 # Solving Equations with Variables on Both Sides

```California
Standards
4.0 Students simplify expressions before
solving linear equations and inequalities in one
variable, such as 3(2x – 5) + 4(x – 2) = 12.
5.0 Students solve multistep problems,
including word problems, involving linear
equations and linear inequalities in one variable
and provide justification for each step.
To solve an equation with variables on both sides,
use inverse operations to "collect" variable terms
on one side of the equation.
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.
Solving Equations with Variables on Both Sides
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
–5n
–5n
2n – 2 =
+2
2n
=
n=4
6
+2
8
1: Move variable to one side
3: multiple/divide
4b + 2 = 3b
4b + 2 = 3b
–3b
–3b
b+2= 0
–2
–2
b = –2
1: Move variable to one side
Check
4b + 2 = 3b
4(–2) + 2
–8 + 2
–6
3(–2)
–6
–6 
0.5 + 0.3y = 0.7y – 0.3
0.5 + 0.3y = 0.7y – 0.3
–0.3y
0.5
+0.3
0.8
–0.3y
= 0.4y – 0.3
+ 0.3
1: Move variable to one side
3: Multiple/divide
= 0.4y
Check
0.5 + 0.3y = 0.7y – 0.3
0.5 + 0.3(2)
2=y
0.5 + 0.6
1.1
0.7(2) – 0.3
1.4 – 0.3
1.1 
1.4a – 3 = 2a + 7
a=5
2.-3r + 9 = -4r + 5
r = -4
3.3d + 8 = 2d – 17
d = -25
4.2x – 7 = 5x – 10
x=1
To solve more complicated
equations, you may need to first
simplify by using the Distributive
Property or combining like terms.
Don’t call me after midnight
1. D= Distributive property
2. C= combine like term
3. M = move variable to one side
5. M = Multiplication/division
3x + 15 – 9 = 2(x + 2)
D: distributive Property
3x + 15 – 9 = 2(x + 2)
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
–2x
–2x
x+6=
4
–6
–6
x = –2
C: Combine like term
M: move variable to one
side
Simplifying Each Side Before
Solving Equations
4 – 6a + 4a = –1 – 5(7 – 2a)
D: Distributive P.
4 – 6a + 4a = –1 –5(7 – 2a)
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – 35 + 10a
4 – 2a = –36 + 10a
+2a
+2a
4
= -36 + 12a
C: Combine like
terms
M: move variable to
One side
4
+ 36
40
= -36 + 12a
+36
= 12a
M: multiply/divide
Check 3x + 15 – 9 = 2(x + 2)
3(–2) + 15 – 9
2(–2 + 2)
–6 + 15 – 9
2(0)
0
0
An identity is an equation that is always
true, no matter what value is substituted
for the variable. The solution set of an
identity is all real numbers. Some
equations are always false. Their solution
sets are empty. In other words, their
solution sets contain no elements.
Infinitely Many Solutions or No Solutions
Solve the equation.
10 – 5x + 1 = 7x + 11 – 12x
10 – 5x + 1 = 7x + 11 – 12x
C: combine like
term
11 – 5x = 11 – 5x
The statement 11 – 5x = 11 – 5x is true for all values
of x. The equation 10 – 5x + 1 = 7x + 11 – 12x is an
identity. All values of x will make the equation true.
In other words, all real numbers are solutions.
Infinitely Many Solutions or No Solutions
Solve the equation.
12x – 3 + x = 5x – 4 + 8x
12x – 3 + x = 5x – 4 + 8x
13x – 3 = 13x – 4
–13x
–13x
–3 = –4 
C: combine like term
M: move variable to
one side
The equation 12x – 3 + x = 5x – 4 + 8x is always
false. There is no value of x that will make the
equation true. There are no solutions.
Writing Math
The empty set can be written as  or {}.
Check It Out! Example 3a
Solve the equation.
4y + 7 – y = 10 + 3y
4y + 7 – y = 10 + 3y
Identify like terms.
3y + 7 = 3y + 10
–3y
–3y
7 = 10
Subtract 3y from both sides.
False statement; the

solution set is .
The equation 4y + 7 – y = 10 + 3y is always
false. There is no value of y that will make the
equation true. There are no solutions.
Check It Out! Example 3b
Solve the equation.
2c + 7 + c = –14 + 3c + 21
2c + 7 + c = –14 + 3c + 21
3c + 7 = 3c + 7
Identify like terms.
Combine like terms on the
left and the right.
The statement 3c + 7 = 3c + 7 is true for all values
of c. The equation 2c + 7 + c = –14 + 3c + 21 is an
identity. All values of c will make the equation true.
In other words, all real numbers are solutions.
Jon and Sara are planting tulip bulbs. Jon has
planted 60 bulbs and is planting at a rate of 44
bulbs per hour. Sara has planted 96 bulbs and
is planting at a rate of 32 bulbs per hour. In
how many hours will Jon and Sara have planted
the same number of bulbs? How many bulbs
will that be?
Person
Bulbs
Jon
60 bulbs plus 44 bulbs per hour
Sara
96 bulbs plus 32 bulbs per hour
Let h represent hours, and write expressions for
the number of bulbs planted.
When is
60
60
plus
bulbs
+
44h
44
bulbs
each
hour
=
the
same
as
96
96
bulbs
+
plus
32
bulbs
?
each
hour
32h
60 + 44h = 96 + 32h To collect the variable terms
on one side, subtract 32h
–32h
–32h
from both sides.
60 + 12h = 96
60 + 12h = 96
–60
– 60
12h = 36
h=3
Since 60 is added to 12h,
subtract 60 from both sides.
Since h is multiplied by 12,
divide both sides by 12 to
undo the multiplication.
After 3 hours, Jon and Sara will have planted the
same number of bulbs. To find how many bulbs they
will have planted in 3 hours, evaluate either
expression for h = 3:
60 + 44h = 60 + 44(3) = 60 + 132 = 192
96 + 32h = 96 + 32(3) = 96 + 96 = 192
After 3 hours, Jon and Sara will each have planted
192 bulbs.
Check It Out! Example 4
Four times Greg's age, decreased by 3 is equal
to 3 times Greg's age, increased by 7. How old
is Greg?
Let g represent Greg's age, and write expressions
for his age.
Four
times
Greg's
age
4g
decreased
by
3
–
3
is
equal
to
=
three
times
Greg's
age
3g
increased
by
+
7
7
.
Check It Out! Example 4 Continued
4g – 3 = 3g + 7
–3g
–3g
g–3=
+3
g=
7
+3
10
Greg is 10 years old.
To collect the variable terms on
one side, subtract 3g from both
sides.
Since 3 is subtracted from g, add
3 to both sides.
Lesson Quiz
Solve each equation.
1. 7x + 2 = 5x + 8 3
2. 4(2x – 5) = 5x + 4 8
3. 6 – 7(a + 1) = –3(2 – a)
4. 4(3x + 1) – 7x = 6 + 5x – 2 all real numbers
5.
1
20 hours
6. A painting company charges \$250 base plus \$16
per hour. Another painting company charges \$210
base plus \$18 per hour. How long is a job for
which the two companies costs are the same?
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