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7.3: Systems of Linear Equations with Three Variables

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• Page ID 15088

Learning Objectives

• Solve systems of three equations in three variables.
• Identify inconsistent systems of equations containing three variables.
• Express the solution of a system of dependent equations containing three variables.

John received an inheritance of \($12,000\) that he divided into three parts and invested in three ways: in a money-market fund paying \(3\%\) annual interest; in municipal bonds paying \(4\%\) annual interest; and in mutual funds paying \(7\%\) annual interest. John invested \($4,000\) more in municipal funds than in municipal bonds. He earned \($670\) in interest the first year. How much did John invest in each type of fund?

Figure \(\PageIndex{1}\): (credit: “Elembis,” Wikimedia Commons)

Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics.

Solving Systems of Three Equations in Three Variables

In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution \((x,y,z)\), which we call an ordered triple. A system in upper triangular form looks like the following:

\[\begin{align*} Ax+By+Cz &= D \nonumber \\[4pt] Ey+Fz &= G \nonumber \\[4pt] Hz &= K \nonumber \end{align*} \nonumber\]

The third equation can be solved for \(z\),and then we back-substitute to find \(y\) and \(x\). To write the system in upper triangular form, we can perform the following operations:

• Interchange the order of any two equations.
• Multiply both sides of an equation by a nonzero constant.
• Add a nonzero multiple of one equation to another equation.

The solution set to a three-by-three system is an ordered triple \({(x,y,z)}\). Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes.

NUMBER OF POSSIBLE SolutionS

Figure \(\PageIndex{2}\) and Figure \(\PageIndex{3}\) illustrate possible solution scenarios for three-by-three systems.

• Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple \({(x,y,z)}\). Graphically, the ordered triple defines a point that is the intersection of three planes in space.
• Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as \(0=0\). Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.
• Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as \(3=0\). Graphically, a system with no solution is represented by three planes with no point in common.

Figure \(\PageIndex{2}\): (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions.

Figure \(\PageIndex{3}\): All three figures represent three-by-three systems with no solution. (a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. (c) All three planes are parallel, so there is no point of intersection.

Example \(\PageIndex{1}\): Determining Whether an Ordered Triple Is a Solution to a System

Determine whether the ordered triple \((3,−2,1)\) is a solution to the system.

\[\begin{align*} x+y+z &= 2 \nonumber \\[4pt] 6x−4y+5z &= 31 \nonumber \\[4pt] 5x+2y+2z &= 13 \nonumber \end{align*} \nonumber\]

We will check each equation by substituting in the values of the ordered triple for \( x,y\), and \(z\).

\[\begin{array}{rrr} { \text{} \nonumber \\[4pt] x+y+z=2 \nonumber \\[4pt] (3)+(−2)+(1)=2 \nonumber \\[4pt] \text{True}} & {6x−4y+5z=31 \nonumber \\[4pt] 6(3)−4(−2)+5(1)=31 \nonumber \\[4pt] 18+8+5=31 \nonumber \\[4pt] \text{True} } & { 5x+2y+2z = 13 \nonumber \\[4pt] 5(3)+2(−2)+2(1)=13 \nonumber \\[4pt] 15−4+2=13 \nonumber \\[4pt] \text{True}} \end{array}\]

The ordered triple \((3,−2,1)\) is indeed a solution to the system.

How to: Given a linear system of three equations, solve for three unknowns

• Pick any pair of equations and solve for one variable.
• Pick another pair of equations and solve for the same variable.
• You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
• Back-substitute known variables into any one of the original equations and solve for the missing variable.

Example \(\PageIndex{2}\): Solving a System of Three Equations in Three Variables by Elimination

Find a solution to the following system:

\[\begin{align} x−2y+3z=9 \; &(1) \nonumber \\[4pt] −x+3y−z=−6 \; &(2) \nonumber \\[4pt] 2x−5y+5z=17 \; &(3) \nonumber \end{align} \nonumber\]

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate \(x\) by adding equations (1) and (2).

\[\begin{align} x−2y+3z=9 \; \; &(1) \nonumber \\[4pt] \underline{−x+3y−z=−6 }\; \; &(2) \nonumber \\[4pt] y+2z=3 \;\; &(4) \nonumber \end{align} \nonumber\]

The second step is multiplying equation (1) by \(−2\) and adding the result to equation (3). These two steps will eliminate the variable \(x\).

