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**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

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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,

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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.

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**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

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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).

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**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

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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.

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**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.

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**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).

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**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.

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**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.

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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

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**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.

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**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

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