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How do solve the following linear system?: $6 x - 4 y = - 1, 3 x - y = 4$ $6x-4y=-1 , 3 x-y=4$ ?

1 Answer

Jun 1, 2016

Given
[1] $XXX 6 x - 4 y = - 1$ $color(white)("XXX")6x-4y=-1$
[2] $XXX 3 x - y = 4$ $color(white)("XXX")3x-y=4$

Subtract 2 times [2] from [1]
[1] $XXXxXX 6 x - 4 y = - 1$ $color(white)("XXXxXX")6x-4y=-1$
$2 \times$ $2xx$ [2] $color(white)("XXX")underline(6x-2y=color(white)("X"x)8)$
[3] $color(white)("XXXXXxXX")-2y=-9$

Divide both sides of [3] by $(-2)$
[4] $color(white)("XXXXXXXXX")y=9/2$

Substitute $9/2$ for $y$ in [1]
[5] $color(white)("XXXXXX")6x-4(9/2)=-1$

Simplify
[6] $color(white)("XXXXXX")6x-18=-1$

[7] $color(white)("XXXXXX")6x=17$

[8] $color(white)("XXXXXX")x=17/6$

So the solution is $(x,y)=(17/6,9/2)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As always you should verify this result.

In this case substitute $17/6$ for $x$ and $9/2$ for $y$ in [2]

$color(white)("XXX")3x-y$

$color(white)("XXXXXX")=3(17/6)-9/2$

$color(white)("XXXXXX")=17/2-9/2$

$color(white)("XXXXXX")=8/2$

$color(white)("XXXXXX")=4$ (which agrees with [2] so our work is valid)

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