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Transcript
Solving Equations With
Variables on Both Sides
Section 2-4
Goals
Goal
• To solve equations with
variables on both sides.
• To identify equations that
are identities or have no
solution.
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.
Vocabulary
• Identity
Equations With Variables on
Both Sides
To solve an equation with variables on both sides, use
inverse operations to "collect" variable terms on one
side of the equation.
Helpful Hint
Equations are often easier to solve when the variable
has a positive coefficient. Keep this in mind when
deciding on which side to "collect" variable terms.
Example: Solving Equations
with Variables on Both Sides
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
–5n
–5n
2n – 2 =
+2
2n
=
To collect the variable terms on one
side, subtract 5n from both sides.
6
+2
8
Since n is multiplied by 2, divide both
sides by 2 to undo the multiplication.
n=4
Your Turn:
Solve 4b + 2 = 3b.
4b + 2 = 3b
–3b
–3b
b+2= 0
–2 –2
b = –2
To collect the variable terms on one
side, subtract 3b from both sides.
Your Turn:
Solve 0.5 + 0.3y = 0.7y – 0.3.
0.5 + 0.3y = 0.7y – 0.3
–0.3y –0.3y
0.5
= 0.4y – 0.3
+0.3
+ 0.3
0.8
= 0.4y
2=y
To collect the variable terms
on one side, subtract 0.3y
from both sides.
Since 0.3 is subtracted from
0.4y, add 0.3 to both sides
to undo the subtraction.
Since y is multiplied by 0.4,
divide both sides by 0.4 to
undo the multiplication.
Using the Distributive
Property
To solve more complicated equations, you may
need to first simplify by using the Distributive
Property or combining like terms.
Example: Simplifying Both
Sides
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
4 – 6a + 4a = –1 –5(7 – 2a)
Distribute –5 to the
expression in parentheses.
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – 35 + 10a
4 – 2a = –36 + 10a
+36
+36
40 – 2a =
+ 2a
40
=
10a
+2a
12a
Combine like terms.
Since –36 is added to 10a,
add 36 to both sides.
To collect the variable
terms on one side, add
2a to both sides.
Example: Continued
40 = 12a
Since a is multiplied by 12,
divide both sides by 12.
Your Turn:
Solve
.
1
Distribute to the expression in
2
parentheses.
3=b–1
+1
+1
4=b
To collect the variable terms on
one side, subtract 1 b from
2
both sides.
Since 1 is subtracted from
b, add 1 to both sides.
Your Turn:
Solve 3x + 15 – 9 = 2(x + 2).
3x + 15 – 9 = 2(x + 2) Distribute 2 to the expression
3x + 15 – 9 = 2(x) + 2(2)
in parentheses.
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
–2x
–2x
x+6=
–6
x = –2
4
–6
Combine like terms.
To collect the variable terms
on one side, subtract 2x
from both sides.
Since 6 is added to x, subtract
6 from both sides to undo
the addition.
Summary
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.
Conditional Equations &
Contradictions
A conditional equation is an equation that is true for some values
of the variable and false for other values of the variable. Normal
equations with a finite number of solutions.
A contradiction is an equation that is false for every replacement
value of the variable.
Example:
Solve the equation 4x – 8 = 4x – 1.
4x – 8 = 4x – 1
–8=–1
Subtract 4x from both sides.
This is a false statement so the equation is a contradiction.
The solution set is the empty set, written { } or .
Contradictions
WORDS
NUMBERS
ALGEBRA
Contradiction
When solving an equation, if you get a false
equation, the original equation is a contradiction,
and it has no solutions.
1=1+2
1=3
x= x+3
–x –x
0=3
Identities
An identity is an equation that is satisfied for all values of the
variable for which both sides of the equation are defined.
Example:
Solve the equation 5x + 3 = 2x + 3(x + 1).
5x + 3 = 2x + 3(x + 1).
5x + 3 = 2x + 3x + 3
Distribute to remove parentheses.
5x + 3 = 5x + 3
Combine like terms.
3=3
Subtract 5x from both sides.
This is a true statement for all real numbers x.
The solution set is all real numbers.
Identities
WORDS
Identity
When solving an equation, if you get an
equation that is always true, the original
equation is an identity, and it has
infinitely many solutions.
NUMBERS
2+1=2+1
3=3
ALGEBRA
2+x=2+x
–x
–x
2=2
Summary
1) A conditional equation is an equation in one variable that
has a finite (normally one solution) number of solutions.
2) An identity is an equation that is true for all values of the
variable (ie. the variable is eliminated and results in a true
statement). An equation that is an identity has infinitely
many solutions.
3) A contradiction is an equation that is false for any value
of the variable (ie. the variable is eliminated and results in
a false statement). It has no solutions.
Example: Identity
Solve 10 – 5x + 1 = 7x + 11 – 12x.
10 – 5x + 1 = 7x + 11 – 12x
10 – 5x + 1 = 7x + 11 – 12x
11 – 5x = 11 – 5x
+ 5x
+ 5x
11
= 11 
Identify like terms.
Combine like terms on the left and the right.
Add 5x to both sides.
True statement.
The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All
values of x will make the equation true. All real numbers are
solutions.
Example: Contradiction
Solve 12x – 3 + x = 5x – 4 + 8x.
12x – 3 + x = 5x – 4 + 8x
12x – 3 + x = 5x – 4 + 8x
13x – 3 = 13x – 4
–13x
–13x
–3 =
–4
Identify like terms.
Combine like terms on the left and the right.
Subtract 13x from both sides.

