Title: Lesson 1–1
Author: Teri Willard
Last saved by: Glencoe Mathematics

Lesson 10–7

Example 1  Linear–Quadratic System
Solve the system of equations.

 +  = 1
3y – 2x = 6

You can use a graphing calculator to help visualize the relationships of the graphs of the equations and predict the number of solutions.
Since an ellipse is easy to graph, sketch this ellipse on grid paper. Rewrite 3y – 2x = 6 as y = x + 2 and graph the line on the same axes. The graph indicates that the ellipse and line intersect in two points. So the system has two solutions.

Use substitution to solve the system.

   +  = 1	First equation in the system
   +  = 1	Substitute x + 2 for y.
4x^2 + 9 = 36	Multiply each side by 36.
 4x^2 + 9= 36	Expand the binomial square.
    4x^2 + 4x^2 + 24x + 36 = 36	Distributive Property
    8x^2 + 24x = 0	Simplify and subtract 36 from each side.
      8x(x + 3) = 0	Factor.
8x = 0       or     x + 3 = 0	Zero Product Property
x = 0                      x = –3	Solve each equation for x.

Now solve for y.
y = x + 2	y = x + 2	Equation for y in terms of x
= (0) + 2	= (–3) + 2	Substitute the x–values.
= 2	= 0	Simplify.

The solutions to the system are (0, 2) and (–3, 0). Based on the graph, these solutions are reasonable.

Example 2  Quadratic–Quadratic System
Solve the system of equations.
y = –2x^2
y = 2x^2 – 4

A graph indicates that the two parabolas intersect in two points. So, this system has two solutions. 

Use the elimination method to solve.

Rewrite the equations.
y = –2x^2                 y + 2x^2 = 0
y = 2x^2 – 4            y – 2x^2 = –4

  y + 2x^2 =   0	First equation
  y – 2x^2 = –4	Second equation
2y          = –4	Add.
           y = –2	Divide each side by 2.

Now solve for x.

  y = –2x^2 	First equation
–2 = –2x^2 	Substitute –2 for y.
  1 = x^2	Divide each side by –2.
±1 = x	Take the square root of each side.

The solutions are (–1, –2) and (1, –2). 

Example 3 System of Quadratic Inequalities
Solve the system of inequalities by graphing.	
16y^2 + 36x^2  576
y  x – 2

The graph of 16y^2 + 36x^2 ≤ 576 is the ellipse 
16y^2 + 36x^2 = 576. Write the equation in standard form.

16y^2 + 36x^2 = 576	Original equation
 +  = 1	Divide each side by 576.

Graph  +  = 1. The inequality  +   1 is the interior of the ellipse shaded in yellow.

The graph of y  x – 2 is all points above the graph of y = x – 2. This region is shaded blue.

The intersection of these two regions, shaded green, represents the solution of the system of inequalities.