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How do you solve the following system?: $x + 3 - 2 y = 7, x = - 3 x + 4 y - 17$ $x + 3-2y=7 , x=-3x + 4y-17$

1 Answer

Mar 15, 2018

$(x, y) = (- \frac{25}{2}, - \frac{33}{4})$ $(x,y)=(-25/2,-33/4)$

Explanation:

Given
[1] $XXX x + 3 - 2 y = 7$ $color(white)("XXX")x+3-2y=7$
[2] $XXX x = - 3 x + 4 y - 17$ $color(white)("XXX")x=-3x+4y-17$

Re-arrange [1] and [2] into standard form to get (respectively)
[3] $XXX x - 2 y = 4$ $color(white)("XXX")x-2y=4$
[4] $XXX 4 x - 4 y = - 17$ $color(white)("XXX")4x-4y=-17$

Since this was asked under "Systems Using Substitution"
Re-write [3] as $x$ $x$ in terms of $y$ $y$
[5] $XXX x = 2 y + 4$ $color(white)("XXX")x=2y+4$

Then substitute $(2 y + 4)$ $(2y+4)$ for $x$ $x$ in [4]
[6] $XXX 4 (2 y + 4) - 4 y = - 17$ $color(white)("XXX")4(2y+4)-4y=-17$

Simplifying
[7] $XXX 8 y + 16 - 4 y = - 17$ $color(white)("XXX")8y+16-4y=-17$

[8] $XXX 4 y = - 33$ $color(white)("XXX")4y=-33$

[9] $XXX y = - \frac{33}{4}$ $color(white)("XXX")y=-33/4$

Substituting $(- \frac{33}{4})$ $(-33/4)$ for $y$ $y$ in [3]
[10] $XXX x - 2 (- \frac{33}{4}) = 4$ $color(white)("XXX")x-2(-33/4)=4$

Simplifying
[11] $XXX x + \frac{33}{2} = \frac{8}{2}$ $color(white)("XXX")x+33/2=8/2$

[12] $XXX x = - \frac{25}{2}$ $color(white)("XXX")x=-25/2$

[These values may look a bit ugly, but substituting back into equations [1] and [2] verify that they are correct.]

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