Consider the area between the graphs http://euler.vaniercollege.qc.ca/web...168a3bf431.png and http://euler.vaniercollege.qc.ca/web...6e724df531.png. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals

where http://euler.vaniercollege.qc.ca/web...ea9131f391.png , http://euler.vaniercollege.qc.ca/web...7973f0e8c1.png, http://euler.vaniercollege.qc.ca/web...61d577bdc1.png and

http://euler.vaniercollege.qc.ca/web...8785a207f1.png

http://euler.vaniercollege.qc.ca/web...77bbb02e81.png

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First off, to find a, b and c I believe we have to find the intercepts, and to do that we need to make the equations equal to each other and solve for x. But I'm having problems because of the y values when trying to isolate them.

$\displaystyle x + 3y = 36$

$\displaystyle y = - \frac {x}{3}+12$

$\displaystyle x+4=y^2$

$\displaystyle y = \sqrt{x+4}$

I'm stuck there. The square root poses a problem for me.

$\displaystyle - \frac{x}{3} + 12 = \sqrt{x+4} $

How can I isolate and solve for x?