\[\begin{align} −2x+4y−6z=−18\; &(1) \;\;\;\; \text{ multiplied by }−2 \nonumber \\[4pt] \underline{2x−5y+5z=17} \; & (3) \nonumber \\[4pt]−y−z=−1 \; &(5) \nonumber \end{align} \nonumber\]

In equations (4) and (5), we have created a new two-by-two system. We can solve for \(z\) by adding the two equations.

\[\begin{align} y+2z=3 \; &(4) \nonumber \\[4pt] \underline{−y−z=−1} \; & (5) \nonumber \\[4pt] z=2 \; & (6) \nonumber \end{align} \nonumber\]

Choosing one equation from each new system, we obtain the upper triangular form:

\[\begin{align} x−2y+3z=9 \; &(1) \nonumber \\[4pt] y+2z =3 \; &(4) \nonumber \\[4pt] z=2 \; &(6) \nonumber \end{align} \nonumber\]

Next, we back-substitute \(z=2\) into equation (4) and solve for \(y\).

\[\begin{align} y+2(2) &=3 \nonumber \\[4pt] y+4 &= 3 \nonumber \\[4pt] y &= −1 \nonumber \end{align} \nonumber\]

Finally, we can back-substitute \(z=2\) and \(y=−1\) into equation (1). This will yield the solution for \(x\).

\[\begin{align} x−2(−1)+3(2) &= 9 \nonumber \\[4pt] x+2+6 &=9 \nonumber \\[4pt] x &= 1 \nonumber \end{align} \nonumber\]

The solution is the ordered triple \((1,−1,2)\). See Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\)

Example \(\PageIndex{3}\): Solving a Real-World Problem Using a System of Three Equations in Three Variables

In the problem posed at the beginning of the section, John invested his inheritance of \($12,000\) in three different funds: part in a money-market fund paying \(3\%\) interest annually; part in municipal bonds paying \(4\%\) annually; and the rest in mutual funds paying \(7\%\) annually. John invested \($4,000\) more in mutual funds than he invested in municipal bonds. The total interest earned in one year was \($670\). How much did he invest in each type of fund?

To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:

\[\begin{align} x &= \text{amount invested in money-market fund} \nonumber \\[4pt] y &= \text{amount invested in municipal bonds} \nonumber \\[4pt] z &= \text{amount invested in mutual funds} \nonumber \end{align} \nonumber\]

The first equation indicates that the sum of the three principal amounts is \($12,000\).

\[x+y+z=12,000 \nonumber\]

We form the second equation according to the information that John invested \($4,000\) more in mutual funds than he invested in municipal bonds.

\[z=y+4,000 \nonumber\]

The third equation shows that the total amount of interest earned from each fund equals \($670\).

\[0.03x+0.04y+0.07z=670 \nonumber\]

Then, we write the three equations as a system.

\[\begin{align} x+y+z &=12,000 \nonumber \\[4pt] −y+z &= 4,000 \nonumber \\[4pt] 0.03x+0.04y+0.07z &= 670 \nonumber \end{align} \nonumber\]

To make the calculations simpler, we can multiply the third equation by \(100\). Thus,

\[\begin{align} x+y+z &=12,000 \; &(1) \nonumber \\[4pt] −y+z &= 4,000 \; &(2) \nonumber \\[4pt] 3x+4y+7z &= 67,000 \; &(3) \nonumber \end{align} \nonumber\]

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

\[\begin{align} x+y+z &= 12,000 \nonumber \\[4pt] 3x+4y+7z &= 67,000 \nonumber \\[4pt] −y+z &= 4,000 \nonumber \end{align} \nonumber\]

Step 2. Multiply equation (1) by \(−3\) and add to equation (2). Write the result as row 2.

\[\begin{align} x+y+z &= 12,000 \nonumber \\[4pt] y+4z &= 31,000 \nonumber \\[4pt] −y+z &= 4,000 \nonumber \end{align} \nonumber\]

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

\[\begin{align} x+y+z &= 12,000 \nonumber \\[4pt] y+4z &= 31,000 \nonumber \\[4pt] 5z &= 35,000 \nonumber \end{align} \nonumber\]

Step 4. Solve for \(z\) in equation (3). Back-substitute that value in equation (2) and solve for \(y\). Then, back-substitute the values for \(z\) and \(y\) into equation (1) and solve for \(x\).

\[\begin{align} 5z &= 35,000 \nonumber \\[4pt] z &= 7,000 \nonumber \\[4pt] \nonumber \\[4pt] y+4(7,000) &= 31,000 \nonumber \\[4pt] y &=3,000 \nonumber \\[4pt] \nonumber \\[4pt] x+3,000+7,000 &= 12,000 \nonumber \\[4pt] x &= 2,000 \nonumber \end{align} \nonumber\]

John invested \($2,000\) in a money-market fund, \($3,000\) in municipal bonds, and \($7,000\) in mutual funds.

