EXAMPLE 1 Solve a linear-quadratic system by graphing Solve the system using a graphing calculator....

EXAMPLE 1 Solve a linear-quadratic system by graphing

Solve the system using a graphing calculator.

y2 – 7x + 3 = 0 Equation 1

2x – y = 3 Equation 2

SOLUTION

STEP 1 Solve each equation for y.

y2 – 7x + 3 = 0

y2 = 7x – 3

y = + 7x – 3 Equation 1

2x – y = 3

– y = – 2x + 3

y = 2x – 3 Equation 2


STEP 2

Graph the equations y = y = and y = 2x – 3

7x – 3, 7x – 3,–

Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs of and y = 2x – 3 intersect at (0.75, 21.5). The graphs of and y = 2x – 3 intersect at (4, 5).

y = – 7x – 3,

y = 7x – 3,


ANSWER

The solutions are (0.75, – 1.5) and (4, 5). Check the solutions by substituting the coordinates of the points into each of the original equations.

EXAMPLE 2 Solve a linear-quadratic system by substitution

Solve the system using substitution.

x2 + y2 = 10 Equation 1

y = – 3x + 10 Equation 2

SOLUTION

Substitute –3x + 10 for y in Equation 1 and solve for x.x2 + y2 = 10

x2 + (– 3x + 10)2 = 10x2 + 9x2 – 60x + 100 = 10

10x2 – 60x + 90 = 0x2 – 6x + 9 = 0

(x – 3)2 = 0x = 3

Equation 1

Substitute for y.

Expand the power.

Combine like terms.

Divide each side by 10.

Perfect square trinomial

Zero product property

EXAMPLE 2 Solve a linear-quadratic system by substitution

y = – 3(3) + 10 = 1

To find the y-coordinate of the solution, substitute x = 3 in Equation 2.

ANSWER

The solution is (3, 1).

CHECK You can check the solution by graphing the equations in the system. You can see from the graph shown that the line and the circle intersect only at the point (3, 1).

GUIDED PRACTICE for Examples 1 and 2

1. x2 + y2 = 13y = x – 1

SOLUTION

y = x – 1Equation 1

Equation 2

Step each equation for y.

x2 + y2 = 13y2 = 13 – x2

+y = – √13 – x2

Equation 1

y = x – 1Equation 2

x2 + y2 = 13STEP 1


Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs of

y = √13 – x2 and y = x – 1. intersect at (3, 2). The graphs

y = √13 – x2 and y = x – 1. intersect at (–2,–3).

The solutions are (3,2) and(–2, –3). Check the solutions by substituting the coordinates of the points into each of the original equations.

y = x – 1.

Graph the equation y = √13 – x2 y = – √13 – x2, and

STEP 2


2.

x2 + 8y2 – 4 = 0y = 2x + 7


8y2 = – x2 + 4

Equation 1

y = 2x + 1

Equation 2

Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphs does not intersect at any point so, no solutions.

y = – x2 + 48

STEP 2 Graph the equation

and y = 2x + 7.

y = x2 + 48

–


3.

y2 + 6x – 1 = 0

y = – 04x + 2.6


y2 + 6x – 1 = 0

y = +– √ – 6 x +1

y = – 0.4x + 2.6

Equation 2Equation 1

STEP 2 Graph the equation y = √ – 6 x +1 , y = √ – 6 x +1–

and y = – 0.4x + 2.6

Use the calculator’s intersect feature to find the coordinates of the intersection points. The graphsintersect at the points (–1.57, 3.23) and (–22.9, 11.8).


SOLUTION

4.

y = 0.5x – 3

x2 + 4y2 – 4 = 0

Substitute 0.5x – 3 for y in Equation 2 and solve for x.x2 + 4y2 – 4 = 0x2 + 4 (0.5x – 3)2 – 4 = 0x2 + y (0.25x2 – 3x + 9) – 4 = 02x2 – 12x + 32 = 0x2 – 6x + 16 = 0

Equation 2

Substitute for y.

Expand the power.

Combine like terms.

Divide each side by 2.

This equation has no solution.


5.

y2 – 2x – 10 = 0 y = x 1 – –

SOLUTION

Substitute – x – 1 for y in Equation 1 and solve for x.y2 – 2x2 – 10 = 0

(– x – 1)2 – 2x – 10 = 0x2 + 1 + 2x – 2x – 10 = 0

x2 – 9 = 0x2 = 9

Equation 1

Substitute for y.

Expand the power.

Combine like terms.

Add 9 to each side.

x = ±3 simplify.


To find the y-coordinate of the solution, substitute x = 3 and x = 3 in equation 2.

y = –3 –1 = – 4

y = 3 –1 = 2

The solutions are (3, –4), and (–3, 2)

ANSWER


6.

y = 4x – 8

9x2 – y2 – 36 = 0

SOLUTION

Substitute 4x – 8 for y in Equation 2 and solve for x.9x2 – y2 – 36 = 0

9x2 – (4x – 8)2 – 36 = 0

9x2 –(16x2 – 64x + 64) – 36 = 0

– 7x2 + 64x – 100 = 0

7x2 – 64x + 100 = 0

Equation 2

Substitute for y.

Expand the power.

Combine like terms.

Divide each side by –1.

(7x – 50) (x –2) = 0 simplify.


7x – 50 = 0 or x – 2 = 0

x = 750

or x = 2

To find the y-coordinate of the solution, substitute.

x = 750

and x = 2 in Equation 1

y = 4(2) – 8 = 0 y = – 8 =750 144

7

The solutions are (2, 0), and ,

ANSWER

750 144

7

EXAMPLE 1 Solve a linear-quadratic system by graphing Solve the system using a graphing calculator....

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