How do you solve the system $2x+7y=8$ and $x+5y=7$ using substitution?

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Answer 1 · 2015-06-07T01:44:41

$x=-3$ and $y=2$ .

Solve $x+5y=7$ for $x$ .

$x=7-5y$

Substitute $7-5y$ for $x$ into the other equation.

$2(7-5y)+7y=8$ =

$14-10y+7y=8$ '

$14-3y=8$

Subtract $14$ from both sides.

$-3y=-6$

Divide both sides by $-3$ .

$y=2$

Substitute $2$ for $y$ in the first equation.

$x+5(2)=7$ =

$x+10=7$

Subtract $10$ from both sides.

$x=-3$

Check by substituting $-3$ for $x$ and $2$ for $y$ into both equations.

$2x+7y=8$ =

$2(-3)+7(2)=8$ =

$-6+14=8$ =

$8=8$ Check.

$x+5y=7$ =

$-3+5(2)=7$ =

$-3+10=7$ =

$7=7$ Check.

How do you solve the system 2x+7y=82x+7y=8 and x+5y=7x+5y=7 using substitution?