The sides of a right triangle have a relationship as depicted below, where x is the length of the longer leg in inches.The area of this triangle, A, is equal to the perimeter of this triangle.Create a system of equations to model this situation. Determine if there are any solutions, and, if possible, whether or not they are viable.How many total possible solutions are there to this system? Of any possible solutions, how many are viable solutions for this situation?

The equation to represent the area of the triangle would be:
y = 1/2(x²) - (7/2)x

The equation to represent the perimeter of the triangle would be:
y = 3x - 6

The solutions to the system would be (12, 30) or (1, -3).  The only viable solution is (12, 30).

Explanation
The area of a triangle is found using the formula
A = 1/2bh

For our triangle, b = x and h = x-7, so we have:
A = 1/2(x)(x-7)
A = 1/2(x²-7x)
A = 1/2(x²) - (7/2)x

We will replace A with y, so we have:
y = 1/2(x²) - (7/2)x

The perimeter of a triangle is found by adding together all sides, so we have:
P = (x-7) + x + (x+1)

Combining like terms we get:
P = 3x - 6

We will replace P with y, so we have:
y = 3x - 6

Since both equations have y isolated on one side, it will be easy to use substitution to solve the system:

3x - 6 = 1/2(x²) - (7/2)x

It's easier to work with whole numbers, so we will multiply everything by 2:
6x - 12 = x² - 7x

We want all of the variables on one side, so we will subtract 6x:
6x - 12 - 6x = x² - 7x - 6x
-12 = x² - 13x

When solving quadratics, we want the equation equal to 0, so we will add 12:
-12+12 = x² - 13x + 12
0 = x² - 13x + 12

This is easy to factor, as there are factors of 12 that sum to -13; -12(-1) = 12 and -12+-1 = -13:
0 = (x-12)(x-1)

Using the zero product property, we know that either x-12=0 or x-1=0; therefore x=12 or x=1.

Putting these back into our equation for perimeter (the simplest one) we have:
y = 3(12)-6 = 36-6 = 30; (12, 30);
y = 3(1) - 6 = 3 - 6 = -3; (1, -3)

We cannot have a negative perimeter, so the only viable solution is (12, 30).