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How do solve the following linear system?: $9 x - 6 y = 3, - 3 x - 8 y = - 9$ $9 x-6y =3 , -3x -8y = -9$ ?

1 Answer

Feb 3, 2016

$(x, y) = (\frac{13}{15}, \frac{4}{5})$ $(x,y) = (13/15,4/5)$

Explanation:

Given
[equation 1] $XXX 9 x - 6 y = 3$ $color(white)("XXX")9x-6y=3$
[equation 2] $XXX - 3 x - 8 y = - 9$ $color(white)("XXX")-3x-8y=-9$

If we multiply [equation 2] by $3$ $3$ then the coefficient of $x$ $x$ will be of the same magnitude as in [equation 1]
[equation 3] $XXX - 9 x - 24 y = - 27$ $color(white)("XXX")-9x-24y=-27$

Adding [equation 1] and [equation 3]
[equation 4] $XXX - 30 y = - 24$ $color(white)("XXX")-30y=-24$

Dividing [equation 4] by $(- 30)$ $(-30)$ gives
[equation 5] $XXX y = \frac{24}{30} = \frac{4}{5}$ $color(white)("XXX")y= 24/30 = 4/5$

Substituting $\frac{4}{5}$ $4/5$ for $y$ $y$ in [equation 1]
[equation 6] $XXX 9 x - 6 \times \frac{4}{5} = 3$ $color(white)("XXX")9x-6xx4/5 =3$

[equation 7] $XXX 9 x = 3 + \frac{24}{5} = \frac{39}{5}$ $color(white)("XXX")9x=3+24/5=39/5$

Dividing [equation 7] by $9$ $9$ gives
[equation 8] $XXX x = \frac{39}{45} = \frac{13}{15}$ $color(white)("XXX")x=39/45=13/15$

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