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How do solve the following linear system?: $- 9 x + 6 y = - 2, 3 x - y = 7$ $-9x + 6y = -2 , 3x-y=7$ ?

1 Answer

Mar 20, 2017

$(x, y) = (\frac{40}{9}, \frac{19}{3})$ $(x,y)=(40/9,19/3)$

Explanation:

Given
[1] $XXX - 9 x + 6 y = - 2$ $color(white)("XXX")-9x+6y=-2$
[2] $XXX 3 x - y = 7$ $color(white)("XXX")3x-y=7$

Multiplying [2] by $3$ $3$
[3] $XXX 9 x - 3 y = 21$ $color(white)("XXX")9x-3y=21$

Adding [1] and [3]
[4] $XXX 3 y = 19$ $color(white)("XXX")3y=19$

Dividing [4] by $3$ $3$
[5] $XXX y = \frac{19}{3}$ $color(white)("XXX")y=19/3$

Substituting $\frac{19}{4}$ $19/4$ for $y$ $y$ in [2]
[6] $XXX 3 x - \frac{19}{3} = 7$ $color(white)("XXX")3x-19/3=7$

Adding $\frac{19}{4}$ $19/4$ to both sides of [6] and simplifying the right side
[7] $XXX 3 x = \frac{40}{3}$ $color(white)("XXX")3x=40/3$

Dividing [7] by $3$ $3$
[8] $XXX x = \frac{40}{9}$ $color(white)("XXX")x=40/9$

You might want to simplify this as $(x, y) = (3 \frac{4}{9}, 6 \frac{1}{3})$ $(x,y)=(3 4/9,6 1/3)$
however
leaving these values in their improper fraction forms will probably make the verification easier (...and you always do this; right?)

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