Solve Systems of Equations by Substitution: example 5.18

Solve the system by substitution. ${\begin{matrix} 4 x + 2 y = 4 \\ 6 x - y = 8 \end{matrix}$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system by substitution. ${\begin{matrix} x - 4 y = −4 \\ −3 x + 4 y = 0 \end{matrix}$

How did it go?


Solve the first equation for y.
Substitute −2x + 2 for y in the second equation.
Replace the y with −2x + 2.
Solve the equation for x.

Substitute $x = \frac{5}{4}$ into 4x + 2y = 4 to find y.
The ordered pair is $(\frac{5}{4}, - \frac{1}{2}) .$
Check the ordered pair in both equations. $\begin{matrix} \begin{matrix} 4 x + 2 y & = & 4 \\ 4 (\frac{5}{4}) + 2 (- \frac{1}{2}) & \overset{?}{=} & 4 \\ 5 - 1 & \overset{?}{=} & 4 \\ 4 & = & 4 ✓ \end{matrix} & \begin{matrix} 6 x - y & = & 8 \\ 6 (\frac{5}{4}) - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{15}{4} - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{16}{2} & \overset{?}{=} & 8 \\ 8 & = & 8 ✓ \end{matrix} \end{matrix}$
	The solution is $(\frac{5}{4}, - \frac{1}{2}) .$


Solve the first equation for y.
Substitute −2x + 2 for y in the second equation.
Replace the y with −2x + 2.
Solve the equation for x.

Substitute $x = \frac{5}{4}$ into 4x + 2y = 4 to find y.
The ordered pair is $(\frac{5}{4}, - \frac{1}{2}) .$
Check the ordered pair in both equations. $\begin{matrix} \begin{matrix} 4 x + 2 y & = & 4 \\ 4 (\frac{5}{4}) + 2 (- \frac{1}{2}) & \overset{?}{=} & 4 \\ 5 - 1 & \overset{?}{=} & 4 \\ 4 & = & 4 ✓ \end{matrix} & \begin{matrix} 6 x - y & = & 8 \\ 6 (\frac{5}{4}) - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{15}{4} - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{16}{2} & \overset{?}{=} & 8 \\ 8 & = & 8 ✓ \end{matrix} \end{matrix}$
	The solution is $(\frac{5}{4}, - \frac{1}{2}) .$