10-8 Solve Linear-Quadratic Systems
Name
Date
x y 11
Solve: y x2 2x 5
{
y x 11
x 11 x2 2x 5
2
x x 6 0
(x 3)(x 2) 0
x 3 0 or x 2 0
x 3 or
x 2
Solve x y 11 for y using the Subtraction Property of Equality.
Substitute x 11 for y in y x2 2x 5.
Use Addition and Subtraction Properties of Equality to write the resulting
quadratic equation in standard form.
Factor.
Apply the Zero-Product Property.
Use the Addition and Subtraction Properties of Equality.
To find y, substitute the values of x in either equation.
If x 3, then y 3 11 8.
If x 2, then y (2) 11 13.
(3, 8) is a solution.
(2, 13) is a solution.
11
6 4 2 0
x y 11
?
2 13 11
11 11 True
(3,8)
y
x y 11
?
3 8 11
11 11 True
8
6
4
2
y x 2 2x 5
Check: (2, 13)
y x2 2x 5
?
13 (2)2 2(2) 5
?
13 4 4 5
13 13 True
x
Check: (3, 8)
y x2 2x 5
?
8 (3)2 2(3) 5
?
8965
8 8 True
y
20
18
16
14
(2,13) 12
10
x
2 4 6 8 10
Copyright © by William H. Sadlier, Inc. All rights reserved.
Solve each system of equations. Check your solutions on a separate
sheet of paper. Check students’ work.
1. y 4
y 3x2 5x 4
{
5x 4 4
3x2 5x 0
x(3x 5) 0
x 0 or 3x 5 0
3x2
2. y 2
y 5x2 3x 2
3x 2 2
5x2 3x 0
x(5x 3) 0
x 0 or 5x 3 0
x
(35, 2), (0, 2)
(53, 4), (0, 4)
{
x2 4x 4 1
(x2 4x 5) 0
a 1, b 4, c 5
b2 4ac 16 20 4
no solution
{
5x2
5
x 3
4. y 1
y x2 4x 4
3. y 1
y x2 4x 6
{
3
5
5. x y 1
y x2 5x 7
{
y 1 x
x2 5x 7 1 x
x2 6x 8 0
(x 2)(x 4) 0
x 2; y 1 2 1
x 4; y 1 4 3
(2, 1), (4, 3)
Lesson 10-8, pages 266–269.
x2 4x 6 1
(x2 4x 5) 0
a 1, b 4, c 5
b2 4ac 42 4(1)(5)
4
no solution
6. x y 2
y x2 6x 8
{
yx2
x2 6x 8 x 2
x2 7x 10 0
(x 2)(x 5) 0
x 2; y 2 2 0
x 5; y 5 2 3
(2, 0), (5, 3)
Chapter 10 261
For More Practice Go To:
Solve each system of equations. Check your solutions on a separate
sheet of paper. Check students’ work.
{
y 9x 4
9x 4 3x2 15x 7
3x2 6x 3 0
3(x 1)2 0
x 1; y 9 4 5
8. 6x y 12
y 4x2 10x 4
{
y 2x 8
2x 8 4x2 8x 20
4x2 6x 28 0
a 4, b 6, c 28
b2 4ac 36 448 , 0
no solution
{
y 6x 12
6x 12 4x2 10x 4
4x2 16x 16 0
4(x 2)2 0
x 2; y 12 12 0
y 12 3x
12 3x 3x2 11x 16
3x2 14x 28 0
a 3, b 14, c 28
b2 4ac 196 336 , 0
no solution
11. 4x y 25
y 5 3x 5x2 3x 10
12. 5x y 20
y 3x 5 3x2 2x 20
(2, 0)
(1, 5)
10. 4x 2y 16
y 4x2 8x 20
9. 3x y 12
y 3x2 11x 16
{
{
{
y 5x 20
y 3x2 x 25
5x 20 3x2 x 25
0 3x2 6x 45
0 3(x2 2x 15)
0 3(x 5)(x 3)
x 5; y 25 20 45
x 3; y 15 20 5
y 4x 25
y 5x2 6x 15
4x 25 5x2 6x 15
0 5x2 10x 40
0 5(x2 2x 8)
0 5(x 4)(x 2)
x 4; y 16 25 41
x 2; y 8 25 17
(3, 5), (5, 45)
(2, 17), (4, 41)
13. The length of a rectangle is 1 foot more than
twice the width. The area of the rectangle is
two times the square of the width, plus three
times the width, less 14 square feet. What is
the width of the rectangle?
Make a drawing. Let x width, 2x 1 length;
Area x(2x 1) 2x2 3x 14;
2x2 x 2x2 3x 14; 2x 14; 7 x
The rectangle has a width of 7 ft.
14. The equation of a line is y k. The equation
of a parabola is y 2x2 4x 7. What value
of k will give the system of equations exactly
one solution? two solutions? no solutions?
Use logical reasoning. The vertex of the parabola is:
4
b
x 1; y 2(1)2 4(1) 7 5.
4
2a
If y 5, the line intersects only at the vertex (one
solution). If y . 5, the line intersects the parabola
twice (two solutions). If y , 5, the line never
intersects the parabola (no solution).
15. Find the x-coordinates of the points of intersection of the system: y 2x2 3x 8
y x2 2x 6
{
2x2 3x 8 x2 2x 6; 3x2 5x 14 0; a 3, b 5, and c 14.
( )2
( )(
)
x 5 5 4 3 14 ; x 5 193 ; x 5 193 or x 5 193
(
)
6
6
6
23
262 Chapter 10
Copyright © by William H. Sadlier, Inc. All rights reserved.
7. 9x y 4
y 3x2 15x 7