Systems of equations finding right triangles I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."
I know that the area is $A = \frac12 b  h$ and perimeter is $P = a + b + h$.
Using this, would the system look something like?
$a + b + c = 24$
$\frac12b c = 24$
(I'm using $c$ for consistency)
Then continue on using substitution?

You might be more comfortable using $a$ and $b$ for the two "legs" of the triangle, and $c$ for the hypotenuse. 
Then, exactly as you did, we obtain
$$a+b+c=24 \qquad \text{and}\qquad \frac{1}{2}ab=24.\qquad\qquad(1)$$ 
By the Pythagorean Theorem, we have
$$a^2+b^2=c^2.\qquad\qquad(2)$$
Now what? One approach (but not the only one) is to start by writing $c=24-(a+b)$, square both sides, and use Equation $(2)$ to eliminate $c$ and obtain a simple equation that involves only $a$ and $b$.  We have 
$$c^2=(24-(a+b))^2=24^2-48(a+b)+a^2+2ab+b^2=a^2+b^2.$$
There is a fair bit of cancellation. Note that $2ab=96$. So we get 
$$48(a+b)=24^2+96$$
and therefore $a+b=14$.  
Now we could write $b=14-a$ and substitute into $ab=48$ to get a quadratic equation in $a$. But the following is I think prettier. From $(a+b)^2=196$ and $4ab=192$ we conclude by subtraction that $(a-b)^2=4$, so $a-b=\pm 2$.  We can decide now that without loss of generality $a>b$. So $a-b=2$. From $a+b=14$, by adding and dividing by $2$, we find that $a=8$ and $b=6$. And of course $c=24-(a+b)=10$.