Exercise \(\PageIndex{1}\)

Solve the system of equations in three variables.

\[\begin{align} 2x+y−2z &= −1 \nonumber \\[4pt] 3x−3y−z &= 5 \nonumber \\[4pt] x−2y+3z &= 6 \nonumber \end{align} \nonumber\]

\((1,−1,1)\)

Identifying Inconsistent Systems of Equations Containing Three Variables

Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as \(3=7\) or some other contradiction.

Example \(\PageIndex{4}\): Solving an Inconsistent System of Three Equations in Three Variables

Solve the following system.

\[\begin{align} x−3y+z &=4 \label{4.1}\\[4pt] −x+2y−5z &=3 \label{4.2} \\[4pt] 5x−13y+13z &=8 \label{4.3} \end{align} \nonumber\]

Looking at the coefficients of \(x\), we can see that we can eliminate \(x\) by adding Equation \ref{4.1} to Equation \ref{4.2}.

\[\begin{align} x−3y+z = 4 &(1) \nonumber \\[4pt] \underline{−x+2y−5z=3} & (2) \nonumber \\[4pt] −y−4z =7 & (4) \nonumber \end{align} \nonumber\]

Next, we multiply equation (1) by \(−5\) and add it to equation (3).

\[\begin{align} −5x+15y−5z =−20 & (1) \;\;\;\;\; \text{multiplied by }−5 \nonumber \\[4pt] \underline{5x−13y+13z=8} &(3) \nonumber \\[4pt] 2y+8z=−12 &(5) \nonumber \end{align} \nonumber\]

Then, we multiply equation (4) by 2 and add it to equation (5).

\[\begin{align} −2y−8z=14 & (4) \;\;\;\;\; \text{multiplied by }2 \nonumber \\[4pt] \underline{2y+8z=−12} & (5) \nonumber \\[4pt] 0=2 & \nonumber \end{align} \nonumber\]

The final equation \(0=2\) is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.

Exercise \(\PageIndex{2}\)

Solve the system of three equations in three variables.

\[\begin{align} x+y+z &= 2 \nonumber \\[4pt] y−3z &=1 \nonumber \\[4pt] 2x+y+5z &=0 \nonumber \end{align} \nonumber\]

No solution.

Expressing the Solution of a System of Dependent Equations Containing Three Variables

We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line.

Example \(\PageIndex{5}\): Finding the Solution to a Dependent System of Equations

Find the solution to the given system of three equations in three variables.

\[\begin{align} 2x+y−3z &= 0 &(1) \nonumber \\[4pt] 4x+2y−6z &=0 &(2) \nonumber \\[4pt] x−y+z &= 0 &(3) \nonumber \end{align} \nonumber\]

First, we can multiply equation (1) by \(−2\) and add it to equation (2).

\[\begin{align} −4x−2y+6z =0 & (1) \;\;\;\;\; \text{multiplied by }−2 \nonumber \\[4pt] \underline{4x+2y−6z=0} & (2) \nonumber \\[4pt] 0=0& \nonumber \end{align} \nonumber\]

We do not need to proceed any further. The result we get is an identity, \(0=0\), which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by \(−2\), and adding it to equation (1). We then perform the same steps as above and find the same result, \(0=0\).

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

\[\begin{align} 2x+y−3z &= 0 \nonumber \\[4pt]x−y+z &= 0 \nonumber \\[4pt] 3x−2z &= 0 \nonumber \end{align} \nonumber\]

We then solve the resulting equation for \(z\).

\[\begin{align} 3x−2z &= 0 \nonumber \\[4pt] z &= \dfrac{3}{2}x \nonumber \end{align} \nonumber\]

We back-substitute the expression for \(z\) into one of the equations and solve for \(y\).

\[\begin{align} 2x+y−3 (\dfrac{3}{2}x) &= 0 \nonumber \\[4pt] 2x+y−\dfrac{9}{2}x &= 0 \nonumber \\[4pt] y &= \dfrac{9}{2}x−2x \nonumber \\[4pt] y &=\dfrac{5}{2}x \nonumber \end{align} \nonumber\]

So the general solution is \(\left(x,\dfrac{5}{2}x,\dfrac{3}{2}x\right)\). In this solution, \(x\) can be any real number. The values of \(y\) and \(z\) are dependent on the value selected for \(x\).

