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Algebra 2
Solving Equations
Lesson 1-4
Goals
Goal
• To solve equations.
• To solve problems by
writing equations.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Essential Question
Big Idea: Solve Equations
• How do you solve an equation?
– Students will apply the Properties of Equality to solve
an equation.
– Students will find all of the values of a variable that
make an equation true.
Vocabulary
•
•
•
•
•
Equation
Solution of an Equation
Inverse Operations
Identity
Literal Equation
Definitions
• Equation – is a mathematical statement that
two algebraic expressions are equivalent.
– Equations always contain an equal sign, but an
expression does not have an equal sign.
– To solve equations you use the properties of
equality and the properties of real numbers.
Properties of Equality
Properties of Equality
Definitions
• Solution of an Equation – is the value or
values of the variable that make the equation
true.
– Example: The solution of 2x – 7 = -12 is x = -2.5.
Check
2(-2.5) – 7 = -12
-5 – 7 = -12
-12 = -12
Definitions
• Inverse Operations – are operations that undo
each other.
– Addition and Subtract are inverse operations.
– Multiplication and Division are inverse operations.
• When solving equations, to isolate the variable,
perform the inverse or opposite of every operation
in the equation on both sides of the equation. Do
inverse operations in the reverse order of
operations.
Solving One-Step
Equations
Example:
a.) 8 + z = – 8
8 + (– 8) + z = – 8 + – 8
z = – 16
(Add –8 to each side)
(Simplify both sides)
Solving Multi-Step Equations
Example:
4p – 11 – p = 2 + 2p – 20
3p – 11 = 2p – 18 (Simplify both sides)
3p + (– 2p) – 11 = 2p + (– 2p) – 18
(Add –2p to both sides)
p – 11 = – 18 (Simplify both sides)
p – 11 + 11 = – 18 + 11
(Add 11 to both sides)
p = – 7(Simplify both sides)
Solving Equations
Example:
5(3 + z) – (8z + 9) = – 4z
15 + 5z – 8z – 9 = – 4z
6 – 3z = – 4z
(Use distributive property)
(Simplify left side)
6 – 3z + 4z = – 4z + 4z
(Add 4z to both sides)
6+z=0
(Simplify both sides)
6 + (– 6) + z = 0 +( – 6)
(Add –6 to both sides)
z = – 6 (Simplify both sides)
Solving Equations
Example:
1
5
x
7
9
1  5
7 x    7
7  9
35
x
9
(Multiply both sides by 7)
(Simplify both sides)
Solving Equations
Example:
8
x6
3
3 8 
3 (Multiply both sides by fraction)
 x    6
8 3 
8
18 9 (Simplify both sides)
x 
8 4
Solving Equations
Example:
3z – 1 = 26
3z – 1 + 1 = 26 + 1
(Add 1 to both sides)
3z = 27
3 z 27

3
3
(Simplify both sides)
z=9
(Simplify both sides)
(Divide both sides by 3)
Solving Equations
Example:
12x + 30 + 8x – 6 = 10
20x + 24 = 10
20x + 24 + (– 24) = 10 + (– 24)
(Simplify left side)
(Add –24 to both sides)
20x = – 14
(Simplify both sides)
20 x  14

20
20
(Divide both sides by 20)
7
x
10
(Simplify both sides)
Your Turn:
Solve:
1. 2  5x 17
x  3
2. 5 y  7  21 2 y
3. 3( y  2)  5 y  8
y2
y  1
4. 5(2  3x)  4 1 2(3  x)
x
5. 3  1
5
x  10
x
11
13
Procedure for Solving Equations
Solving linear equations in one variable
1) Clear the equation of fractions by multiplying both sides
of the equation by the LCD of all denominators in the
equation.
2) Use the distributive property to remove grouping symbols
such as parentheses.
3) Combine like terms on each side of the equation.
4) Use the addition property of equality to rewrite the
equation as an equivalent equation with variable terms on
one side and numbers on the other side.
5) Use the multiplication property of equality to isolate the
variable.
6) Check the proposed solution in the original equation.
Solving Equations
Example:
5
3
• Solve: x  x  2
6
4
• LCD: 12
• Multiply both sides by LCD:
5 
3

12 x   12 x  2 
6 
4

10x  9x  24
x  24
Solving Equations
Example:
7
5
• Solve:
x  3  x 1
24
9
• LCD: 72
• Multiply both sides by LCD:
 7

5

72 x  3   72 x  1
 24

9

21x  216  40x  72
 216  19x  72
 288  19 x
288

x
19
Solving Equations
Example:
3( y  3)
 2y  6
5
5  3( y  3)
 52 y  6 
5
3 y  9  10 y  30
3 y  (3 y )  9  10 y  (3 y )  30
9  (30)  7 y  30  (30)
 21 7 y

7
7
3  y
(Multiply both sides by 5)
(Simplify)
(Add –3y to both sides)
(Simplify; add –30 to both sides)
(Simplify; divide both sides by 7)
(Simplify both sides)
Your Turn:
Solve:
3x 1 x 5
1.

