Solve for x
x=\frac{3y-8}{7}
Solve for y
y=\frac{7x+8}{3}
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-7x-8=-3y
Subtract 3y from both sides. Anything subtracted from zero gives its negation.
-7x=-3y+8
Add 8 to both sides.
-7x=8-3y
The equation is in standard form.
\frac{-7x}{-7}=\frac{8-3y}{-7}
Divide both sides by -7.
x=\frac{8-3y}{-7}
Dividing by -7 undoes the multiplication by -7.
x=\frac{3y-8}{7}
Divide -3y+8 by -7.
3y-8=7x
Add 7x to both sides. Anything plus zero gives itself.
3y=7x+8
Add 8 to both sides.
\frac{3y}{3}=\frac{7x+8}{3}
Divide both sides by 3.
y=\frac{7x+8}{3}
Dividing by 3 undoes the multiplication by 3.
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