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How do you solve by substitution $3 y = - \frac{1}{2} x + 2$ $3y=-1/2x+2$ and $y = - x + 9$ $y=-x+9$ ?

1 Answer

Jun 15, 2015

For the given equations, $(x, y) = (10, - 1)$ $(x,y) = (10,-1)$

Explanation:

[1] $XXXX$ $color(white)("XXXX")$ $3 y = - \frac{1}{2} x + 2$ $3y = -1/2x+2$
[2] $XXXX$ $color(white)("XXXX")$ $y = - x + 9$ $y = -x +9$

While it is not technically necessary, I find it easier to clear the fractions before actually beginning; so multiplying [1] by $2$ $2$
[3] $XXXX$ $color(white)("XXXX")$ $6 y = - x + 4$ $6y = -x+4$

Substituting (from [2]) $(- x + 9)$ $(-x+9)$ for $y$ $y$ in [3]
[4] $XXXX$ $color(white)("XXXX")$ $6 (- x + 9) = - x + 4$ $6(-x+9) = -x+4$

Simplifying
[5] $XXXX$ $color(white)("XXXX")$ $- 6 x + 54 = - x + 4$ $-6x+54 = -x+4$
[6] $XXXX$ $color(white)("XXXX")$ $5 x = 50$ $5x = 50$
[7] $XXXX$ $color(white)("XXXX")$ $x = 10$ $x=10$

Substituting (from [7]) $10$ $10$ for $x$ $x$ in [2]
[8] $XXXX$ $color(white)("XXXX")$ $y = - 10 + 9 = - 1$ $y = -10 +9 = -1$

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