2
3
13
x
7
3
1
2. ( z  5)  (2  z)  4
2
3
z
3.
32x x 5

4
2
x
4.
x 3
4 x2
3
x
7
4
9
2
73
11
Solving Equations
• So far we have solved equations that have a
single solution.
• Equations may also have infinitely many
solutions or no solution.
Linear Equations with No
Solution or Infinite Solutions
• Many equations only have one number as a
solution, but some have no solution
(contradictions) and others have infinite solutions
(identities).
• In trying to solve an equation, if the variable
disappears (same variable & coefficient on both
sides) and the constants that are left make a
statement that is:
– false, the equation has “no solution” (contradiction).
– true, the equation has “all real numbers” as solutions
(identity).
Identity
Example:
5x – 5 = 2(x + 1) + 3x – 7
5x – 5 = 2x + 2 + 3x – 7 (Use distributive property)
5x – 5 = 5x – 5
-5 = -5
(Simplify the right side)
True Statement
Both sides of the equation are identical. Since this
equation will be true for every x that is substituted into
the equation, the solution is “all real numbers.”
Contradiction
Example:
3x – 7 = 3(x + 1)
3x – 7 = 3x + 3
(Use distributive property)
3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides)
–7
=3
False Statement
Since no value for the variable x can be substituted
into this equation that will make this a true statement,
there is “no solution.”
Your Turn:
Solve 5(x – 6) = 3x – 18 + 2x.
5(x – 6) = 3x – 18 + 2x
5x – 30 = 5x – 18
–5x
–5x
–30 ≠ –18
Simplify.
Contradiction
The equation has no solution. The solution set is the
empty set, which is represented by the symbol
.
Your Turn:
Solve 3(2 –3x) = –7x – 2(x –3).
3(2 –3x) = –7x – 2(x –3)
6 – 9x = –9x + 6
+ 9x +9x
Simplify.
6=6 
Identity
The solutions set is all real numbers, or .
Problem Solving
Problem solving is the ability to use information,
tools, and our own skills to achieve a goal.
The process of taking a verbal description of the
problem and developing it into an equation that can be
used to solve the problem is mathematical modeling.
The equation that is developed is the mathematical
model.
Steps for Solving Problems
Step 1: Identify What You Are Looking For - Read the problem carefully.
Typically the last sentence in the problem indicates what it is we wish to
solve for.
Step 2: Give Names to the Unknowns - Assign variables to the unknown
quantities. Choose a variable that is representative of the unknown
quantity it represents. For example, use t for time.
Step 3: Translate into the Language of Mathematics - Determine if each sentence
can be translated into a mathematical statement. If necessary, combine
the statements into an equation that can be solved.
Step 4: Solve the Equation(s) Found in Step 3 - Solve the equation for the variable
and then answer the question posed by the original problem.
Step 5: Check the Reasonableness of Your Answer - Check your answer to be sure
that it makes sense. If it does not, go back and try again.
Step 6: Answer the Question - Write your answer in a complete sentence.
Example:
In a baseball game, the Yankees scored 4 more runs
than the White Sox. A total of 12 runs were scored.
How many runs were scored by each team?
Step 1: Identify. We are looking for the number of runs
scored by each team.
Step 2: Name. Let x represent the number of runs scored by
the White Sox. The number of runs scored by the Yankees is
equal to x + 4.
Example: Continued
Step 3: Translate. Since we know that the total number of
runs is 12, we have
White Sox runs
x
+
Yankees runs
x+4
= 12
Step 4: Solve.
x + x + 4 = 12
2x + 4 = 12
2x = 8
x=4
Combine like terms.
Subtract 4 from both sides.
Divide both sides by 2.
Example: Continued
Step 5: Check. Since x represents the number of runs scored
by the White Sox, the White Sox scored 4 runs. The Yankees
scored x + 4 = 4 + 4 = 8 runs.
4 + 8 = 12 
Step 6: Answer the Question.
The Yankees scored 8 runs and the White Sox
scored 4 runs.
Example:
The product of twice a number and three is the same as
the difference of five times the number and ¾. Find the
number.
Step 1: Identify. Find a number that meets the given conditions.
Step 2: Name. Let x = the unknown number, then “twice a
number” is 2x, “the product of twice a number and three” is 2x · 3,
“five times the number” is 5x, and “the difference of five times the
number and ¾” is 5x – ¾.
Example: Continued
Step 3: Translate. The product of twice a number and three is the
same as the difference of five times the number and ¾. Find the
number.
The product of
is the same as
twice a
number
2x
the difference of
and 3
·
3
=
5 times the
number
5x
and ¾
–
¾
Example: Continued
Step 4: Solve.
2x · 3 = 5x – ¾
6x = 5x – ¾
6x + (– 5x) = 5x + (– 5x) – ¾
x=–¾
(Simplify left side)
(Add –5x to both sides)
(Simplify both sides)
Step 5: Check. Replace “number” in the original statement of
the problem with – ¾. The product of twice – ¾ and 3 is 2(–
¾)(3) = – 4.5. The difference of five times – ¾ and ¾ is 5(– ¾) –
¾ = – 4.5. We get the same results for both portions.
Step 6: Answer the Question. The number is – ¾.
Your Turn:
A car rental agency advertised renting a Buick Century
for $24.95 per day and $0.29 per mile. If you rent this car
for 2 days, how many whole miles can you drive on a
$100 budget?
Solution:
A car rental agency advertised renting a Buick Century
for $24.95 per day and $0.29 per mile. If you rent this car
for 2 days, how many whole miles can you drive on a
$100 budget?
Step 1: Identify. Find how many whole miles can you drive on a
$100 budget.
Step 2: Name. Let x = the number of whole miles driven, then
0.29x = the cost for mileage driven.
.
Solution:
Step 3: Translate. A car rental agency advertised renting a Buick
Century for $24.95 per day and $0.29 per mile. If you rent this car
for 2 days, how many whole miles can you drive on a $100 budget?
Daily costs
mileage costs
plus
2(24.95)
+
maximum budget
is equal to
0.29x
=
100
Solution:
Step 4: Solve.
2(24.95) + 0.29x = 100
49.90 + 0.29x = 100
(Simplify left side)
49.90 – 49.90 + 0.29x = 100 – 49.90 (Subtract 49.90 from both sides)
0.29x = 50.10
0.29 x 50.10

