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How do you solve the following system?: $16 x + 3 y = 7, 9 x + 5 y = - 12$ $16x +3y =7, 9x +5y = -12$

1 Answer

Jan 4, 2016

$(x, y) = (\frac{71}{53}, - \frac{255}{53})$ $(x,y)=(71/53, -255/53)$

Explanation:

Given
[1] $XXX 16 x + 3 y = 7$ $color(white)("XXX")16x+3y=7$
[2] $XXX 9 x + 5 y = - 12$ $color(white)("XXX")9x+5y=-12$

Multiply [1] by $5$ $5$ and [2] by $3$ $3$ to get equations with equal coefficients for $y$ $y$
[3] $XXX 80 x + 15 y = 35$ $color(white)("XXX")80x+15y=35$
[4] $XXX 27 x + 15 y = - 36$ $color(white)("XXX")27x+15y=-36$

Subtract [4] from [3]
[5] $XXX 53 x = 71$ $color(white)("XXX")53x=71$

Divide by $23$ $23$
[6] $XXX x = \frac{71}{53}$ $color(white)("XXX")x=71/53$
$XXX$ $color(white)("XXX")$ Yes; it's going to be ugly; but that doesn't mean it's going to be wrong.

Substitute $\frac{71}{53}$ $71/53$ for $x$ $x$ in [1]
[7] $XXX \frac{16 \times 71}{53} + 3 y = 7$ $color(white)("XXX")(16xx71)/53+3y=7$

Simplify
[8] $XXX \frac{1136}{53} + \frac{159}{53} y = \frac{371}{53}$ $color(white)("XXX")1136/53+159/53y=371/53$

[9] $XXX 159 y = - 765$ $color(white)("XXX")159y=-765$

[10] $XXX y = - \frac{765}{159} = - \frac{255}{23}$ $color(white)("XXX")y=-765/159=-255/23$

I would (strongly) recommend using a calculator or spreadsheet to verify these values in the original equations.

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