Systems of quadratic equations often have more than one solution.

You can have no solutions and the graph would look like this:

No points of

intersection

You can have one solution and the graph might look something like this:

One intersectio

n point.

You could have two solutions and the graph might look something like this:

Two points of

intersection

Solving a System of Non-Linear Equations

The solution(s) are all (x, y) point(s) that make both equations true.

1.Graph and find the intersection point(s) on the graphing calculator.

2.Substitute the value of one variable from one equation into the other equation and solve for that variable, then the other.

3.Eliminate one variable and solve for that variable, then the other.

 Online extra help: Systems

EXAMPLES: Solve each system of equations

1. Solve by graphing or substitution.

*To solve by graphing, graph both equations on the same graph and find the intersection points.

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xy

Substitution: (You can check using a calculator, but elimination is not a good idea)

2x + 1 = x² - 2 Substitute y = 2x + 1 in for y in the second equation

0 = x² - 2x – 3 Write in standard form

0 = (x – 3) (x + 1) Factor

x - 3 = 0 or x + 1=0 Set each factor = 0

x = 3 x = -1 Solve each equation

But these are not points of intersection – we need to find the y-coordinate. Substitute the x value in either equation to find the y.

Solution 1: Solution 2:

x = 3, use y = 2x + 1 x = -1, use y = 2x + 1

y = 2(3) + 1 = 7 y = 2(-1) + 1 = -1

(3, 7) is a solution (-1, -1) is a solution

2. Solve by graphing, elimination, or substitution.

Elimination: (You can check using either of the other methods)

0 = 4x - 8 Subtract equation 1 from equation 2 and eliminate both y & x2

x = 2 Solve for x

But this is not a point, so find the y-coordinate by substitution.

Solution: x = 2, use y = x² + x – 1y = 2² + 2 – 1 = 5(2, 5) is the only point where the quadratics intersect

**You can check this point by graphing both equations on the same graph. (see next slide)

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