UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU5-209

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.

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.

• 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

xb b ac

a=− ± −2 4

2.

Prerequisite Skills

This lesson requires the use of the following skills:

• solving quadratic functions by factoring

• solving quadratic functions by using the quadratic formula

Common Errors/Misconceptions

• miscalculating signs

• incorrectly distributing coefficients

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource U5-210

© Walch Education

Example 1

Solve the given system of equations algebraically.

y x x

y x

= − += − +

2 11 28

3 12

1. Since both equations are equal to y, substitute by setting the equations equal to each other.

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

2. 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.

–3x + 12 = x2 – 11x + 28 Equation

0 = x2 – 8x + 16Set the equation equal to 0 by adding 3x to both sides, and subtracting 12 from both sides.

0 = (x – 4)2 Factor.

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

x = 4

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

y = –3(4) + 12 Substitute 4 for x.

y = 0

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

Guided Practice 5.4.2

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU5-211

3. Check your solution(s) by graphing.

x

y

-8 -6 -4 -2 0 2 4 6 8

-5

0

5

(4, 0)

The equations do indeed intersect at (4, 0); therefore, (4, 0) checks out as the solution to this system.

Example 2

Solve the given system of equations algebraically.

y x x

y x

= + += −

2 13 15

1

2

1. Since both equations are equal to y, substitute by setting the equations equal to each other.

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

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource U5-212

© Walch Education

2. Solve the equation either by factoring or by using the quadratic formula.

Since the equation can be factored easily, choose this method.

x – 1 = 2x2 + 13x + 15 Equation

0 = 2x2 + 12x + 16Set 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.

Next, set the factors equal to 0 and solve.

x + 2 = 0 x + 4 = 0

x = –2 x = –4

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

y = (–2) – 1 Substitute –2 for x.

y = –3

y = (–4) – 1 Substitute –4 for x.

y = –5

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

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU5-213

3. Check your solution(s) by graphing.

x

y

-8 -6 -4 -2 0 2 4 6 8

-5

0

5

(–2, –3)

(–4, –5)

The equations do indeed intersect at (–2, –3) and (–4, –5);

therefore, (–2, –3) and (–4, –5) check out as the solutions to

this system.

Example 3

Solve the given system of equations algebraically.

y x x

y x

= − −= − −

7 25 12

3 30

2

1. Since both equations are equal to y, substitute by setting the equations equal to each other.

–3x – 30 = 7x2 – 25x – 12Substitute –3x – 30 for y in the first equation.

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource U5-214

© Walch Education

2. Solve the equation either by factoring or by using the quadratic formula.

The quadratic equation cannot easily be factored, so use the quadratic formula.

The quadratic formula is xb b ac

a=− ± −2 4

2.

–3x – 30 = 7x2 – 25x – 12 Equation

0 = 7x2 – 22x + 18Set the equation equal to 0 by adding 3x and 30 to both sides; then, apply the quadratic formula.

a = 7, b = –22, c = 18

x =− −( )± −( ) − ( )( )

( )22 22 4 7 18

2 7

2Substitute values for a, b, and c into the quadratic formula.

x =± −22 20

14Simplify.

Since the value under the square root is negative, there are no real solutions to the quadratic. Simplify to find the complex solutions.

xi

=±22 2 5

14

xi

=±11 5

7

UNIT 5 • QUADRATIC FUNCTIONSLesson 4: Solving Systems of Equations

Instruction

CCGPS Analytic Geometry Teacher Resource © Walch EducationU5-215

3. Check your solution(s) by graphing.

xy

-15 -10 -5 0 5 10 15

-30

-20

-10

Since the functions do not intersect, a complex solution is feasible.