How do you solve 7y - 2x = 10 and -3y + x = -3 using substitution?

Oct 17, 2017

See a solution process below:

Explanation:

Step 1: Solve the second equation for $x$:

$- 3 y + x = - 3$

$\textcolor{red}{3 y} - 3 y + x = \textcolor{red}{3 y} - 3$

$0 + x = 3 y - 3$

$x = 3 y - 3$

Step 2: Substitute $\left(3 y - 3\right)$ for $x$ in the first equation and solve for $y$:

$7 y - 2 x = 10$ becomes:

$7 y - 2 \left(3 y - 3\right) = 10$

$7 y - \left(2 \cdot 3 y\right) - \left(2 \cdot - 3\right) = 10$

$7 y - 6 y - \left(- 6\right) = 10$

$7 y - 6 y + 6 = 10$

$\left(7 - 6\right) y + 6 = 10$

$1 y + 6 = 10$

$y + 6 - \textcolor{red}{6} = 10 - \textcolor{red}{6}$

$y + 0 = 4$

$y = 4$

Step 3: Substitute $4$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = 3 y - 3$ becomes:

$x = \left(3 \cdot 4\right) - 3$

$x = 12 - 3$

$x = 9$

The Solution Is: $x = 9$ and $y = 4$ or $\left(9 , 4\right)$