MATH 0322 Intermediate Algebra
Unit 3
Solving Quadratic Equations
by Factoring
Section: 6.6
Solving Quadratic Equations
 Quadratic Equation: an equation of the form
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
where 𝑎, 𝑏, and 𝑐 are real coefficients, but 𝑎 ≠ 0.
• This form is called Standard Form.
• In higher level MATH courses, Quadratic Equations
are also called 2nd degree Polynomial Equations.
• degree of equation: highest power of 𝑥 in equation,
• degree of equation: indicates maximum number of
solutions and factors the equation will have.
• the graph of a Quadratic Equation is called a parabola,
and it is shaped like this……
Solving Quadratic Equations
• Quadratic Equations are one of the most
important equations in our daily lives.
• They are used for:
 Structures/Bridges
 Lenses
 Mirrors
 Sound wave/
Radio signal reception
 Projectile trajectory
 ….
Solving Quadratic Equations
 Practice your Factoring Strategies in Unit 1.
(You will need them.)
 Definition: Zero Product Principle (p471)
If 𝐴 ∙ 𝐵 = 0, then 𝐴 = 0 or 𝐵 = 0.
Example: Find the solution to the equation.
If (𝑥 + 3)(𝑥 − 6) = 0,
then 𝑥 + 3 = 0 or 𝑥 − 6 = 0. Solve for 𝑥.
Check.
𝑥 = −3
𝑥=6
𝑥+3 𝑥−6 =0
𝑥+3 𝑥−6 =0
−𝟑 + 3 −𝟑 − 6 = 0
𝟔+3 𝟔−6 =0
𝟎 −9 = 0
9 𝟎 =0
P
P
Solving Quadratic Equations
 To solve Quadratics Equations by Factoring:
in Standard Form(if necessary).
PRewrite
& Apply Zero Product Principle.
PFactor
PSolve for 𝑥, then Check solutions.3
1)
2)
3)
2
3
− 1 ==33
Example: Solve.
+𝑥 =3 2 − 2
+
2
2
2
2 1 +1=3
2𝑥 + 𝑥 − 3 = 0
1)
9 2 +31 = 3
2) (2𝑥 + 3)( 𝑥 − 1) = 0
2
− =3
4
2
2𝑥 + 3 = 0 or 𝑥 − 1 = 0
9 3
3)
𝑥=1
2𝑥 = −3
− =3
3
2 2
𝑥=−
6
2
=3
2
2𝑥 2
P
1+ 2
P
Solving Quadratic Equations
Practice: Solve by factoring.
𝑥 2 − 6𝑥 + 5 = 0
a)
1)
2)
3)
(𝑥 − 5)(𝑥 − 1) = 0
𝑥−5=0 𝑥−1=0
𝑥=5
𝑥=1
P
P
Your turn.
c) 9𝑥 2 = 30𝑥 − 25
9𝑥 2 − 30𝑥 + 25 = 0
1)
( 3𝑥 − 5)(3𝑥 − 5) = 0
2)
3𝑥 − 5 = 0 3𝑥 − 5 = 0
3)
3𝑥 = 5
3𝑥 = 5
5
5
𝑥=
𝑥=
P
3
P
3
b)
1)
2)
3)
4𝑥 2 = 3𝑥
4𝑥 2 − 3𝑥 = 0
(𝑥)(4𝑥 − 3) = 0
𝑥 = 0 4𝑥 − 3 = 0
4𝑥 = 3
𝑥=0
3
𝑥=4
P
P
𝑥 − 5 𝑥 − 2 = 28
𝑥 2 − 7𝑥 + 10 = 28
𝑥 2 − 7𝑥 − 18 = 0
2) (𝑥 − 9)(𝑥 + 2) = 0
𝑥−9=0 𝑥+2=0
𝑥 = −2
𝑥=9
3)
d)
1)
P
P
Solving Quadratic Equations
Complete and
Practice HW 6.6
MATH 0322 Intermediate Algebra
Unit 3
Quadratic Formula
Section: 11.2
Quadratic Formula
 Quadratic Formula: a formula used to solve
Quadratic Equations in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0.
−(𝑏)± (𝑏)2 −4(𝑎)(𝑐)
𝑥=
2(𝑎)
• Derived using the Complete the Square method.(p789)
• ± indicates equation has two solutions,
which may be distinct(different) or the same.
