How do you solve the following system?: $\frac{1}{9} x - \frac{3}{13} y = 1, \frac{3}{2} x - 8 y = 2$ $1/9x-3/13y=1 , 3/2x - 8y = 2$

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How do you solve the following system?: $\frac{1}{9} x - \frac{3}{13} y = 1, \frac{3}{2} x - 8 y = 2$ $1/9x-3/13y=1 , 3/2x - 8y = 2$

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Answer 1 · 2017-06-08T12:34:32

$y = - 0.25, x = 0$ $color(blue)(y=-0.25,x=0$

Explanation:

$:.1/9x-3/13y=1$ ----(1)

$:.3/2x-8y=2$ ----(2)

multiply $(2) xx 2$

$:.3x-16y=4$

$:.-16y=-3x+4$

$:.16y=3x-4$

$:.color(blue)(y=(3x-4)/16$

substitute $color(blue)(y=(3x-4)/16$ --- in $(2)$

$:.3/2x-8((3x-4)/16)=2$

$:.3/2x-(24x+32)/16=2$

multiply both sides by 16

$:.24x-24x+32=32$

$:.24x-24x=32-32$

$:.0=0$

$:.color(blue)(x=0$

substitute $color(blue)(x=0$ in $(2)$

$:.3/2(color(blue)0)-8y=2$

$:.-8y=2$

multiply both sides by $-1$

$:.8y=-2$

$:.y=-2/8$

$:.color(blue)(y=-0.25$ or $color(blue)( -1/4$

check:

substitute $color(blue)(y=-0.25$ in $(2)$

$:.3/2x-8(color(blue)(-0.25))=2$

$:.3/2x+2=2$

$:.3/2x=2-2$

$:.3/2x=0$

$:.color(blue)(x=0$

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How do you solve the following system?: $\frac{1}{9} x - \frac{3}{13} y = 1, \frac{3}{2} x - 8 y = 2$ $1/9x-3/13y=1 , 3/2x - 8y = 2$

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How do you solve the following system?: 1/9x-3/13y=1 , 3/2x - 8y = 2 19x−313y=1,32x−8y=2 1/9x-3/13y=1 , 3/2x - 8y = 2

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How do you solve the following system?: $\frac{1}{9} x - \frac{3}{13} y = 1, \frac{3}{2} x - 8 y = 2$ $1/9x-3/13y=1 , 3/2x - 8y = 2$