False statement.
The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There
is no value of x that will make the equation true. There are no
solutions.
Your Turn:
Solve 4y + 7 – y = 10 + 3y.
4y + 7 – y = 10 + 3y
4y + 7 – y = 10 + 3y
3y + 7 = 3y + 10
–3y
–3y
7 =
10
Identify like terms.
Combine like terms on the left and the right.
Subtract 3y from both sides.

False statement.
The equation 4y + 7 – y = 10 + 3y is a contradiction. There is
no value of y that will make the equation true. There are no
solutions.
Your Turn:
Solve 2c + 7 + c = –14 + 3c + 21.
2c + 7 + c = –14 + 3c + 21
2c + 7 + c = –14 + 3c + 21
3c + 7 = 3c + 7
–3c
–3c
7=
7
Identify like terms.
Combine like terms on the left and the right.
Subtract 3c both sides.

True statement.
The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All
values of c will make the equation true. All real numbers are
solutions (infinite number of solutions).
Strategy for Problem Solving
1) UNDERSTAND the problem. During this step, become
comfortable with the problem. Some way of doing this
are:
• Read and reread the problem
• Propose a solution and check.
• Construct a drawing.
• Choose a variable to represent the unknown
2) TRANSLATE the problem into an equation.
3) SOLVE the equation.
4) INTERPRET the result. Check the proposed solution in
stated problem and state your conclusion.
Finding an Unknown Number
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.
1.) Understand
Read and reread the problem. If we let
x = the unknown number, then “twice a number” translates to 2x,
“the product of twice a number and three” translates to 2x · 3,
“five times the number” translates to 5x, and
“the difference of five times the number and ¾” translates to 5x – ¾.
Continued
Example continued:
2.) Translate
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
Continued
Example continued:
3.) 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)
4.) Interpret
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.
State: The number is – ¾.
Your Turn:
Example:
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?
1.) Understand
Read and reread the problem. Let’s propose that we drive a total of
100 miles over the 2 days. Then we need to take twice the daily rate
and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29
= 78.90. This gives us an idea of how the cost is calculated, and also
know that the number of miles will be greater than 100. If we let
x = the number of whole miles driven, then
0.29x = the cost for mileage driven
Continued
Continued:
2.) Translate
Daily costs
mileage costs
plus
2(24.95)
+
maximum budget
is equal to
0.29x
=
100
Continued
Continued:
3.) Solve
2(24.95) + 0.29x = 100
49.90 + 0.29x = 100
49.90 – 49.90 + 0.29x = 100 – 49.90
0.29x = 50.10
0.29 x 50.10

0.29
0.29
x  172.75
(Simplify left side)
(Subtract 49.90 from both sides)
(Simplify both sides)
(Divide both sides by 0.29)
(Simplify both sides)
Continued
Continued:
4.) Interpret
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.
State: The maximum number of whole number miles is 172.
Joke Time
• Where do you find a dog with no legs?
• Right where you left him.
• What did E.T.’s Mom say when he got home?
• Where on Earth have you been?
• What happened when the pig pen broke?
• The pigs had to use a pencil.
Assignment
• 2.4 Exercises Pg. 114 – 116: #10 – 48 even