Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

Here are the four systems of equations you saw in the previous exercise. Solve each system. Then, check your solutions by substituting them into the original equations to see if the equations are true.

Use this system of equations to answer 1 - 3.

${\begin{matrix} x + 2 y = 8 \\ x = - 5 \end{matrix}$

1.

What is the $x$ value?

-5

2.

What is the $y$ value?

6.5

3.

Explain your solution.

Compare your answer: Your answer may vary, but here is a sample.

$\begin{matrix} - 5 + 2 y & = & 8 \\ 2 y & = & 13 \\ y & = & 6.5 \end{matrix}$

$\begin{matrix} x + 2 y & = & 8 \\ x + 2 (6.5) & = & 8 \\ x + 13 & = & 8 \\ x & = & - 5 \end{matrix}$

$\begin{matrix} - 5 + 2 (6.5) \\ - 5 + 13 \overset{?}{=} 8 \\ 8 = 8 ✓ & - 5 = - 5 ✓ \end{matrix}$

Use this system of equations to answer 4 - 6.

${\begin{matrix} y = - 7 x + 13 \\ y = - 1 \end{matrix}$

4.

What is the $x$ value?

2

5.

What is the $y$ value?

-1

6.

Explain your solution.

Compare your answer: Your answer may vary, but here is a sample.

$\begin{matrix} - 1 & = & - 7 x + 13 \\ - 14 & = & - 7 x \\ 2 & = & x \end{matrix}$

$\begin{matrix} y & = & - 7 (2) + 13 \\ y & = & - 14 + 13 \\ y & = & - 1 \end{matrix}$

$\begin{matrix} - 1 = - 7 (2) + 13 \\ - 1 \overset{?}{=} - 14 + 13 \\ - 1 = - 1 ✓ & - 1 = - 1 ✓ \end{matrix}$

Use this system of equations to answer questions 7 – 9.

${\begin{matrix} 3 x = 8 \\ 3 x + y = 15 \end{matrix}$

7.

What is the $x$ value?

Compare your answer:

$\frac{8}{3}$

8.

What is the $y$ value?

7

9.

Explain your solution.

Compare your answer: Your answer may vary, but here is a sample.

$\begin{matrix} 3 x & = & 8 \\ x & = & \frac{8}{3} \end{matrix}$

$\begin{matrix} 3 (\frac{8}{3}) + y & = & 15 \\ 8 + y & = & 15 \\ y & = & = 7 \end{matrix}$

$\begin{matrix} 3 (\frac{8}{3}) + 7 = 15 \\ - 8 + 7 \overset{?}{=} 15 & 3 (\frac{8}{3}) \overset{?}{=} 8 \\ 15 = 15 ✓ & 8 = 8 ✓ \end{matrix}$

Use this system of equations to answer questions 10 – 12.

${\begin{matrix} y = 2 x - 7 \\ 4 + y = 12 \end{matrix}$

10.

What is the $x$ value?

7.5

11.

What is the $y$ value?

8

12.

Explain your solution.

Compare your answer: Your answer may vary, but here is a sample.

$\begin{matrix} 4 + y & = & 12 \\ y & = & 8 \end{matrix}$

$\begin{matrix} 8 & = & 2 x - 7 \\ 15 & = & 2 x \\ 7.5 & = & x \end{matrix}$

$\begin{matrix} 8 = 2 (7.5) - 7 \\ 8 = 2 (7.5) - 7 & 4 + y = 12 \\ 8 \overset{?}{=} 15 - 7 & 4 + 8 \overset{?}{=} 12 \\ 8 = 8 ✓ & 12 = 12 ✓ \end{matrix}$

Video: Checking Solutions in Systems

Watch the following video to learn more about systems.

Access multimedia content

Self Check

Solve the following system of equations. Check to see if your solution makes each equation true.

$\left\{\begin{array}{l}3x\;+\;y\;=\;6\\x\;+\;2y\;=\;7\end{array}\right.$

$x = 1$ , $y = 3$
$x = 3$ , $y = -3$
$x = -3$ , $y = 5$
$x = 3$ , $y = 1$

Additional Resources

Solving Systems of Linear Equations

Let’s solve this system of linear equations using the substitution method.

${\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}$

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.

How to 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 from 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.

