How do you solve $8x+7y=-5$ and $x=35-8y$ using substitution?

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Answer 1 · 2017-03-22T02:09:49

See the entire solution process below:

Step 1) Because the second equation is already solved for $x$ we can substitute $35 - 8y$ for $x$ in the first equation and solve for $y$ :

$8x + 7y = -5$ becomes:

$8(35 - 8y) + 7y = -5$

$(8 xx 35) - (8 xx 8y) + 7y = -5$

$280 - 64y + 7y = -5$

$280 - 57y = -5$

$-color(red)(280) + 280 - 57y = -color(red)(280) - 5$

$0 - 57y = -285$

$-57y = -285$

$(-57y)/color(red)(-57) = (-285)//color(red)(-57)$

$(color(red)(cancel(color(black)(-57)))y)/cancel(color(red)(-57)) = 5$

$y = 5$

Step 2) Now, substitute $5$ for $y$ in the second equation and calculate $x$ :

$x = 35 - 8y$ becomes:

$x = 35 - (8 xx 5)$

$x = 35 - 40$

$x = -5$

The solution is: $x = -5$ and $y = 5$ or $(-5, 5)$

How do you solve 8x+7y=-5 8x+7y=-5 and x=35-8yx=35-8y using substitution?