Question 883590

{{{8x-7y=-51}}}...........eq.1
{{{12x+13y=41}}}..........eq.2
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how do we know what number to multiply by to cancel out the {{{x}}}: 

first look what are coefficients of the {{{x}}} in eq.1 and eq.2, you see these are {{{8}}} and {{{12}}}
now, you know that {{{24}}} is divisible by {{{8}}} and by {{{12}}}; so make it as common  coefficients of the {{{x}}} in eq.1 and eq.2


{{{8x-7y=-51}}}...........eq.1.............multiply by {{{3}}}
{{{12x+13y=41}}}..........eq.2.............multiply by {{{2}}}
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{{{24x-21y=-153}}}...........eq.1
{{{24x+26y=82}}}..........eq.2
___________________________...............now subtract eq.2 from eq.1

{{{24x-21y=-153}}}...........eq.1
-
{{{24x+26y=82}}}..........eq.2
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{{{cross(24x)-cross(24x)-21y-26y=-153-82}}}

{{{-47y=-235}}}

{{{y=-235/-47}}}

{{{highlight(y=5)}}}

now, go back to eq.1, plug in {{{y=5}}} and solve for {{{x}}}


{{{8x-7*5=-51}}}...........eq.1

{{{8x-35=-51}}}

{{{8x=-51+35}}}

{{{8x=-16}}}

{{{x=-16/8}}}

{{{highlight(x=-2)}}}

see it on a graph:


{{{ graph( 600, 600, -10, 10, -10, 10, 8x/7+51/7, -12x/13+41/13) }}}