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.
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5.4.2: Solving Systems Algebraically
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
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 x =
-b ± b 2 - 4ac
2a
.
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5.4.2: Solving Systems Algebraically
Common Errors/Misconceptions
• miscalculating signs
• incorrectly distributing coefficients
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5.4.2: Solving Systems Algebraically
Guided Practice
Example 1
Solve the given system of equations algebraically.
ì y = x 2 - 11x + 28
í
î y = -3x + 12
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 1, continued
1. Since both equations are equal to y,
substitute by setting the equations equal
to each other.
–3x + 12 = x2 – 11x + 28 Substitute –3x + 12 for y
in the first equation.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 1, continued
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 + 16
Set the equation equal
to 0 by adding 3x to both
sides, and subtracting
12 from both sides.
0 = (x – 4)2
Factor.
5.4.2: Solving Systems Algebraically
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Guided Practice: Example 1, continued
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
y=0
Substitute 4 for x.
For x = 4, y = 0. Therefore, (4, 0) is the solution.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 1, continued
3. Check your solution(s) by graphing.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 1, continued
The equations do indeed intersect at (4, 0); therefore,
(4, 0) checks out as the solution to this system.
✔
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 1, continued
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5.4.2: Solving Systems Algebraically
Guided Practice
Example 2
Solve the given system of equations algebraically.
ì y = 2x 2 + 13x + 15
í
îy = x - 1
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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 + 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.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
3. Check your solution(s) by graphing.
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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.
✔
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5.4.2: Solving Systems Algebraically
Guided Practice: Example 2, continued
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5.4.2: Solving Systems Algebraically