### Home > AC > Chapter 8 > Lesson 8.2.2 > Problem8-58

8-58.

Solve the following systems of equations using any method. Check your solution if possible.

1. $6x−2y=10$
$3x−y=2$

1. $x−3y=1$
$y=16−2x$

You can use the elimination method.

Multiply the second equation by $−2$.

Add the two equations together, combining like terms.

There is no solution.

$−2(3x−y=2)$
$−6x+2y=−4$

$6x−2y=10$
$−6x+2y=−4$
$0=6$

You can use the substitution method.

Distribute and solve for $x$.

Combine like terms.

Add $48$ to both sides.

Divide by $7$ on both sides.

$x−3(16−2x)=1$
$x−48+6x=1$

$7x−48=1$

$7x=49$