As shown in Figure \(\PageIndex{5}\), two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

Figure \(\PageIndex{5}\)

Q&A: Does the generic solution to a dependent system always have to be written in terms of \(x\)?

No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of \(x\) and if needed \(x\) and \(y\).

Exercise \(\PageIndex{3}\):

\[\begin{align} x+y+z &= 7 \nonumber \\[4pt] 3x−2y−z &= 4 \nonumber \\[4pt] x+6y+5z &= 24 \nonumber \end{align} \nonumber\]

Infinite number of solutions of the form \((x,4x−11,−5x+18)\).

Access these online resources for additional instruction and practice with systems of equations in three variables.

• Ex 1: System of Three Equations with Three Unknowns Using Elimination
• Ex. 2: System of Three Equations with Three Unknowns Using Elimination

Key Concepts

• A solution set is an ordered triple {(x,y,z)} that represents the intersection of three planes in space. See Example \(\PageIndex{1}\).
• A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example \(\PageIndex{2}\).
• Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example \(\PageIndex{3}\).
• A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example \(\PageIndex{4}\).
• Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location.
• A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example \(\PageIndex{5}\).
• Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line.

Contributors

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a  Creative Commons Attribution License 4.0  license. Download for free at  https://openstax.org/details/books/precalculus .

5.2 Solving Systems of Equations by Substitution

Learning objectives.

By the end of this section, you will be able to:

• Solve a system of equations by substitution
• Solve applications of systems of equations by substitution

Be Prepared 5.4

Before you get started, take this readiness quiz.

Simplify −5 ( 3 − x ) −5 ( 3 − x ) . If you missed this problem, review Example 1.136 .

Be Prepared 5.5

Simplify 4 − 2 ( n + 5 ) 4 − 2 ( n + 5 ) . If you missed this problem, review Example 1.123 .

Be Prepared 5.6

Solve for y y : 8 y − 8 = 32 − 2 y 8 y − 8 = 32 − 2 y If you missed this problem, review Example 2.34 .

Be Prepared 5.7

Solve for x x : 3 x − 9 y = −3 3 x − 9 y = −3 If you missed this problem, review Example 2.65 .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in Example 5.13 .

Example 5.13

How to solve a system of equations by substitution.

Solve the system by substitution. { 2 x + y = 7 x − 2 y = 6 { 2 x + y = 7 x − 2 y = 6

Try It 5.25

Solve the system by substitution. { −2 x + y = −11 x + 3 y = 9 { −2 x + y = −11 x + 3 y = 9

Try It 5.26

Solve the system by substitution. { x + 3 y = 10 4 x + y = 18 { x + 3 y = 10 4 x + y = 18

Solve a system of equations by substitution.

• Step 1. Solve one of the equations for either variable.
• Step 2. Substitute the expression from Step 1 into the other equation.
• Step 3. Solve the resulting equation.
• Step 4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
• Step 5. Write the solution as an ordered pair.
• Step 6. Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in Example 5.14 .

Example 5.14

Solve the system by substitution.

{ x + y = −1 y = x + 5 { x + y = −1 y = x + 5

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

Try It 5.27

Solve the system by substitution. { x + y = 6 y = 3 x − 2 { x + y = 6 y = 3 x − 2

Try It 5.28

Solve the system by substitution. { 2 x − y = 1 y = −3 x − 6 { 2 x − y = 1 y = −3 x − 6

If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

Example 5.15

Solve the system by substitution. { 3 x + y = 5 2 x + 4 y = −10 { 3 x + y = 5 2 x + 4 y = −10

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Try It 5.29

Solve the system by substitution. { 4 x + y = 2 3 x + 2 y = −1 { 4 x + y = 2 3 x + 2 y = −1

Try It 5.30

Solve the system by substitution. { − x + y = 4 4 x − y = 2 { − x + y = 4 4 x − y = 2

In Example 5.15 it was easiest to solve for y in the first equation because it had a coefficient of 1. In Example 5.16 it will be easier to solve for x .

