But I can't graph
that negative inside the square root! What's going on here?

Take a look at
the graph:

The lines do not intersect.
Since there is no intersection, then there is no solution. That is,
this is an inconsistent system. My final answer is: no
solution: inconsistent system.

Then the
solution consists of the points (–4,
1) and (1,
–4).

Note the procedure: I solved
one of the equations (the first equation looked easier) for one of the
variables (solving for "y="
looked easier), and
then plugged the resulting expression back into the other equation. This
gave me one equation in one variable (the variable happened to be x),
and a one-variable equation is something I know how to solve. Once I had
the solution values for x,
I back-solved for the correspondingy-values.
I emphasize "corresponding" because you have to keep track of
which y-value
goes with which x-value.
In the example above, the points (–4,
–4) and (1,
1) are not solutions.
Even though I came up with x
= –4 and 1
and y
= –4 and 1,
the x
= –4 did not go with
the y
= –4, and the x
= 1 did not go with
the y
= 1. Warning: You must
match the x-values
and y-values
correctly!

Solve the following
system of equations:

y
= (^{1}/_{2})x – 5 y = x^{2}
+ 2x – 15

Since both equations
are already solved for y,
I'll set them equal and solve for the values of x:

Stapel, Elizabeth.
"Systems of Non-Linear Equations: Intermediate Systems."
Purplemath. Available from http://www.purplemath.com/modules/syseqgen4.htm.
Accessed