0.29
0.29
x  172.75
(Simplify both sides)
(Divide both sides by 0.29)
(Simplify both sides)
Solution:
Step 5: Check. Recall that the original statement of the problem
asked for a “whole number” of miles. If we replace “number of
miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07,
which is over our budget. However, 49.90 + 0.29(172) = 99.78,
which is within the budget.
Step 6: Answer the Question. The maximum number of whole
number miles is 172.
Definitions
• Literal Equation – is an equation with two or
more variables. To solve for one of the variables,
use inverse operations.
– A formula is a type of literal equation.
– Example:
, A=
bh .
Literal Equations
• Literal Equations are equations and formulas
that have several variables (letters).
• Your job, usually, will be to solve the
equation for one of the variables.
• You will do this by following the same
steps you did in solving linear equations.
Literal Equations
Solving for a Variable
Step 1 Locate the variable you are asked to
solve for in the equation.
Step 2 Identify the operations on this
variable and the order in which they
are applied.
Step 3 Use inverse operations to undo
operations and isolate the variable.
Example:
• Solve for w: V = l w h
V  l wh
l w h V
V
w
lh
Rewrite the equation with variable w on the left
divide by l h
Example:
Given the equation
ax + by = c. Solve for y.
ax  by  c
by  c  ax
c  ax
y
b
Subtract ax
Divide by b
Example:
Shoe sizes and foot length are related by the
formula S = 3F - 24, where S represents the shoe
size and F represents the length of the foot, in
inches. Solve the formula for F.
S  3 F  24
3 F  24  S
3 F  S  24
S  24
F 
3
add 24
divide by 3
Your Turn:
9
F  C  32
5
A
a bcd
4
h (a  b )
A
2
C  2 r
Solve for C
Solve for b
Solve for a
Solution:
9
F  C  32
5
9
C  32  F
5
9
C  F  32
5
5 9
5
5
 C   F   32
9 5
9
9
5
160
C F
9
9
Solve for C
Solution:
abcd
A
4
abcd
A
4
4A  a  b  c  d
4A  a  c  d  b
b  4A  a  c  d
Solve for b
Solution:
h(a  b)
A
2
2 A  h(a  b)
2A
ab
h
2A
ab 
h
2A
a
b
h
Solve for a
Assignment:
• Section 1-4, Pg 30 – 32; #1 – 9 all, 10 – 54
even.