Example 5.16

Solve the system by substitution. { x − 2 y = −2 3 x + 2 y = 34 { x − 2 y = −2 3 x + 2 y = 34

We will solve the first equation for x x and then substitute the expression into the second equation.

Try It 5.31

Solve the system by substitution. { x − 5 y = 13 4 x − 3 y = 1 { x − 5 y = 13 4 x − 3 y = 1

Try It 5.32

Solve the system by substitution. { x − 6 y = −6 2 x − 4 y = 4 { x − 6 y = −6 2 x − 4 y = 4

When both equations are already solved for the same variable, it is easy to substitute!

Example 5.17

Solve the system by substitution. { y = −2 x + 5 y = 1 2 x { y = −2 x + 5 y = 1 2 x

Since both equations are solved for y , we can substitute one into the other.

Try It 5.33

Solve the system by substitution. { y = 3 x − 16 y = 1 3 x { y = 3 x − 16 y = 1 3 x

Try It 5.34

Solve the system by substitution. { y = − x + 10 y = 1 4 x { y = − x + 10 y = 1 4 x

Be very careful with the signs in the next example.

Example 5.18

Solve the system by substitution. { 4 x + 2 y = 4 6 x − y = 8 { 4 x + 2 y = 4 6 x − y = 8

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 5.35

Solve the system by substitution. { x − 4 y = −4 −3 x + 4 y = 0 { x − 4 y = −4 −3 x + 4 y = 0

Try It 5.36

Solve the system by substitution. { 4 x − y = 0 2 x − 3 y = 5 { 4 x − y = 0 2 x − 3 y = 5

In Example 5.19 , it will take a little more work to solve one equation for x or y .

Example 5.19

Solve the system by substitution. { 4 x − 3 y = 6 15 y − 20 x = −30 { 4 x − 3 y = 6 15 y − 20 x = −30

We need to solve one equation for one variable. We will solve the first equation for x .

Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Try It 5.37

Solve the system by substitution. { 2 x − 3 y = 12 −12 y + 8 x = 48 { 2 x − 3 y = 12 −12 y + 8 x = 48

Try It 5.38

Solve the system by substitution. { 5 x + 2 y = 12 −4 y − 10 x = −24 { 5 x + 2 y = 12 −4 y − 10 x = −24

Look back at the equations in Example 5.19 . Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example 5.20

Solve the system by substitution. { 5 x − 2 y = −10 y = 5 2 x { 5 x − 2 y = −10 y = 5 2 x

The second equation is already solved for y , so we can substitute for y in the first equation.

Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Try It 5.39

Solve the system by substitution. { 3 x + 2 y = 9 y = − 3 2 x + 1 { 3 x + 2 y = 9 y = − 3 2 x + 1

Try It 5.40

Solve the system by substitution. { 5 x − 3 y = 2 y = 5 3 x − 4 { 5 x − 3 y = 2 y = 5 3 x − 4

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5.

How to use a problem solving strategy for systems of linear equations.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose variables to represent those quantities.
• Step 4. Translate into a system of equations.
• Step 5. Solve the system of equations using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example 5.21

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It 5.41

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Try It 5.42

The sum of two number is −6. One number is 10 less than the other. Find the numbers.

In the Example 5.22 , we’ll use the formula for the perimeter of a rectangle, P = 2 L + 2 W .

Example 5.22

The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width.

Try It 5.43

The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle.

Try It 5.44

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle.

For Example 5.23 we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle.

Example 5.23

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It 5.45

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Try It 5.46

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Example 5.24

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $10,000 + $40 for each training session. How many training sessions would make the salary options equal?

Try It 5.47

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Try It 5.48

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

• Instructional Video-Solve Linear Systems by Substitution
• Instructional Video-Solve by Substitution

Section 5.2 Exercises

Practice makes perfect.