• Solutions to a Quadratic Equation are 𝑥-intercepts on the graph.
• 𝑏 2 − 4𝑎𝑐 is called the discriminant.
 if 𝑏 2 − 4𝑎𝑐 = 0, one repeated real solution exists.(1 𝑥-intercept)
 if 𝑏 2 − 4𝑎𝑐 > 0, two distinct real solutions exist.(2 𝑥-intercepts)
(Read p793 to know when they are rational or irrational.)
 if 𝑏 2 − 4𝑎𝑐 < 0, two distinct non-real solutions exist (𝑎 ± 𝑏𝑖).
Quadratic Formula
Methods used to solve Quadratic Equations:
(Know when one is more efficient to use than the others.)
• Factoring Method
(when 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is factorable)
• Square Root Method
(when 𝑏 = 0)
• Quadratic Formula
(when 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is not easily factored or
is not factorable)
Quadratic Formula
 Practice: Solve. 3𝑥 2 + 𝑥 − 2 = 0
 Standard Form 3𝑥 2 + 𝑥 − 2 = 0
−(𝑏) ± (𝑏)2 −4 𝑎 𝑐
 Quadratic Formula 𝑥 =
2(𝑎)
−( 1 ) ± ( 1 )2 −4 3 −2
𝑥=
1 5
2
2( 3 )
− + =
6 6
3
−1 ± 5
1 5
−1 ± 25
=
= − ±
=
1 5
6
6 6
6
− − = −1
6 6
Solutions are rational, so Factoring Method more efficient.
3𝑥 2 + 𝑥 − 2 = 0
3𝑥 − 2 𝑥 + 1 = 0
3𝑥 − 2 = 0
𝑥+1=0
2
𝑥 = −1
𝑥=
P
P
3
P
Quadratic Formula
 Practice: Solve. 2𝑥 2 = 6𝑥 − 1
 Standard Form 2𝑥 2 − 6𝑥 + 1 = 0
 Quadratic Formula
−(𝑏) ± (𝑏)2 −4 𝑎 𝑐
𝑥=
2(𝑎)
−(−6) ± ( −6)2 −4 2 1
𝑥=
2( 2 )
6 ± 2 7 36 12 7 3
6 ± 28
6 ± 41∙ 7
7
=
= ±
=
=
= ±
24
24
4
4
4
2
2
3
7 3
7
𝑥= +
,,, −
2
2
2
2
P
 Know what the discriminant says about the solutions?
Quadratic Formula
 Practice: Solve. 4𝑥 2 − 8𝑥 = 𝑥 2 − 7
 Standard Form 3𝑥 2 − 8𝑥 + 7 = 0
 Quadratic Formula
−(𝑏) ± (𝑏)2 −4 𝑎 𝑐
𝑥=
2(𝑎)
−(−8) ± ( −8)2 −4 3 7
𝑥=
2( 3 )
8 ± −20 8 ± −41∙ 5 8 ± 2𝑖 5 48 12𝑖 5 4
5
=
=
=
= ±
= ±𝑖
3
3
6
6
6
6
6
3
3
4
5 4
5
𝑥 = +𝑖
,,, − 𝑖
3
3
3
3
P
 Know what the discriminant says about the solutions?
Quadratic Formula
 Practice: Suppose a projectile follows a parabolic
trajectory given by the function
𝑓 𝑥 = −0.0064𝑥 2 + 2𝑥 + 3
where 𝑥 is horizontal distance traveled in meters
and 𝑓(𝑥) is the height along the path.
Find 𝑥 when the projectile strikes the ground.
(Round to nearest whole number.)
 Hint: When it strikes the ground, the height 𝑓(𝑥) is zero.
Quadratic Formula
 Solution:
𝑓 𝑥 = −0.0064𝑥 2 + 2𝑥 + 3
0 = −0.0064𝑥 2 + 2𝑥 + 3
−( 2 ) ± ( 2 )2 −4 −0.0064
𝑥=
2(−0.0064 )
Factoring Method?
Square Root Method?
Quadratic Formula?
3
−2 ± 4.0768
𝑥=
−0.0128
−2
4.0768
𝑥=
+
−0.0128 −0.0128
𝑥 = 156.25 + (−157.74)
𝑥 = −1.49
?
