EXAMPLE 3 Solve a quadratic system by elimination

EXAMPLE 3 Solve a quadratic system by elimination Solve the system by elimination. 9 x 2 + y 2 – 90 x + 216 = 0 Equation 1 Equation 2 x 2 – y 2 – 16 = 0 SOLUTION Add the equations to eliminate the y 2 - term and obtain a quadratic equation in x. 9 x 2 + y 2 – 90 x + 216 = 0 x 2 – y 2 – 16 = 0 Add. 10 x 2 – 90 x + 200 = 0 Divide each side by 10. x 2 – 9 x + 20 = 0 Factor (x – 4)(x – 5) = 0 Zero product property x = 4 or x = 5

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 (longdistance 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 x 2 – y 2 – 16 x + 32 = 0 – x 2 + y 2 – 8 y + 8 = 0 Equation 1 Equation 2 SOLUTION STEP 1 Add the equations to eliminate the x 2 - and y 2 - terms. x 2 – y 2 – 16 x + 32 = 0 – x 2 + y 2 – 8 y + 8 = 0 Add. – 16 x – 8 y + 40 = 0 Solve for y. y = – 2 x + 5

EXAMPLE 4 Solve a real-life quadratic system STEP 2 Substitute – 2 x + 5 for y in Equation 1 and solve for x. Equation 1 x 2 – y 2 – 16 x + 32 = 0 x 2 – ( 2 x + 5)2 – 16 x + 32 = 0 Substitute for y. Simplify. 3 x 2 – 4 x – 7 = 0 Factor. (x + 1)(3 x – 7) = 0 7 Zero product property x = – 1 or x = 3

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

for Examples 3 and 4 GUIDED PRACTICE Solve the system. 7. – 2 y 2 + x + 2 = 0 x 2 + y 2 – 1 = 0 SOLUTION – 2 y 2 + x + 2 = 0 x 2 + y 2 – 1 – 2 y 2 + 2 x 2 + x =0 x+2=0 Multiply 2 nd equation by 2 to eliminate y 2 term and obtain quadratic equation. – 2=0 =0 Add.

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

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

GUIDED PRACTICE for Examples 3 and 4 2 x 2 – 16 x – 30 = 0 2(x – 5)(x – 3) = 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. x 2 + 4 y 2 + 4 x + 8 y = 8 y 2 – x + 2 y = 5 SOLUTION x 2 + 4 y 2 + 4 x + 8 y = 8 4 y 2 + 4 x 8 y = 20 Multiply 2 nd equation by 4 to obtain a quadratic equation. Add. x 2 + 8 x = – 12 (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).

GUIDED PRACTICE for Examples 3 and 4 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. Equation 1 x 2 – y 2 – 12 x + 18 = 0 Equation 2 y 2 – x 2 – 4 y + 2 = 0 10. SOLUTION STEP 1 Add the equations to eliminate the x 2 - and y 2 - terms. x 2 – y 2 – 12 x + 18 = 0 – x 2 + y 2 – 4 y + 2 = 0 Add. – 12 x – 4 y + 20 = 0

GUIDED PRACTICE for Examples 3 and 4 y = – 3 x + 5 Solve for y. STEP 2 Substitute – 3 x + 5 for y in equation 1 and solved for x. x 2 – y 2 – 12 x + 18 = 0 x 2 – (– 3 x + 5)2 – 12 x + 18 = 0 8 x 2 – 18 x + 7 = 0 (2 x – 1)(4 x – 7) = 0 7 1 x= or x = 2 4 Equation 1 Substitute for y. Simplify. Factor. Zero product property

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