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ −2 x + 2 y = 6 y = −3 x + 1 { −2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 3 y = 3 y = − x + 3 { 2 x + 3 y = 3 y = − x + 3

{ 2 x + 5 y = −14 y = −2 x + 2 { 2 x + 5 y = −14 y = −2 x + 2

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 3 x − 2 y = 6 y = 2 3 x + 2 { 3 x − 2 y = 6 y = 2 3 x + 2

{ −3 x − 5 y = 3 y = 1 2 x − 5 { −3 x − 5 y = 3 y = 1 2 x − 5

{ 2 x + y = 10 − x + y = −5 { 2 x + y = 10 − x + y = −5

{ −2 x + y = 10 − x + 2 y = 16 { −2 x + y = 10 − x + 2 y = 16

{ 3 x + y = 1 −4 x + y = 15 { 3 x + y = 1 −4 x + y = 15

{ x + y = 0 2 x + 3 y = −4 { x + y = 0 2 x + 3 y = −4

{ x + 3 y = 1 3 x + 5 y = −5 { x + 3 y = 1 3 x + 5 y = −5

{ x + 2 y = −1 2 x + 3 y = 1 { x + 2 y = −1 2 x + 3 y = 1

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ y = 2 x − 8 y = 3 5 x + 6 { y = 2 x − 8 y = 3 5 x + 6

{ y = − x − 1 y = x + 7 { y = − x − 1 y = x + 7

{ 4 x + 2 y = 8 8 x − y = 1 { 4 x + 2 y = 8 8 x − y = 1

{ − x − 12 y = −1 2 x − 8 y = −6 { − x − 12 y = −1 2 x − 8 y = −6

{ 15 x + 2 y = 6 −5 x + 2 y = −4 { 15 x + 2 y = 6 −5 x + 2 y = −4

{ 2 x − 15 y = 7 12 x + 2 y = −4 { 2 x − 15 y = 7 12 x + 2 y = −4

{ y = 3 x 6 x − 2 y = 0 { y = 3 x 6 x − 2 y = 0

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x + 16 y = 8 − x − 8 y = −4 { 2 x + 16 y = 8 − x − 8 y = −4

{ 15 x + 4 y = 6 −30 x − 8 y = −12 { 15 x + 4 y = 6 −30 x − 8 y = −12

{ y = −4 x 4 x + y = 1 { y = −4 x 4 x + y = 1

{ y = − 1 4 x x + 4 y = 8 { y = − 1 4 x x + 4 y = 8

{ y = 7 8 x + 4 −7 x + 8 y = 6 { y = 7 8 x + 4 −7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 15. One number is 3 less than the other. Find the numbers.

The sum of two numbers is 30. One number is 4 less than the other. Find the numbers.

The sum of two numbers is −26. One number is 12 less than the other. Find the numbers.

The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width.

The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width.

The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width.

The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Maxim has been offered positions by two car dealers. The first company pays a salary of $10,000 plus a commission of $1,000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $ 14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Everyday Math

When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system { 15 e + 30 c = 435 30 e + 40 c = 690 { 15 e + 30 c = 435 30 e + 40 c = 690 for e e , the number of calories she burns for each minute on the elliptical trainer, and c c , the number of calories she burns for each minute of circuit training.

Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system { 56 s = 70 t s = t + 1 2 { 56 s = 70 t s = t + 1 2 .

• ⓐ for t t to find out how long it will take Tina to catch up to Stephanie.
• ⓑ what is the value of s s , the number of hours Stephanie will have driven before Tina catches up to her?

Writing Exercises

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing. ⓑ by substitution. ⓒ Which method do you prefer? Why?

Solve the system of equations { 3 x + y = 12 x = y − 8 { 3 x + y = 12 x = y − 8 by substitution and explain all your steps in words.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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Equations and Inequalities (Pre-Algebra - Unit 3) | All Things Algebra®

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This Equations and Inequalities Unit Bundle includes guided notes, homework assignments, four quizzes, a study guide, and a unit test that cover the following topics:

• One-Step Equations

• Rational Equations

• Two-Step Equations

• Solving Equations by Square Roots

• Multi-Step Equations (Variables on One Side)

• Multi-Step Equations (Variables on Both Sides)

• Special Solutions: No Solution and Infinite Solution

• Solving Equations by Clearing Fractions

• Translating and Solving Equations

• Equation Word Problems

• Representing and Graphing Inequalities

• One- and Two-Step Inequalities

• Multi-Step Inequalities

• Translating and Solving Inequalities

• Inequality Word Problems

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(1) Links to Instructional Videos: Links to videos of each lesson in the unit are included. Videos were created by fellow teachers for their students using the guided notes and shared in March 2020 when schools closed with no notice.  Please watch through first before sharing with your students. Many teachers still use these in emergency substitute situations. (2) Editable Assessments: Editable versions of each quiz and the unit test are included. PowerPoint is required to edit these files. Individual problems can be changed to create multiple versions of the assessment. The layout of the assessment itself is not editable. If your Equation Editor is incompatible with mine (I use MathType), simply delete my equation and insert your own.