−2
4.0768
𝑥=
−
−0.0128 −0.0128
𝑥 = 156.25 − (−157.74)
P
𝑥 = 313.99 = 314𝑚
Quadratic Formula
Complete and
Practice HW 11.2
MATH 0322 Intermediate Algebra
Unit 3
Point-Slope Form
Section: 3.5
Point-Slope Form
 What is the formula for slope, as discussed in class?
𝑦2 −𝑦1
𝑚=
, with 𝑥2 − 𝑥1 ≠ 0
𝑥2 −𝑥1
• Multiply both sides by the denominator and simplify.
𝑦2 − 𝑦1
1
𝑚 𝑥2 − 𝑥1 =
𝑥2 − 𝑥1
1 𝑥2 − 𝑥1
𝑚 𝑥2 − 𝑥1 = 𝑦2 − 𝑦1
𝑦2 − 𝑦1 = 𝑚 𝑥2 − 𝑥1
With 𝑥2 , 𝑦2 fixed, this
results in…..
Point-Slope Form of equation of a non-vertical line:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
Point-Slope Form
Forms of lines and their names are:
Equation
Name
1)
2)
3)
4)
5)
𝐴𝑥 + 𝐵𝑦 = 𝐶
Standard Form
𝑥=𝑎
Vertical Line Form
Horizontal Line Form
𝑦=𝑏
Slope-Intercept Form
𝑦 = 𝑚𝑥 + 𝑏
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Point-Slope Form
Point-Slope Form
Point-Slope Form 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) is used:
• to write equation of a line having slope 𝑚
and point (𝑥1 , 𝑦1 ),
** (Only substitute values for 𝑥1 , 𝑦1 , and 𝑚.)
• to write equation of a line 𝐿1 that is Parallel or
Perpendicular to another line 𝐿2 ,
(Parallel: 𝑚1 = 𝑚2 , lines have different points.)
(Perpendicular: 𝑚1 = −
1
𝑚2
, lines have one common point.)
• to graph lines using 𝑥1 , 𝑦1 , and 𝑚.
Point-Slope Form
 Practice:
• What slope is parallel to 𝑚 = −2?
P
• What slope is perpendicular to 𝑚 = 5?
𝑚 = −2
1
𝑚=−
5
P
• What slope is perpendicular to 𝑚 =
4
𝑚=
3
P
3
− ?
4
Point-Slope Form
 Practice: Write the equation of a line, first in
Point-Slope form, then in Slope-Intercept form,
having 𝑚 = −2 and passing through (3,1).
1) Write your form.
2) Substitute values
for 𝑥1 , 𝑦1 , 𝑚.
3) Simplify to write in
Point-Slope Form.
Slope-Intercept Form.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − 1 = −2(𝑥 − ( 3))
𝑦 − 1 = −2(𝑥 − 3)
P
𝑦 − 1 = −2𝑥 + 6
𝑦 = −2𝑥 + 7
P
Point-Slope Form
 Your turn: Write the equation of a line, first in
Point-Slope form, then in Slope-Intercept form,
having 𝑚 = 5 and passing through (1,2).
1) Write your form.
2) Substitute values
for 𝑥1 , 𝑦1 , 𝑚.
3) Simplify to write in
Point-Slope Form.
Slope-Intercept Form.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − 2 = 5 (𝑥 − ( 1 ))
𝑦 − 2 = 5(𝑥 − 1)
P
𝑦 − 2 = 5𝑥 − 5
𝑦 = 5𝑥 − 3
P
Point-Slope Form 4
 Practice: Write the equation of a line, first in
Point-Slope form, then in Slope-Intercept form,
passing through (4, −3) and (−2,6).
(𝒙𝟏 , 𝒚𝟏 )
(𝒙𝟐 , 𝒚𝟐 )
1) Need a slope.
3
(𝑦2 − 𝑦1 ) ((6) − (−3)) 9
=
=
=−
𝑚=
2
(𝑥2 − 𝑥1 ) ((−2) − (4)) −6
2) Substitute values
for 𝑥1 , 𝑦1 , 𝑚.
3) Simplify to write in
Point-Slope Form.
Slope-Intercept Form.
𝑦 − −3 =
𝑦+3=
𝑦+3=
𝑦=
3
− (𝑥
2
− ( 4 ))
P
3
− (𝑥 − 4)
2
3
− 𝑥+6
2
3
− 𝑥+3
2
P
Point-Slope Form 4
 Your turn: Write the equation of a line, first in
Point-Slope form, then in Slope-Intercept form,
passing through (−2, −1) and (−1, −6).
