How to Solve Nonlinear Systems on the SAT

Welcome to SAT Climb. When a system mixes a line with a parabola, it looks like two problems stacked together. It's really one move, and once you see it, nonlinear systems stop being scary and start being some of the most predictable points on the test.

Here's the whole idea. One equation is a line, the other is a parabola, and both are set equal to y. If they're both equal to y, they're equal to each other. So set the two right sides equal, move everything to one side, and you're left with a single quadratic. Solve that quadratic and you've solved the system. Let's try one. y equals x squared, and y equals x plus two. Where do these two graphs meet?

Set the two right sides equal. x squared equals x plus two. Now move the entire right side across, so x squared minus x minus two equals zero. That factors into x minus two, times x plus one. So x is two, or x is negative one. Two answers, because a line usually cuts a parabola in two places. Last step, find y, and use the line, it's the easy one. At x equals two, y is four. At x equals negative one, y is one. The graphs meet at two, four, and at negative one, one.

Sometimes the SAT doesn't want the points, it just asks how many there are. Count them with the discriminant, b squared minus four a c, from the quadratic you collected. Positive means two solutions, the line crosses. Zero means exactly one, the line just touches, it's tangent. Negative means none, the line misses completely. Same parabola. The line y equals four x minus four gives x squared minus four x plus four, a perfect square, discriminant zero, so it touches at one point. The line y equals x minus one gives a discriminant of negative three, so it never meets the parabola at all.

Your turn. y equals x squared minus three, and y equals two x. Set them equal, x squared minus three equals two x. Move it over, x squared minus two x minus three equals zero. That factors into x minus three, times x plus one. So x is three or negative one. Put each back into the line, and the meeting points are three, six, and negative one, negative two.

Three traps. One, stopping at a single answer. A line and a parabola usually meet twice, so find both roots. Two, solving for x and forgetting y, plug each x back into the line to finish the point. Three, a sign slip when you move terms, subtract the entire right side, and remember the discriminant counts the solutions, positive is two, zero is one, negative is none. Set them equal, make one quadratic, and solve it all the way.

Nonlinear systems, solved. Set the equations equal, collect everything into one quadratic, and solve it, or let the discriminant count the solutions for you. Start practicing free at satclimb.com. Your SAT is closer than you think.