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Solving Absolute Value Equations and Inequalities

As we saw earlier in the Negative Numbers and Absolute Value section, an absolute value  (designated by |  |) means take the positive value of whatever is between the two bars. The absolute value is always positive, so you can think of it as the distance from 0 . As an example, $ \left| 3 \right|=3$ and $ \left| {-3} \right|=3$. It’s as simple as that!

(Note that we also address absolute values here in the Piecewise Functions section  and  here in the Rational Functions, Equations, and Inequalities section .)

Solving Absolute Value Equations

Solving absolute equations  isn’t too difficult; just have to separate the equation into two different equations (once we isolate the absolute value), since we don’t if what’s inside the absolute value is  positive  or  negative . Then, make the expression on the right-hand side (without the variables) both positive and negative and solve each equation; typically, we will get two answers . We must  check our answer s , since we may get  extraneous solutions (solutions that don’t work). .

There are a few cases with absolute value equations or inequalities where we don’t have to even solve! One is when we have isolated the absolute value, and it is set equal to a negative number , such as $ \left| {x-5} \right|=-4$, or $ \left| {x-5} \right|\le -4$, for example. Since an absolute value can never be negative , we have  no solution for this case. The other is when the absolute value is greater than a negative number, such as  $ \left| {x-5} \right|>-4$  for example. In this case our answer is all real numbers , since an absolute value is always positive.

Note that we can always solve absolute value equations and inequalities graphically , as shown below.

Here are some problems:

Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases , just to be sure we cover all the possibilities. We then must check for extraneous solutions , possible solutions that don’t work.

For example, when the expression $ 3x-2$ is negative, the absolute value of that expression is the negation of it, or $ -3x+2$, to make it positive in the equation. Play around with some numbers and you’ll see this!

When we get all the possible answers, check for extraneous solutions , since we’re dealing with absolute value. We found two answers that worked: $ \displaystyle x=\frac{3}{2}$ and $ x=-1$. You can also put the equation in your graphing calculator to check your answers!

Here’s another way to approach the absolute value problem above, using number lines :

Now draw number lines for each absolute value, and then for the whole equation above . We see for the last number line that for $ <-2$, we’ll use $ 2-3x$ and $ -x-2$, between $ -2$ and $ \displaystyle \frac{2}{3}$, we’ll use $ 2-3x$ and $ x+2$ , and $ \displaystyle >\frac{2}{3}$, we’ll use $ 3x-2$ and $ x+2$.

After solving for $ x$ in the original equation, we have to check to make sure each value we get for $ x$ falls into the correct interval of the number line. For example, $ \displaystyle -\frac{3}{2}$ isn’t  $ <-2$,  so we have to “throw it away”. $ -1$ is between $ -2$ and $ \displaystyle \frac{2}{3}$, so it works, and $ \displaystyle \frac{3}{2}$ is $ \displaystyle >\frac{2}{3}$, so it works.

We have $ \displaystyle x=\frac{3}{2}$ and $ x=-1$.   √

Solving Absolute Value Inequalities

Note that we learned about Linear Inequalities here .

When dealing with absolute values and inequalities (just like with absolute value equations), we have to separate the inequality into two different ones , if there are any variables inside the absolute value bars.

First, get the absolute value all by itself on the left (remember to reverse the inequality sign when multiplying or dividing by a negative number). Now, separate the equations. We get the first equation by just taking away the absolute value sign away on the left. The easiest way to get the second equation is to take the absolute value sign away on the left, and do two things on the right :  reverse the inequality sign , and change the sign of everything on the right (even if we have variables over there).

We also have to think about whether or not to use “ or ” or “ and ” between the two new equations. The way I remember this is that with a $ >\,\text{or}\,\,\ge  $ sign, you can remember “gore”: greater than uses “or” . With a  $ <\,\text{or}\,\,\le  $ sign, think “land”: less than uses “and” .

GORE: Greater Than ­­­uses OR LAND:  Less Than uses AND

Note that statement with “or” is a disjunction , which means that it works if only one (or both) parts are true. A statement with “and” is a conjunction , which means it only works if both parts are true.

And again, if we get something like $ \left| {x+3} \right|<0$ (or a negative number), there is no solution , and something like $ \left| {x+3} \right|\ge 0$ (or a negative number), there are infinite solutions (all real numbers).

