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Introduction Previously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus on solving a quadratic-linear system algebraically. When doing so, substitution is often the best choice. Substitution is the replacement of a term of an equation by another that is known to have the same value. 1 5.4.2: Solving Systems Algebraically

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Page 1: Introduction

IntroductionPreviously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus on solving a quadratic-linear system algebraically. When doing so, substitution is often the best choice. Substitution is the replacement of a term of an equation by another that is known to have the same value.

1

5.4.2: Solving Systems Algebraically

Page 2: Introduction

Key Concepts• When solving a quadratic-linear system, if both

functions are written in function form such as “y =” or “f(x) =”, set the equations equal to each other.

• When you set the equations equal to each other, you are replacing y in each equation with an equivalent expression, thus using the substitution method.

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5.4.2: Solving Systems Algebraically

Page 3: Introduction

Key Concepts

• You can solve by factoring the equation or by using

the quadratic formula, a formula that states the

solutions of a quadratic equation of the form

ax2 + bx + c = 0 are given by

3

5.4.2: Solving Systems Algebraically

Page 4: Introduction

Common Errors/Misconceptions• miscalculating signs • incorrectly distributing coefficients

4

5.4.2: Solving Systems Algebraically

Page 5: Introduction

Guided PracticeExample 1Solve the given system of equations algebraically.

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5.4.2: Solving Systems Algebraically

Page 6: Introduction

Guided Practice: Example 1, continued1. Since both equations are equal to y,

substitute by setting the equations equal to each other.

6

5.4.2: Solving Systems Algebraically

–3x + 12 = x2 – 11x + 28 Substitute –3x + 12 for y in the first equation.

Page 7: Introduction

Guided Practice: Example 1, continued2. Solve the equation either by factoring or

by using the quadratic formula. Since a (the coefficient of the squared term) is 1, it’s simplest to solve by factoring.

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5.4.2: Solving Systems Algebraically

–3x + 12 = x2 – 11x + 28 Equation

0 = x2 – 8x + 16

Set the equation equal to 0 by adding 3x to both sides, and subtracting 12 from both sides.

0 = (x – 4)2 Factor.

Page 8: Introduction

Guided Practice: Example 1, continued

Substitute the value of x into the second equation of the system to find the corresponding y-value.

For x = 4, y = 0. Therefore, (4, 0) is the solution.

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5.4.2: Solving Systems Algebraically

x – 4 = 0 Set each factor equal to 0 and solve.

x = 4

y = –3(4) + 12 Substitute 4 for x.y = 0

Page 9: Introduction

Guided Practice: Example 1, continued3. Check your solution(s) by graphing.

9

5.4.2: Solving Systems Algebraically

Page 10: Introduction

Guided Practice: Example 1, continuedThe equations do indeed intersect at (4, 0); therefore, (4, 0) checks out as the solution to this system.

10

5.4.2: Solving Systems Algebraically

✔

Page 11: Introduction

Guided Practice: Example 1, continued

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5.4.2: Solving Systems Algebraically

http://www.walch.com/ei/00255
Page 12: Introduction

Guided PracticeExample 2Solve the given system of equations algebraically.

12

5.4.2: Solving Systems Algebraically

Page 13: Introduction

Guided Practice: Example 2, continued1. Since both equations are equal to y,

substitute by setting the equations equal to each other.

13

5.4.2: Solving Systems Algebraically

x – 1 = 2x2 + 13x + 15 Substitute x – 1 for y in the first equation.

Page 14: Introduction

Guided Practice: Example 2, continued2. Solve the equation either by factoring or

by using the quadratic formula. Since the equation can be factored easily, choose this method.

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5.4.2: Solving Systems Algebraically

x – 1 = 2x2 + 13x + 15 Equation

0 = 2x2 + 12x + 16

Set the equation equal to 0 by subtracting x from both sides and then adding 1 to both sides.

0 = 2(x + 2)(x + 4) Factor.

Page 15: Introduction

Guided Practice: Example 2, continuedNext, set the factors equal to 0 and solve.

Substitute each of the values you found for x into the second equation of the system to find the corresponding y-value.

15

5.4.2: Solving Systems Algebraically

x + 2 = 0 x + 4 = 0

x = –2 x = –4

y = (–2) – 1 Substitute –2 for x.

y = –3

Page 16: Introduction

Guided Practice: Example 2, continued

For x = –2, y = –3, and for x = –4, y = –5. Therefore, (–2, –3) and (–4, –5) are the solutions to the system.

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5.4.2: Solving Systems Algebraically

y = (–4) – 1 Substitute –4 for x.

y = –5

Page 17: Introduction

Guided Practice: Example 2, continued3. Check your solution(s) by graphing.

17

5.4.2: Solving Systems Algebraically

Page 18: Introduction

Guided Practice: Example 2, continuedThe equations do indeed intersect at (–2, –3) and (–4, –5); therefore, (–2, –3) and (–4, –5) check out as the solutions to this system.

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5.4.2: Solving Systems Algebraically

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Page 19: Introduction

Guided Practice: Example 2, continued

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5.4.2: Solving Systems Algebraically

http://www.walch.com/ei/00256

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