Solve the system by substitution: ${\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix}$

Solution

Step 1 - Solve one of the equations for either variable.

We’ll solve the first equation for y.

$\begin{matrix} {\begin{matrix} 2 x + y = 7 \\ x - 2 y = 6 \end{matrix} \end{matrix}$

$\begin{matrix} 2 x + y & = & 7 \\ y & = & 7 - 2 x \end{matrix}$

Step 2 - Substitute the expression from Step 1 into the other equation.

$\begin{matrix} x - 2 y & = & 6 \\ x - 2 (7 - 2 x) & = & 6 \end{matrix}$

Step 3 - Solve the resulting equation.

Now we have an equation with just 1 variable. We know how to solve this!

$\begin{matrix} x - 2 (7 - 2 x) & = & 6 \\ x - 14 + 4 x & = & 6 \\ 5 x & = & 20 \\ x & = & 4 \end{matrix}$

Step 4 - Substitute the solution from Step 3 into one of the original equations to find the other variable.

We’ll us the first equation and replace $x$ with 4.

$\begin{matrix} 2 x + y & = & 7 \\ 2 (4) + y & = & 7 \\ 8 + y & = & 7 \\ y & = & - 1 \end{matrix}$

Step 5 - Write the solution as an ordered pair.

The ordered pair is $(x, y)$ .

$(4, - 1)$

Step 6 - Check that the ordered pair is a solution to both original equations.

Substitute $x = 4$ , $y = - 1$ into both equations and make sure they are both true.

$\begin{matrix} r v r 2 x + y = 7 & x - 2 y = 6 \\ 2 (4) + (- 1) \overset{?}{=} 7 & 4 - 2 (- 1) \overset{?}{=} 6 \\ 7 = 7 ✓ & 6 = 6 ✓ \end{matrix}$

Both equations are true.

$(4, - 1)$ is the solution to the system.

Try it

Try It: Solving Systems of Linear Equations

Solve the system by substitution, then choose the correct ordered pair. ${\begin{matrix} - 2 x + y = - 11 \\ x + 3 y = 9 \end{matrix}$

Multiple Choice:

$(9, - 11)$

$(1, 6)$

$(6, 1)$

$(- 11, 9)$

$(6, 1)$

This is the solution.

Step 1 - Solve the first equation for $y$ to isolate a variable.Solve the first equation for $y$ to isolate a variable.

$\begin{matrix} - 2 x + y & = & - 11 \\ - 2 x + 2 x + y & = & - 11 + 2 x \\ y & = & - 11 + 2 x \end{matrix}$

Step 2 - Replace the $y$ in the second equation with the expression $- 11 + 2 x$ .Replace the $y$ in the second equation with the expression $- 11 + 2 x$ .

$\begin{matrix} x + 3 y & = & 9 \\ x + 3 (- 11 + 2 x) & = & 9 \\ x - 33 + 6 x & = & 9 \end{matrix}$

Step 3 - Solve the equation with just one variable.Solve the equation with just one variable.

Now we have an equation with just 1 variable. We know how to solve this!

$\begin{matrix} 7 x & = & 42 \\ x & = & 6 \end{matrix}$

Step 4 - Use the first equation and replace $x$ with 6.Use the first equation and replace $x$ with 6.

$\begin{matrix} - 2 x + y & = & - 11 \\ - 2 (6) + y & = & - 11 \\ - 12 + y & = & - 11 \\ y & = & 1 \end{matrix}$

Step 5 - The ordered pair is $(x, y)$ .The ordered pair is $(x, y)$ .

$(6, 1)$

Step 6 - Substitute $x = 6$ , $y = 1$ to check the solution in both equations.Substitute $x = 6$ , $y = 1$ to check the solution in both equations.

$\begin{matrix} - 2 x + y = - 11 \\ - 2 (6) + (1) \overset{?}{=} - 11 & x + 3 y = 9 \\ - 12 + 1 \overset{?}{=} - 11 & (6) + 3 (1) \overset{?}{=} 9 \\ - 11 = - 11 ✓ & 9 = 9 ✓ \end{matrix}$

2.3.2 Checking Solutions in Systems

Activity

Video: Checking Solutions in Systems

Additional Resources

Solving Systems of Linear Equations

Solution

Try It: Solving Systems of Linear Equations