(𝒙𝟏 , 𝒚𝟏 )
(𝒙𝟐 , 𝒚𝟐 )
1) Need a slope.
(𝑦2 − 𝑦1 ) ((−6) − (−1)) −5
=
𝑚=
=
= −5
(𝑥2 − 𝑥1 ) ((−1) − (−2))
1
2) Substitute values
for 𝑥1 , 𝑦1 , 𝑚.
3) Simplify to write in
Point-Slope Form.
Slope-Intercept Form.
𝑦 − −1 = −5(𝑥 − (−2))
P
𝑦 + 1 = −5(𝑥 + 2)
𝑦 + 1 = −5𝑥 − 10
P
𝑦 = −5𝑥 − 11
Point-Slope Form 4
 Practice: Find the slope of a line that is:
a) Parallel to 7𝑥 + 2𝑦 = −6
Write in Slope-Intercept form.
7𝑥 + 2𝑦 = −6
2𝑦 = −7𝑥 − 6
7
𝑦 =− 𝑥−3
2
7
𝑚=−
2
P
b) Perpendicular to 7𝑥 + 2𝑦 = −6.
Need negative reciprocal of slope.
2
𝑚=
7
P
Quadratic Formula
Complete and
Practice HW 3.5
MATH 0322 Intermediate Algebra
Unit 3
Solving Systems of
Linear Equations
(Graphing Method)
Section: 4.1
Solving Systems of Linear Equations
(Graphing Method)
 Chapter 4 studies techniques used to solve a
System of Linear Equations.
*College Algebra advances this study to a technique called the Gaussian Method.
It uses a Matrix and 3 Row Operations to solve a Linear System.
 What is a System of Linear Equations?
• A set of 2 or more Linear Equations in 2 or more variables.
Example:
𝑥 + 𝑦 = −5
−2𝑥 + 𝑦 = 1
Your goal: Solve a Linear System & find its solution.
Straight
lines do 3 Intersect
things, so
3 types of solutions.
Intersect once
many times
Never intersect
𝑦
Eqt#2
𝑦
Eqt#1
𝑥
One
solution
𝑦
Eqt#1
Eqt#2
𝑥
Infinitely
many solutions
Eqt#2
Eqt#1
𝑥
No
solution
Solving Systems of Linear Equations
(Graphing Method)
 A solution to a Linear System must satisfy all
equations in the system.
 A Linear System having infinitely many solutions
is said to have dependent equations.
The solution set for dependent equations is
written in Set Notation.
Example: (𝑥, 𝑦)
Equation #1 or
Equation #2 goes here.
 A Linear System having no solution is called an
Inconsistent System.
 Its solution set is the called the empty set, ∅.
Solving Systems of Linear Equations
(Graphing Method)
 Practice: Determine whether (−3,1) is a
solution to the given Linear System.
−𝑥 + 4𝑦 = 7
2𝑥 + 𝑦 = −1
P
O
− −𝟑 + 4 𝟏 = 7
2 −𝟑 + 𝟏 = −5
1) Substitute to check Equation #1.
2) Substitute to check Equation #2.
Not a solution.
3) Is (−3,1) a solution?
P
Solving Systems of Linear Equations
(Graphing Method)
 Your turn: Determine whether (2, −5) is a
solution to the given Linear System.
3𝑥 + 2𝑦 = −4
−2𝑥 − 𝑦 = 1
P
P
3 𝟐 + 2 −𝟓 = −4
−2 𝟐 − −𝟓 = 1
1) Substitute to check Equation #1.
2) Substitute to check Equation #2.
Yes
3) Is (2, −5) a solution?
P
Solving Systems of Linear Equations
(Graphing Method)
 Practice: Solve by the Graphing Method.
−2𝑥 + 𝑦 = 7
2𝑥 + 𝑦 = −1
1) Rewrite in
Slope-Intercept
form.
2) Graph.
P
P
?
𝑦−2
= −𝟐
2𝑥 ++7 𝟑 =
7
?
𝑦=
− 1𝟑 =
2 −2𝑥
−𝟐 +
−1
𝑦
10
(−𝟐, 𝟑)
−10
10
3) Determine solution,
then Check.
(−2,3)
P
−10
𝑥
Solving Systems of Linear Equations
(Graphing Method)
 Your turn: Solve by the Graphing Method.
