end{bmatrix} C = \begin{bmatrix} 6 \\ 4 \end{bmatrix}[/Tex] AB = C Find Inverse of A: [Tex]A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -2 \\ -1 & 3 \end{bmatrix} =\begin{bmatrix} \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot (-2) \\ \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot 3 \end{bmatrix} = \begin{bmatrix} 0.2 & 0.4 \\ 0.2 & -0.6 \end{bmatrix}[/Tex] Compute B = A − 1 C: [Tex]B = \begin{bmatrix} 0.2 & 0.4 \\ 0.2 & -0.6 \end{bmatrix} \begin{bmatrix} 6 \\ 4 \end{bmatrix} \begin{bmatrix} 0.2 \cdot 6 + 0.4 \cdot 4 \\ 0.2 \cdot 6 – 0.6 \cdot 4 \end{bmatrix} = \begin{bmatrix} 2.8 \\ -1.2 \end{bmatrix}[/Tex] Answer: x = 2.8, y = -1.2

Problem 3: Solve the system using Gaussian elimination.

• 2x + 5y = 14

Form the Augmented Matrix: [Tex]\left[ \begin{array}{cc|c} 1 & 3 & 7 \\ 2 & 5 & 14 \end{array} \right] [/Tex] Row Operations: Subtract 2×R 1 from R 2 : [Tex]\left[ \begin{array}{cc|c} 1 & 3 & 7 \\ 0 & -1 & 0 \end{array} \right] [/Tex] Solve for y: From R 2 : y = 0 ​Substitute y = 0 in R 1 to find x: x + 3(0) = 7⟹x = 7 Solution: x = 7, y = 0

Problem 4: Solve the system using the inverse matrix method.

• x − 2y = -3

​Matrix Form: [Tex]A = \begin{bmatrix} 4 & 1 \\ 1 & -2 \end{bmatrix} B = \begin{bmatrix} x \\ y \end{bmatrix} C = \begin{bmatrix} 9 \\ -3 \end{bmatrix} [/Tex] Find Inverse of A: [Tex]A^{-1} = \frac{1}{(-4-1)} \begin{bmatrix} -2 & -1 \\ -1 & 4 \end{bmatrix} =\begin{bmatrix} \frac{1}{-5} \cdot (-2) & \frac{1}{-5} \cdot (-1) \\ \frac{1}{-5} \cdot (-1) & \frac{1}{-5} \cdot 4 \end{bmatrix} = \begin{bmatrix} 0.4 & 0.2 \\ 0.2 & -0.8 \end{bmatrix}[/Tex] Compute B = A − 1 C: [Tex]B = \begin{bmatrix} 0.4 & 0.2 \\ 0.2 & -0.8 \end{bmatrix} \begin{bmatrix} 9 \\ -3 \end{bmatrix} \begin{bmatrix} 0.4 \cdot 9 + 0.2 \cdot (-3) \\ 0.2 \cdot 9 – 0.6 \cdot (-3) \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}[/Tex] Solution: x = 3, y = -0

Problem 5: Solve the system using Cramer’s Rule.

• 2x + 3y = 8

Form the Coefficient Matrix A and its Determinant:​ [Tex]A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} ,\text{det}(A) = (2 \cdot (-1)) – (3 \cdot 1) = -2 – 3 = -5 [/Tex] Calculate Determinants for x and y: ​ [Tex]A_x = \begin{bmatrix} 8 & 3 \\ 1 & -1 \end{bmatrix} ,\text{det}(A_x) = (8 \cdot (-1)) – (3 \cdot 1) = -8 – 3 = -11[/Tex] [Tex]A_y = \begin{bmatrix} 2 & 8 \\ 1 & 1 \end{bmatrix} ,\text{det}(A_y) = (2 \cdot 1) – (8 \cdot 1) = 2 – 8 = -6[/Tex] Apply Cramer’s Rule: [Tex]x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-11}{-5} = 2.2, y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-6}{-5} = 1.2[/Tex] ​ Solution: x = 2.2, y = 1.2

Problem 6: Solve the system using Gaussian elimination.

• x + y + z = 10
• 2x − y + z = 8
• − x + 3y − 2z = 4

Form the Augmented Matrix: [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 2 & -1 & 1 & 8 \\ -1 & 3 & -2 & 4 \end{array}\right][/Tex] Row Operations: [Tex]R_2 = R_2 – 2R_1, R_3 = R_3 + R_1[/Tex] ​ [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & -3 & -1 & -12 \\ 0 & 4 & -1 & 14 \end{array}\right][/Tex] [Tex]R_3 = R_3 + \frac{4}{3}R_2[/Tex] [Tex]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & -3 & -1 & -12 \\ 0 & 0 & -2 & -4 \end{array}\right][/Tex] Solve for z, y, and x: [Tex]From ?_3: z=2, From R_2:-3y – 1(2) = -12 \implies y = 4, From R_1:x + 1(4) + 1(2) = 10 \implies x = 4[/Tex] Solution: x = 4, y = 4, z = 2

Practice Problems on Systems of Equations Using Matrices

Problem 1: Solve the system using Gaussian elimination

• x + 2y − z = 4
• 3x − y + 2z = 5
• 2x + y − z = 1

Problem 2: Use the inverse matrix method to solve

• 4x + 3y = 18
• − x + 2y = 1

Problem 3: Use Gaussian elimination to solve.

• 5x + 2y = 13

Problem 4: Apply Cramer’s Rule to solve.

Problem 5: Solve the system using Gaussian elimination:

Problem 6: Solve the system using Gauss-Jordan elimination:

• 3x + 2y − z = 4
• x − y + 4z = 3
• 2x + y + 3z = 5

​ Answer Key

• x = 1, y = 1, z = -1
• x = 3, y = 2
• x = 2, y = 1
• x = 1, y = 2
• x = 1, y = 4, z = 5
• x = 2, y = − 1, z = 1
• Types of Matrices
• Elements of Matrix
• Algebra of Matrices
• Matrix Operations
• Solve Systems of Linear Equations by Graphing

FAQs: Systems of Equations Using Matrices

What are the benefits of using matrices to solve linear equations.

Matrices allow for systematic and compact representation and manipulation of systems of linear equations, making them ideal for computer algorithms and handling large-scale problems.

What should I do if a system of equations has no solution or infinitely many solutions?

If a system has no solution, it is inconsistent. If it has infinitely many solutions, it is dependent. In both cases, special techniques or additional constraints may be needed to analyze the system further.

Can matrices be used for non-linear systems?

No, matrices are specifically designed for linear systems. Non-linear systems require other methods such as numerical algorithms or graphical analysis.

How does the determinant of a matrix affect its solvability?

The determinant indicates whether a matrix is invertible. A non-zero determinant means the matrix is invertible, and the system has a unique solution. A zero determinant means the matrix is not invertible, and the system may have no solution or infinitely many solutions.

What are the differences between Gaussian elimination and Gauss-Jordan elimination?

Gaussian elimination transforms the matrix into row echelon form, allowing for back-substitution to find solutions. Gauss-Jordan elimination goes further by transforming the matrix into reduced row echelon form, providing solutions directly without back-substitution.

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