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

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.

Solve a real-life quadratic system EXAMPLE 4 Solve a real-life quadratic system x2 – y2 – 16x + 32 = 0 Equation 1 – x2 + y2 – 8y + 8 = 0 Equation 2 SOLUTION STEP 1 Add 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.

Solve a real-life quadratic system EXAMPLE 4 Solve a real-life quadratic system STEP 2 Substitute – 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 Substitute for y. 3x2 – 4x – 7 = 0 Simplify. (x + 1)(3x – 7) = 0 Factor. x = – 1 or x = 73 Zero product property

( ). ( ). EXAMPLE 4 Solve a real-life quadratic system STEP 3 Substitute for x in y = – 2x + 5 to find the solutions (–1, 7) and , 73 13 ( ). ANSWER Because the ship is east of the y - axis, it is at , 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 x2 + y2 – 1 = 0 Multiply 2nd equation by 2 to eliminate y2 term and obtain quadratic equation. –2y2 + x + 2 = 0 2y2 + 2x2 – 2 = 0 2x2 + x = 0 Add.

GUIDED PRACTICE for Examples 3 and 4 x(2x + 1) = 0 -1 x = 0 or x = 2 Factor x = 0 or x = -1 2 Zero product property 2 3. When x = 0, y = ± 1. When x = , y = ± -1

GUIDED PRACTICE for Examples 3 and 4 Solve the system. 8. x2 + y2 – 16x + 39 = 0 x2 – y2 – 9 = 0 SOLUTION x2 + y2 – 16x + 39 = 0 x2 – y2 – 9 = 0 2x2 – 16x – 30 = 0

GUIDED PRACTICE for Examples 3 and 4 2x2 – 16x – 30 = 0 Factor x = 3 or x = 5 Zero product property When x = 3, y = 0. When x = 5, y = ± 4

GUIDED PRACTICE for Examples 3 and 4 Solve the system. 9. x2 + 4y2 + 4x + 8y = 8 y2 – x + 2y = 5 SOLUTION x2 + 4y2 + 4x + 8y = 8 4y2 + 4x  8y =  20 Multiply 2nd equation by 4 to obtain a quadratic equation. x2 + 8x = –12 Add. (x + 2) (x + 6) = 0 Factor x = – 2 or x = – 6 Zero product property

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).

Add the equations to eliminate the x2 - and y2 - terms. 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 Equation 1 y2 – x2 – 4y + 2 = 0 Equation 2 SOLUTION STEP 1 Add the equations to eliminate the x2 - and y2 - terms. x2 – y2 – 12x + 18 = 0 – x2 + y2 – 4y + 2 = 0 –12x – 4y + 20 = 0 Add.

Substitute –3x + 5 for y in equation 1 and solved for x. GUIDED PRACTICE for Examples 3 and 4 y = – 3x + 5 Solve for y. STEP 2 Substitute –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 Substitute for y. 8x2 –18x + 7 = 0 Simplify. (2x – 1)(4x – 7) = 0 Factor. x = or x = 74 1 2 Zero product property

) ( ). ). ( EXAMPLE 4 Solve a real-life quadratic system STEP 3 Substitute for x in y = – 3x + 5 to find the solutions and, 7 2 1 , ) ( 74 –1 ). 4 ). ANSWER Because the ship is south of the x - axis, it is at , 74 ( –1 4