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

Activity

Solve each system without graphing.

Use this system of equations to answer questions 1 – 3.

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

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.

1.

What is the $x$ value?

6

2.

What is the $y$ value?

2

3.

Explain your solution.

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

$\begin{matrix} 5 x - 2 y & = & 26 \\ 5 (y + 4) - 2 y & = & 26 \\ 5 y + 20 - 2 y & = & 26 \\ 3 y + 20 & = & 26 \\ 3 y & = & 6 \\ y & = & = 2 \end{matrix}$

$\begin{matrix} 2 + 4 & = & x \\ 6 & = & x \end{matrix}$

$\begin{matrix} 5 (6) - 2 (2) = 26 \\ 30 - 4 \overset{?}{=} 26 & 2 + 4 \overset{?}{=} 6 \\ 26 = 26 ✓ & 6 = 6 ✓ \end{matrix}$

Use this system of equations to answer 4 - 6.

${\begin{matrix} 2 m - 2 p = - 6 \\ p = 2 m + 10 \end{matrix}$

4.

What is the $m$ value?

-7

5.

What is the $p$ value?

-4

6.

Explain your solution.

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

$\begin{matrix} 2 m - 2 p & = & - 6 \\ 2 m - 2 (2 m + 10) & = & - 6 \\ 2 m - 4 m - 20 & = & - 6 \\ - 2 m - 20 & = & - 6 \\ - 2 m & = & 14 \\ m & = & - 7 \end{matrix}$

$\begin{matrix} p & = & 2 m + 10 \\ p & = & 2 (- 7) + 10 \\ p & = & - 14 + 10 \\ p & = & - 4 \end{matrix}$

$\begin{matrix} 2 (- 7) - 2 (- 4) = - 6 & - 4 = 2 (- 7) + 10 \\ - 14 + 8 \overset{?}{=} - 6 & - 4 \overset{?}{=} (- 14) + 10 \\ - 6 = - 6 ✓ & - 4 = - 4 ✓ \end{matrix}$

Use this system of equations to answer 7 - 9.

${\begin{matrix} 2 d = 8 f \\ 18 - 4 f = 2 d \end{matrix}$

7.

What is the $d$ value?

6

8.

What is the $f$ value?

Compare your answer:

$\frac{3}{2}$

9.

Explain your solution.

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

$\begin{matrix} 2 d & = & 8 f \\ d & = & 4 f \end{matrix}$

$\begin{matrix} 18 - 4 f & = & 2 d \\ 18 - 4 f & = & 2 (4 f) \\ 18 - 4 f & = & 8 f \\ 18 & = & 12 f \\ \frac{18}{12} & = & f \\ \frac{3}{2} & = & f \end{matrix}$

$\begin{matrix} 2 d & = & 8 (\frac{3}{2}) \\ 2 d & = & 12 \\ d & = & 6 \end{matrix}$

$\begin{matrix} 18 - 4 f = 2 d \\ 2 d = 8 f & 18 - 4 (\frac{3}{2}) = 2 (6) \\ 2 (6) = 8 (\frac{3}{2}) & 18 - (2) (3) = 12 \\ 12 \overset{?}{=} (4) (3) & 18 - 6 = 12 \\ 12 = 12 ✓ & 12 = 12 ✓ \end{matrix}$

Use this system of equations to answer 10 - 12.

${\begin{matrix} w + \frac{1}{7} z = 4 \\ z = 3 w - 2 \end{matrix}$

10.

What is the $w$ value?

3

11.

What is the $z$ value?

7

12.

Explain your solution.

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

$\begin{matrix} w + \frac{1}{7} z & = & 4 \\ w + \frac{1}{7} (3 w - 2) & = & 4 \\ 7 w + 3 w - 2 & = & 28 \\ 10 w - 2 & = & 28 \\ 10 w & = & 30 \\ w & = & 3 \end{matrix}$

$\begin{matrix} z & = & 3 w - 2 \\ z & = & 3 (3) - 2 \\ z & = & 9 - 2 \\ z & = & 7 \end{matrix}$

$\begin{matrix} w + \frac{1}{7} z = 4 & z = 3 w - 2 \\ 3 + \frac{1}{7} (7) = 4 & 7 = 3 (3) - 2 \\ 3 + 1 \overset{?}{=} 1 = 4 & 7 \overset{?}{=} 1 = 9 - 2 \\ 4 = 4 ✓ & 7 = 7 ✓ \end{matrix}$

Are you ready for more?

Extending Your Thinking

Solve this system with four equations.

Use this system of equations to answer questions 1 – 4.

