EXAMPLE 3 Solve a quadratic system by elimination

Solve the system by elimination.9x2 + y2 – 90x + 216 = 0 Equation 1

x2 – y2 – 16 = 0 Equation 2

SOLUTION

9x2 + y2 – 90x + 216 = 0 x2 – y2 – 16 = 0

10x2 – 90x + 200 = 0 Add.

x2 – 9x + 20 = 0 Divide each side by 10.

(x – 4)(x – 5) = 0 Factor

x = 4 or x = 5 Zero product property

Add the equations to eliminate the y2 - term and obtain a quadratic equation in x.

EXAMPLE 3 Solve a quadratic system by elimination

When x = 4, y = 0. When x = 5, y = ±3.

ANSWER

The solutions are (4, 0), (5, 3), and (5, 23), as shown.

EXAMPLE 4 Solve a real-life quadratic system

Navigation

A ship uses LORAN (long-distance radio navigation) to find its position.Radio signals from stations A and B locate the ship on the blue hyperbola, and signals from stations B and C locate the ship on the red hyperbola. The equations of the hyperbolas are given below. Find the ship’s position if it is east of the y - axis.

EXAMPLE 4 Solve a real-life quadratic system

x2 – y2 – 16x + 32 = 0 Equation 1

– x2 + y2 – 8y + 8 = 0 Equation 2

SOLUTION

STEP 1Add the equations to eliminate the x2 - and y2 - terms.x2 – y2 – 16x + 32 = 0– x2 + y2 – 8y + 8 = 0

– 16x – 8y + 40 = 0 Add.

y = – 2x + 5 Solve for y.

EXAMPLE 4 Solve a real-life quadratic system

STEP 2Substitute – 2x + 5 for y in Equation 1 and solve for x.

x2 – y2 – 16x + 32 = 0 Equation 1

x2 – (2x + 5)2 – 16x + 32 = 0

3x2 – 4x – 7 = 0Substitute for y.

Simplify.

(x + 1)(3x – 7) = 0 Factor.

Zero product propertyx = – 1 or x =73

EXAMPLE 4 Solve a real-life quadratic system

ANSWER

Because the ship is east of the y - axis, it is at ,73

13( ).

STEP 3Substitute for x in y = – 2x + 5 to find the solutions (–1, 7) and ,

73

13( ).

GUIDED PRACTICE for Examples 3 and 4

Solve the system.

7.

–2y2 + x + 2 = 0

x2 + y2 – 1 = 0

SOLUTION

–2y2 + x + 2 = 0

2y2 + 2x2 – 2 = 0

–2y2 + x + 2 = 0

x2 + y2 – 1 = 0Multiply 2nd equation by 2 to eliminate y2 term and obtain quadratic equation.

2x2 + x = 0 Add.

GUIDED PRACTICE for Examples 3 and 4

x(2x + 1) = 0 Factor

Zero product propertyx = 0 or x =-12

23.When x = 0, y = ± 1. When x = , y = ±-1

2

GUIDED PRACTICE for Examples 3 and 4

Solve the system.

8.

x2 + y2 – 16x + 39 = 0

x2 – y2 – 9 = 0

SOLUTION

2x2 – 16x – 30 = 0

x2 + y2 – 16x + 39 = 0

x2 – y2 – 9 = 0

GUIDED PRACTICE for Examples 3 and 4

2(x – 5)(x – 3) = 0 Factor

Zero product propertyx = 3 or x = 5

When x = 3, y = 0. When x = 5, y = ± 4

2x2 – 16x – 30 = 0

GUIDED PRACTICE for Examples 3 and 4

Solve the system.

9.

x2 + 4y2 + 4x + 8y = 8y2 – x + 2y = 5

SOLUTION

x2 + 8x = –12

x2 + 4y2 + 4x + 8y = 8

4y2 + 4x 8y = 20

Multiply 2nd equation by 4 to obtain a quadratic equation.

(x + 2) (x + 6) = 0

Add.

Factor

Zero product propertyx = – 2 or x = – 6

GUIDED PRACTICE for Examples 3 and 4

When x = –2, y = –3. When x = – 6, y = – 1

ANSWER

The solutions are (– 6, 1), (– 2, –3), and (–2, 1).

GUIDED PRACTICE for Examples 3 and 4

10.

WHAT IF? In Example 4, suppose that a ship’s LORAN system locates the ship on the two hyperbolas whose equations are given below. Find the ship’s location if it is south of the x-axis.

x2 – y2 – 12x + 18 = 0

y2 – x2 – 4y + 2 = 0

Equation 1

Equation 2

SOLUTION

STEP 1Add the equations to eliminate the x2 - and y2 - terms.x2 – y2 – 12x + 18 = 0– x2 + y2 – 4y + 2 = 0

–12x – 4y + 20 = 0 Add.

GUIDED PRACTICE for Examples 3 and 4

STEP 2Substitute –3x + 5 for y in equation 1 and solved for x.

x2 – y2 – 12x + 18 = 0 Equation 1

x2 – (– 3x + 5)2 – 12x + 18 = 0

8x2 –18x + 7 = 0

Substitute for y.

Simplify.

(2x – 1)(4x – 7) = 0 Factor.

Zero product property

y = – 3x + 5 Solve for y.

x = or x =74

12

EXAMPLE 4 Solve a real-life quadratic system

STEP 3Substitute for x in y = – 3x + 5 to find the solutions and,

72

12

, )( 74

–1( ).,4

).ANSWER

Because the ship is south of the x - axis, it is at ,74( –1

4