1. Solve a linear-quadratic system by graphing
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
2x – y = 3
y2 = 7x – 3
– y = – 2x + 3
y = x – 3
Equation 1
y = 2x – 3
Equation 2

2. EXAMPLE 1
Solve a linear-quadratic system by graphing
STEP 2
Graph the equations y = y = and y = 2x – 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,

3. EXAMPLE 1
Solve a linear-quadratic system by graphing
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.

4. Solve a linear-quadratic system by substitution
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
Equation 1
x2 + (– 3x + 10)2 = 10
Substitute for y.
x2 + 9x2 – 60x = 10
Expand the power.
10x2 – 60x + 90 = 0
Combine like terms.
x2 – 6x + 9 = 0
Divide each side by 10.
(x – 3)2 = 0
Perfect square trinomial
x = 3
Zero product property

5. EXAMPLE 2
Solve a linear-quadratic system by substitution
To find the y-coordinate of the solution, substitute x = 3 in Equation 2.
y = – 3(3) + 10 = 1
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).

6. Step each equation for y.
GUIDED PRACTICE
for Examples 1 and 2
1. x2 + y2 = 13
y = x – 1
SOLUTION
x2 + y2 = 13
STEP 1
Equation 1
y = x – 1
Equation 2
Step each equation for y.
x2 + y2 = 13
y2 = 13 – x2 +
y = – √
13 – x2
y = x – 1
Equation 1
Equation 2

7. GUIDED PRACTICE
for Examples 1 and 2
y = x – 1.
Graph the equation
y = √
13 – x2 – ,
and
STEP 2
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
(–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.

8. Solve each equation for y.
GUIDED PRACTICE
for Examples 1 and 2 2.
x2 + 8y2 – 4 = 0
y = 2x + 7
STEP 1
Solve each equation for y.
8y2 = – x2 + 4
y = 2x + 1
y = –
x2 + 4 8
Equation 1
Equation 2
STEP 2
Graph the equation
and
y = 2x + 7.
y =
x2 + 4 8 –
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.

9. Solve each equation for y.
GUIDED PRACTICE
for Examples 1 and 2 3.
y2 + 6x – 1 = 0
y = – 04x + 2.6
STEP 1
Solve each equation for y.
y2 + 6x – 1 = 0
y = – 0.4x + 2.6
y = + – √
– 6 x +1
Equation 1
Equation 2
STEP 2
Graph the equation
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 graphs
intersect at the points
(–1.57, 3.23) and (–22.9, 11.8).

10. Substitute 0.5x – 3 for y in Equation 2 and solve for x.
GUIDED PRACTICE
for Examples 1 and 2 4.
y = 0.5x – 3
x2 + 4y2 – 4 = 0
SOLUTION
Substitute 0.5x – 3 for y in Equation 2 and solve for x.
x2 + 4y2 – 4 = 0
Equation 2
x2 + 4 (0.5x – 3)2 – 4 = 0
Substitute for y.
x2 + y (0.25x2 – 3x + 9) – 4 = 0
Expand the power.
2x2 – 12x + 32 = 0
Combine like terms.
x2 – 6x + 16 = 0
Divide each side by 2.
This equation has no solution.

11. Substitute – x – 1 for y in Equation 1 and solve for x.
GUIDED PRACTICE
for Examples 1 and 2 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
Equation 1
(– x – 1)2 – 2x – 10 = 0
Substitute for y.
x x – 2x – 10 = 0
Expand the power.
x2 – 9 = 0
Combine like terms.
x2 = 9
Add 9 to each side.
x = ±3
simplify.

12. GUIDED PRACTICE
for Examples 1 and 2
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

13. Substitute 4x – 8 for y in Equation 2 and solve for x.
GUIDED PRACTICE
for Examples 1 and 2 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
Equation 2
9x2 – (4x – 8)2 – 36 = 0
Substitute for y.
9x2 –(16x2 – 64x + 64) – 36 = 0
Expand the power.
– 7x2 + 64x – 100 = 0
Combine like terms.
7x2 – 64x = 0
Divide each side by –1.
(7x – 50) (x –2) = 0
simplify.

14. To find the y-coordinate of the solution, substitute.
GUIDED PRACTICE
for Examples 1 and 2
7x – 50 = 0 or x – 2 = 0
x = 7 50
or x = 2
To find the y-coordinate of the solution, substitute.
x = 7 50
and x = 2
in Equation 1
y = – 8 = 7 50
144
y = 4(2) – 8 = 0
The solutions are (2, 0), and ,
ANSWER 7 50
144