Solve Systems of Equations by Substitution: example 5.15

Solve the system by substitution. ${\begin{matrix} 3 x + y = 5 \\ 2 x + 4 y = −10 \end{matrix}$

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. ${\begin{matrix} 4 x + y = 2 \\ 3 x + 2 y = −1 \end{matrix}$

How did it go?

Solve for y. Substitute into the other equation.
Replace the y with −3x + 5.
Solve the resulting equation for x.

Substitute x = 3 into 3x + y = 5 to find y.

The ordered pair is (3, −4).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} 3 x + y & = & 5 \\ 3 \cdot 3 + (- 4) & \overset{?}{=} & 5 \\ 9 - 4 & \overset{?}{=} & 5 \\ 5 & = & 5 ✓ \end{matrix} & \begin{matrix} 2 x + 4 y & = & - 10 \\ 2 \cdot 3 + 4 (- 4) & = & - 10 \\ 6 - 16 & \overset{?}{=} & - 10 \\ - 10 & = & - 10 ✓ \end{matrix} \end{matrix}$
	The solution is (3, −4).

Solve for y. Substitute into the other equation.
Replace the y with −3x + 5.
Solve the resulting equation for x.

Substitute x = 3 into 3x + y = 5 to find y.

The ordered pair is (3, −4).
Check the ordered pair in both equations: $\begin{matrix} \begin{matrix} 3 x + y & = & 5 \\ 3 \cdot 3 + (- 4) & \overset{?}{=} & 5 \\ 9 - 4 & \overset{?}{=} & 5 \\ 5 & = & 5 ✓ \end{matrix} & \begin{matrix} 2 x + 4 y & = & - 10 \\ 2 \cdot 3 + 4 (- 4) & = & - 10 \\ 6 - 16 & \overset{?}{=} & - 10 \\ - 10 & = & - 10 ✓ \end{matrix} \end{matrix}$
	The solution is (3, −4).