How do you solve by substitution: 3x-8y=-7 and -5x-6y=3?

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Answer 1 · 2018-05-27T18:46:18

$y = -34/38 and x = -269/57$

$x = (-7 + 8y)/3$

-5(x) - 6y = 3
-5((-7 + 8y)/3) - 6y = 3

(35)/3 + (40y)/3 - 6y = 3
35 + 40y -2y = 1
38y = -34
y = -34/38

x = (-7 + 8(-34/38))/3
x = -269/57

Tony B

Answer 2 · 2018-05-27T20:45:43

$y=+13/29$

I will let you finish off

Given:
$+3x-8y=-7" ".........................Equation(1)$
$-5x-6y=3" ".............................Equation(2)$

Consider $Eqn(1)$

Add $8y$ to both sides

$3x=8y-7$

Divide both sides by 3

$color(red)(x=8/3y-7/3)" "...........................Equation(1_a)$
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Using $Eqn(1_a)$ substitute for $x$ in $Eqn(2)$

$color(green)(-5(color(red)(x))-6y=3 color(white)("ddd")->color(white)("ddd")-5(color(red)((8y)/3-7/3))-6y=3 )$

$color(green)(color(white)("ddddddddddddddd")->color(white)("ddd")-(40y)/3+35/3color(white)("dd")-6y=3 )$

Put everything over 3 to make calculation more straight forward.

$color(green)(color(white)("ddddddddddddddd")->color(white)("ddd")-(40y)/3+35/3color(white)("dd")-(18y)/3=9/3 )$

Multiply both sides by 3

$color(green)(color(white)("ddddddddddddddd")->color(white)("ddd")-40y+35color(white)("dd")-18y=9$

$-58y=-26$

$y=+13/29$
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I will let you finish off