Question 583100
Ill do one substitution and one elimination that way you can see the difference you should be able to get the other 2 just fine
{{{7/4y=7y+9}}}
{{{7=28y^2+36y}}}
{{{0=28y^2+36y-7}}}
*[invoke quadratic "y", 28, 36, -7]
y=.1716
y=-1.4573
{{{x=7y+9}}}
x=10.2012
x=-1.2011
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2x-5y=36
5x+7y=-27
im going to eliminate the ys
multiply top equation by 7 and bottom by 5
{{{7(2x-5y)=7(36)}}}
{{{5(5x+7y)=5(-27)}}}
.
{{{14x-35y=252}}}
{{{25x+35y=-135}}} add the equations together
.
{{{14x+25x-35y+35y=252-135}}}  add like terms
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{{{39x=117}}}  divide by 39
.
{{{highlight(x=3)}}}
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{{{2x-5y=36}}} imput x value and solve for y
{{{2(3)-5y=36}}}
{{{6-5y=36}}}
{{{-5y=30}}}
{{{highlight(y=-6)}}}