𝑦 = 3𝑥 + 2
3𝑥 − 𝑦 = −2
𝑦 = 3𝑥 − 5
−3𝑥 + 𝑦 = −5
𝑦
1) Rewrite in
Slope-Intercept
form.
2) Graph.
10
Parallel lines
−10
10
3) Determine solution.
No Solution.
P
−10
𝑥
Solving Systems of Linear Equations
(Graphing Method)
 Without graphing, how can you tell if a
Linear System in Slope-Intercept form has:
? One solution
*Lines intersect once.
𝑚1 ≠ 𝑚2
? Infinitely many
solutions
*Lines are the same.
𝑚1 = 𝑚2 and 𝑏1 = 𝑏2
? No solution
*Lines are parallel.
𝑚1 = 𝑚2 , but 𝑏1 ≠ 𝑏2
Solving Systems of Linear Equations
(Graphing Method)
 Your turn: Without a graphing calculator, what can
you determine from the Linear System below?
−130𝑥 − 2𝑦 = −1780
−357𝑥 + 3𝑦 = 546
P
o A)
𝑦 = −65𝑥 + 890
𝒎𝟏 ≠ 𝒎𝟐
𝑦 = 119𝑥 + 182
It has only one solution?
o B) It has infinitely many solutions?
o C) It has no solution?
Pros/Cons of Graphing Method
 Pros: works well if solution is an integer,
it allows you to “see” if a solution exists.
 Cons: difficult to determine exact value of solution
if it is very large or is a fraction or decimal.
Graphing Method
Complete and
Practice HW 4.1
MATH 0322 Intermediate Algebra
Unit 3
Solving Systems of
Linear Equations
(Substitution Method)
Section: 4.2
Solving Systems of Linear Equations
(Substitution Method)
This section introduces the Substitution Method:
an algebraic method that substitutes the value
of a variable from one equation into the other
equation to solve a Linear System.
To solve a Linear System using this method:
1) Isolate a variable.(simplest one)
2) Substitute into other equation, solve 1st solution.
3) Back-substitute into an original equation,
solve 2nd solution, then Check.
Solving Systems of Linear Equations
(Substitution Method)
 3 types
of results:
One solution
𝑦
Graphing
Method
Substitution
Method
Infinitely many
solutions
𝑦
𝑥
Step 2 & 3
produce single
value for 𝑥, 𝑦.
No solution
𝑦
𝑥
Step 2
produces 𝑐 = 𝑐.
(𝑐 is a constant)
𝑥
Step 2
produces 𝑐 ≠ 𝑐.
(𝑐 is a constant)
Solving Systems of Linear Equations
(Substitution Method)
 Answer the following questions for the
given Linear System.
𝑥 = 2𝑦 − 7
−3𝑥 = 𝑦 + 1
1) Which equation has the simplest variable
to isolate in Step 1 of our strategy? Equation #1
2) Which variable is that? 𝑥-variable
P
P
Solving Systems of Linear Equations
(Substitution Method)
 Answer the following questions for the
given Linear System.
4𝑥 − 𝑦 = 1
−𝑥 + 𝑦 = −6
1) Which equation has the simplest variable
to isolate in Step 1 of our strategy? Equation #2
2) Which variable is that? 𝑦-variable
P
P
Solving Systems of Linear Equations
(Substitution Method)
 Answer the following questions for the
given Linear System.
4𝑥 + 2𝑦 = −10
3𝑥 − 5𝑦 = 12
1) Which equation has the simplest variable
to isolate in Step 1 of our strategy? Equation #1
2) Which variable is that? 𝑦-variable
3) Why? Solving for 𝑦 does not produce fractions.
P
Eqt#1 for 𝒚
Eqt#1 for 𝒙
4𝑥 + 2𝑦 = −10
4𝑥 + 2𝑦 = −10
2𝑦 = −4𝑥 − 10
4𝑥 = −2𝑦 − 10
P
𝑦 = −2𝑥 − 5
1
5
𝑦=− 𝑥−
2
2
Solving Systems of Linear Equations
(Substitution Method)
 Practice: Solve by the Substitution Method.
𝑥 = 2𝑦 − 7
−2𝑥 = −𝑦 + 5
1) Isolate a variable.
2) Substitute into other
equation, then solve.
3) Back-substitute to solve,
then check.