Also, remember to use open brackets for inequalities that aren’t inclusive ($ <$ and $ <$) and closed brackets for inequalities that are inclusive and include the boundary point ($ \le $ and $ \ge $).

Here are some examples:

There are examples of rational functions with absolute values here in the Rational Functions, Equations, and Inequalities section .

Graphs of Absolute Value Functions

Note that you can put absolute values in your Graphing Calculator (and even graph them!) by hitting MATH, scroll right to NUM , and then hitting 1 (abs) or ENTER .

Absolute Value functions typically look like a V (upside down if the absolute value is negative), where the point at the V is called the vertex . For the absolute value parent function, the vertex is at $ \left( {0,0} \right)$.

We looked at absolute value parent functions and their transformations in the  Absolute Value Transformations  section , and absolute value functions as piecewise equations here in the Piecewise Functions section .

Note that the general form for the absolute value function is  $ f\left( x \right)=a\left| {x-h} \right|+k$, where $ \left( {h,k} \right)$ is the vertex. If $ a$ is positive, the function points down (like a V ); if $ a$ is negative, the function points up (like an upside-down V ). Here’s a graph of the parent function, and also a transformation:

Without using a t-chart, we can see that the vertex is at $ \left( {-2,1} \right)$ and the graph is upside-down because of the negative sign. It’s also stretched vertically by a factor of 3 and horizontal by a factor of $ \displaystyle \frac{1}{2}$ (or stretched vertically by a factor of 6 ); thus, other points down can be drawn by going back and forth 1 and down 6 .

You can solve absolute value equations and equalities with graphing ; here are some examples of solving inequalities:

Applications of Absolute Value Functions

Absolute Value Functions are in many applications , especially in those involving V-shaped paths and margin of errors , or tolerances . Here are some examples absolute value “word” problems that you may see:

Here are examples that are absolute value inequality applications . Use this rule of thumb : the absolute value of a difference is usually on the left-hand side, the amount that differs or varies is usually on the right-hand side, with a $ <$ or $ \le $ sign in between.

Learn these rules, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the  Mathway  site, where you can register for the  full version  (steps included) of the software.  You can even get math worksheets.

You can also go to the  Mathway  site here , where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to  Solving Radical Equations and Inequalities – you’re ready!

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Unit 1: Linear equations and inequalities

About this unit.

Linear equations and inequalities are the foundation of many advanced math topics, such as functions, systems, matrices, and calculus. Learn how to master them and unlock new possibilities for your future studies and careers in engineering, finance, computer science, and more.

Solving equations with one unknown

• Equations with parentheses (Opens a modal)
• Multi-step equations review (Opens a modal)
• Number of solutions to equations (Opens a modal)
• Worked example: number of solutions to equations (Opens a modal)
• Equations with parentheses Get 3 of 4 questions to level up!
• Equations with parentheses: decimals & fractions Get 3 of 4 questions to level up!
• Number of solutions to equations Get 3 of 4 questions to level up!

Solutions to linear equations

• Intro to the coordinate plane (Opens a modal)
• Solutions to 2-variable equations (Opens a modal)
• Worked example: solutions to 2-variable equations (Opens a modal)
• Creativity break: Why is creativity important in algebra? (Opens a modal)
• Interpreting points in context of graphs of systems (Opens a modal)
• Solutions to 2-variable equations Get 3 of 4 questions to level up!
• Interpret points relative to a system Get 3 of 4 questions to level up!

Multi-step linear inequalities

• Inequalities with variables on both sides (Opens a modal)
• Inequalities with variables on both sides (with parentheses) (Opens a modal)
• Multi-step inequalities (Opens a modal)
• Multi-step linear inequalities Get 3 of 4 questions to level up!

Compound inequalities

• Compound inequalities: OR (Opens a modal)
• Compound inequalities: AND (Opens a modal)
• A compound inequality with no solution (Opens a modal)
• Double inequalities (Opens a modal)
• Compound inequalities review (Opens a modal)
• Compound inequalities Get 3 of 4 questions to level up!

Modeling with linear equations and inequalities

• Comparing linear rates example (Opens a modal)
• Comparing linear rates word problems Get 3 of 4 questions to level up!

Absolute value equations

• Intro to absolute value equations and graphs (Opens a modal)
• Solving absolute value equations (Opens a modal)
• Absolute value equations Get 3 of 4 questions to level up!