${\begin{matrix} 3 x + 2 y - z + 5 w = 20 \\ y = 2 z - 3 w \\ z = w + 1 \\ 2 w = 8 \end{matrix}$

1.

What is the $x$ value?

3

2.

What is the $y$ value?

-2

3.

What is the $z$ value?

5

4.

What is the $w$ value?

4

Self Check

Solve the systems of equations by substitution.

$\left\{\begin{array}{l}2x\;+\;y\;=\;-4\\\;3x\;-\;2y\;=\;-6\end{array}\right.$

$x = -6$
$y = 6$
$x = -4$
$y = 0$
$x = 0$
$y = 3$
$x = -2$
$y = 0$

Additional Resources

Reinforcing the Substitution Method

Solve the system by substitution:

${\begin{matrix} 4 x + 2 y = 4 \\ 6 x + - y = 8 \end{matrix}$

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

Step 1 - Solve the first equation for $y$ .

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

Step 2 - Substitute $- 2 x + 2$ for $y$ in the second equation.

$\begin{matrix} 6 x - y & = & 8 \\ 6 x - (- 2 x + 2) & = & 8 \end{matrix}$

Step 3 - Solve the equation for $x$ .

$\begin{matrix} 6 x - (- 2 x + 2) & = & 8 \\ 6 x + 2 x - 2 & = & 8 \\ 8 x - 2 & = & 8 \\ 8 x & = & 10 \\ x & = & \frac{10}{8} \\ x & = & \frac{5}{4} \end{matrix}$

Step 4 - Substitute $x = \frac{5}{4}$ into $4 x + 2 y = 4$ to find $y$ .

$\begin{matrix} 4 x + 2 y & = & 4 \\ 4 (\frac{5}{4}) + 2 y & = & 4 \\ 5 + 2 y & = & 4 \\ 2 y & = & - 1 \\ y & = & \frac{- 1}{2} \end{matrix}$

Step 5 - Check the ordered pair. $(\frac{5}{4}, \frac{- 1}{2})$ in both equations.

$\begin{matrix} 6 x - y = 8 \\ 6 (\frac{5}{4}) - (- \frac{1}{2}) \overset{?}{=} 8 \\ 4 x + 2 y = 4 & \frac{30}{4} - (- \frac{1}{2}) \overset{?}{=} 8 \\ 4 (- \frac{1}{2}) + 2 (\frac{5}{4}) \overset{?}{=} 4 & \frac{15}{2} - (- \frac{1}{2}) \overset{?}{=} 8 \\ 5 - 1 \overset{?}{=} 4 & \frac{15}{2} + \frac{1}{2} \overset{?}{=} 8 \\ 4 = 4 ✓ & 8 = 8 ✓ \end{matrix}$

Try it

Try It: Reinforcing the Substitution Method

1.

Solve the system by substitution:

${\begin{matrix} x - 4 y = - 4 \\ - 3 x + 4 y = 0 \end{matrix}$

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

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

$\begin{matrix} x - 4 y & = & - 4 \\ x - 4 y + 4 y & = & - 4 + 4 y \\ x & = & - 4 + 4 y \end{matrix}$

Step 2 - Replace the $x$ in the second equation with the expression $- 4 + 4 y$ .

$\begin{matrix} - 3 x + 4 y & = & 0 \\ - 3 (- 4 + 4 y) + 4 y & = & 0 \end{matrix}$

Step 3 - Solve the equation with just one variable.

$\begin{matrix} - 3 (- 4 + 4 y) + 4 y & = & 0 \\ 12 - 12 y + 4 y & = & 0 \\ - 8 y & = & - 12 \\ y & = & \frac{3}{2} \end{matrix}$

Step 4 - Use the first equation and replace $y$ with $\frac{3}{2}$ .

$\begin{matrix} x - 4 y & = & - 4 \\ x - 4 (\frac{3}{2}) & = & - 4 \\ x - 6 & = & - 4 \\ x & = & 2 \end{matrix}$

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

$(2, \frac{3}{2})$

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

$\begin{matrix} x - 4 y = - 4 \\ (2) - 4 (\frac{3}{2}) \overset{?}{=} - 4 \\ 2 - 6 \overset{?}{=} - 4 \\ - 4 = - 4 ✔ \end{matrix}$

$\begin{matrix} - 3 x + 4 y = 0 \\ - 3 (2) - 4 (\frac{3}{2}) \overset{?}{=} 0 \\ - 6 + 6 \overset{?}{=} 0 \\ 0 = 0 ✔ \end{matrix}$

2.3.3 Solving Equations Using Substitution

Activity

Are you ready for more?

Extending Your Thinking

Additional Resources

Reinforcing the Substitution Method

Try It: Reinforcing the Substitution Method