𝑥 = 2𝑦 − 7
−2𝑥
−2Eqt#𝟐
2𝑦 −
7 = −𝑦 + 5
−4𝑦 + 14 = −𝑦 + 5
−3𝑦 + 14 = 5
−3𝑦 = −9
𝑦 = 3?
Eqt#𝟏 𝑥 = 2( 3 ) − 7
𝑥 =6−7
𝑥 = −1?
Eqt#𝟏
Solving Systems of Linear Equations
(Substitution Method)
 Your turn: Solve by Substitution Method.
−5𝑦 = 4𝑥 − 1
𝑦 = −𝑥 + 3
1) Isolate a variable.
2) Substitute into other
equation, then solve.
3) Back-substitute to solve,
then check.
𝑦 = −𝑥 + 3
−𝑥 −5𝑦
+ 3 = 4𝑥 − 1
−5Eqt#𝟏
5𝑥 − 15 = 4𝑥 − 1
𝑥 − 15 = −1
𝑥 = 14?
Eqt#𝟐
Eqt#𝟐
𝑦 = − 14 + 3
𝑦 = −11?
Solving Systems of Linear Equations
(Substitution Method)
 Your turn: Solve by Substitution Method.
−4𝑥 + 𝑦 = −11
2𝑥 − 3𝑦 = 5
1) Isolate a variable.
2) Substitute into
other equation,
then solve.
Eqt#𝟏
Eqt#𝟏
−4𝑥 + 𝑦 = −11
𝑦 = 4𝑥 − 11
2𝑥 − 3(4𝑥 − 11) = 5
2𝑥 − 12𝑥 + 33 = 5
−10𝑥 + 33 = 5
−10𝑥 = −28
14
𝑥=
5
Solving Systems of Linear Equations
(Substitution Method)
 Your turn: Solve by Substitution Method.
−4𝑥 + 𝑦 = −11
2𝑥 − 3𝑦 = 5
3) Back-substitute to
solve, then check.
Eqt#𝟏
14
)+𝑦 =
5
56 𝟏
− (𝟓)+ 𝑦(𝟓)=
5
𝟏
−4(
−11
−11(𝟓)
−56 + 5𝑦 = −55
5𝑦 = 1
𝑥=
14
5
?
𝑦=
1
5
?
Solving Systems of Linear Equations
(Substitution Method)
 Your turn: Solve by Substitution Method.
−4𝑥 + 𝑦 = −11
2𝑥 − 3𝑦 = 5
3) Back-substitute to
solve,
then
check.
1
14
Eqt#𝟏
−4( ) + ( ) = −11
5
56
−
5
5
1
+
5
55
−
5
= −11
= −11
−11 = −11
𝑥=
Eqt#𝟐
14
5
2
P
14
5
𝑦=
1
5
P
1
− 3( ) = 5
5
28
3
− =5
5
5
25
=5
5
5=5
Solving Systems of Linear Equations
(Substitution Method)
 Your turn: Solve by Substitution Method.
6𝑥 + 2𝑦 = 7
𝑦 − 1 = −3𝑥
1) Isolate a variable.
2) Substitute into
other equation,
then solve.
Eqt#𝟐
Eqt#𝟏
𝑦 − 1 = −3𝑥
𝑦 = −3𝑥 + 1
6𝑥 + 2(−3𝑥 + 1) = 7
6𝑥 − 6𝑥 + 2 = 7
2=7
𝑐≠𝑐
System has no solution.
P
Substitution Method
Complete and
Practice HW 4.2
MATH 0322 Intermediate Algebra
Unit 3
Solving Systems of
Linear Equations
(Addition Method)
Section: 4.3
Solving Systems of Linear Equations
(Addition Method)
 This section introduces the Addition Method:
an algebraic method that eliminates a variable by
adding opposites to solve a Linear System.
 The goal of this method is to reduce the Linear
System to only 1 equation in 1 variable so that a
solution may be reached.
 To do this:
• the system must be in Standard Form,
• and have opposite 𝑥 or 𝑦 terms,
• so equations can be added to combine like terms
into 1 equation in 1 variable.
(opposite 𝑥 or 𝑦 terms are eliminated to zero.)
Solving Systems of Linear Equations
(Addition Method)
How to solve a Linear System using this method:
(*Make sure system is in Standard Form.*)
1) Reduce the system: add equations to eliminate
opposite 𝑥 or 𝑦 terms.
