### Home > AC > Chapter 7 > Lesson 7.3.1 > Problem7-87

7-87.

Solve the following systems of equations. Remember to check your solution in both equations to make sure it is the point of intersection.

1. \begin{aligned}[t] &y = 2x -3 \\ &x - y = -4 \end{aligned}

One way to solve this system of equations is by using the substitution method.

$y=2x-3$. Substitute $y$:
$\begin{array}{l} x - y = -4 \\ x - \left(2x-3\right) = -4 \end{array}$

Solve for $x$.

Distribute the left side.

$x-2x+3=-4$

Combine like terms.

$-x+3=-4$

Subtract $3$ on both sides.

$-x=-7$

Use the solution you found for $x$ and substitute it into either original equation and solve for $y$.

$x = 7$

$y = 11$

1. \begin{aligned}[t] &y - x = -2 \\ &-3y + 2x = 14 \end{aligned}

One way to solve this system of equations is by using the elimination method.

Multiply the top equation by $2$.

$\begin{array}{l} 2\left(y - x = - 2\right)\\ 2y - 2x = -2 \end{array}$

Add the two equations together by combining like terms.

$\begin{array}{l} 2y - 2x = -4 \\ -3y + 2x = 14 \\ -y + 0 = 10 \end{array}$

Solve for $y$.

When you find $y$, you can now substitute the value into either original equation and solve for $x$.

Don't forget to check your solution!