*Multiply equation(s) by a constant to form opposites if needed.
2) Solve remaining equation for 1st solution.
3) Back-substitute into an original equation,
solve for 2nd solution, then Check.
Solving Systems of Linear Equations
(Addition Method)
 3 types
of results:
One solution
𝑦
Graphing
Method
Infinitely many
solutions
𝑦
𝑥
No solution
𝑦
𝑥
𝑥
Substitution
Method
Step 2 & 3
produce single
value for 𝑥, 𝑦.
Step 2
produces 𝑐 = 𝑐.
(𝑐 is a constant)
Step 2
produces 𝑐 ≠ 𝑐.
(𝑐 is a constant)
Addition
Method
Step 2 & 3
produce single
value for 𝑥, 𝑦.
Step 1
produces 0 = 0.
Step 1
produces 0 ≠ 0.
Solving Systems of Linear Equations
(Addition Method)
 Answer the following questions for the
given Linear System.
−3𝑥 − 2𝑦 = −7
−3𝑥 + 2𝑦 = 1
P
1) Is the Linear System in the proper
Yes
form to begin Step 1 of our strategy?
2) What form is that? Standard Form 𝐴𝑥 + 𝐵𝑦 = 𝐶
3) For Step 1 of our strategy, which
𝑦-variable
variables, 𝑥 or 𝑦, are opposites?
P
P
Solving Systems of Linear Equations
(Addition Method)
 Answer the following questions for the
given Linear System.
3𝑦 = −𝑥 + 1 (−𝟐) 𝑥 + 3𝑦 = 1
2𝑥 + 5𝑦 = −4
2𝑥 + 5𝑦 = −4
−2𝑥 − 6𝑦 = −2
2𝑥 + 5𝑦 = −4
P
? Is the Linear System in the proper
No
form to begin Step 1 of our strategy?
? Which equation must be rewritten?
Equation #1
? For Step 1 of our strategy, which
variable can be made to be
𝑥-variable
opposites easier, 𝑥 or 𝑦?
? How?
Multiply Equation#1 by −2.
P
P
P
Solving Systems of Linear Equations
(Addition Method)
 Practice: Solve by the Addition Method.
𝑥 − 𝑦 = −7
−2𝑥 = −𝑦 + 5
𝑥 − 𝑦 = −7
Step 1) + −2𝑥 + 𝑦 = 5
−𝑥 = −2
Step 2)
𝑥=2
?Standard Form?
1) Reduce system:
opposites 𝑦-terms, Step 3) Eqt#𝟏 2 − 𝑦 = −7
add equations.
−𝑦 = −9
𝑦=9
2) Solve for 1st solution.
3) Back-substitute for
(𝟐) − (𝟗 ) = −7
2nd solution, then check.
−2 𝟐 = − 𝟗 + 5
P
P
Solving Systems of Linear Equations
(Addition Method)
 Practice: Solve by the Addition Method.
4𝑥 − 8𝑦 = 36
4𝑥 = 36 + 8𝑦
(−𝟒) 𝑥 − 2𝑦 = 9
𝑥 − 2𝑦 = 9
?Standard Form?
4𝑥 − 8𝑦 = 36
Step 1) + −4𝑥 + 8𝑦 = −36
1) Reduce system:
form opposite 𝑥-terms,
0=0
add equations.
Infinitely many
solutions
P
Solving Systems of Linear Equations
(Addition Method)
 Practice: Solve by the Addition Method.
−4𝑥 + 𝑦 = −11
−12𝑥 + 3𝑦 = −33
2𝑥 − 3𝑦 = 5
Step 1) +
2𝑥 − 3𝑦 = 5
?Standard Form? Yes
−10𝑥 = −28
Step 2)
14
1) Reduce system:
𝑥=
5
make opposite 𝑦-terms,
14
Step 3) Eqt#𝟐 2
− 3𝑦 = 5
add equations.
5
𝟏
28
2) Solve for 1st solution.
(𝟓)
−(𝟓)3𝑦 = 5(𝟓)
𝟏5
3) Back-substitute for
28 − 15𝑦 = 25
2nd solution, then check.
−15𝑦 = −3
1
𝑦=
(𝟑)
P
P
5
Addition Method
Complete and
Practice HW 4.3
Keep moving forward and
finish strong on the Final Exam.