Solving Equations
Most of algebra involves either simplifying expressions (by
writing equivalent expressions) or solving equations (by writing
equivalent equations). We can use the addition and multiplication
principles to produce equivalent equations, like x = 5 from which
the solution  in this case, 5 is obvious. Here we will find
that more complicated equations can be solved by using both
principles together and by using the commutative, associative,
and distributive laws to write equivalent expressions. An
important strategy for solving a new look like a problem we
already know how to solve. This is precisely the approach taken
in this section.
What is new in this section appears in the early steps of each
example. Without a solid grasp of how and when to use the
addition and multiplication principles, the problems in this
section will seem much more difficult than they really are.
Using both the Addition and Multiplication Principles
In the expression 5 + 3x the variable x is multiplied by 3 and
then 5 is added. To reverse these steps, we first subtract 5 and
then divide by 3. Thus, to solve 5 + 3x = 17 we first subtract 5
from each side and then divide both sides by 3.
Example 1
Solve: 5 + 3x = 17
Solution
We have
5 + 3x = 17
5 + 3x  5 = 17  5 Using the addition principle: subtracting
5 from both sides (adding 5)
5 + ( 5) + 3x = 12 Using a commutative law. Try to perform
this step mentally.
First isolate the xterm
3x = 12 Simplifying
Using the multiplication principle: dividing both sides by 3
multiplying by
Then isolate x
x = 4 Simplifying
Check:
We use the rules for order of operations:
Find the product , and then add.
The solution is 4.
Multiplication by a negative number and subtraction are
handled in much the same way.
Example 2
Solve 5x 6 = 16.
Solution
In 5x 6 we multiply first and then subtract. To reverse
these steps, we first add 6 and then divide by 5.
5x 6 = 16
5x 6 + 6 = 16 + 6 Adding 6 to both sides.
Dividing both sides by 5
Simplifying
Check:
The soultion is .
Example 3
Solve 45  t = 13.
Solution
45  t = 13
45  t  45 = 13  45 Substracting 45 from both sides.
Try
to do these steps mentally
t = 32 Try to go directly to this step
(1)(t) = (1)(32) Multiply both sides by 1 (Dividing by 1
would also work)
t = 32
Check:
The soultion is 32.
As our skills improve, many of the steps can be streamlined.
Example 4
Solve 16.3  7.2y = 8.18
Solution
We have
16.3  7.2y = 8.18
 7.2y = 8.18 16.3 Subtracting 16.3 from both sides. We
write the subtraction of 16.3 on the right side and remove 16.3
from the left side.
 7.2y = 24.48
Dividing both sides by 7.2.We write the division by 7.2 on the
right side and remove the from the left side.
y = 3.4
Check:
The solution is 3.4.
Combining Like Terms
If like terms appear on the same side of an equation, we
combine them and then solve. Should like terms appear on both
sides of an equation, we can use the addition principle to
rewrite all like terms on one side
Example 5
Solve:
a) 3x + 4x = 14
b) 2x  4 = 3x + 1
c) 6x + 5 7x = 10  4x + 7
d) 2  5(x + 5) = 3(x  2)  1
Solution
a) 3x + 4x = 14
7x = 14 Combining like terms
Dividing both sides by 7
x = 2
The check is left to the student. The solution is 2.
To solve 2x  4 = 3x + 1 we must first write only variable
terms on one side and only constant terms on the other. This can
be done by adding 4 to both sides, to get all constant terms on
the right, and 3x to both sides, to get all variable terms on the
left. We can add 4 first, or 3x first, or do both in one step.
2x  4 = 3x + 1
2x  4 + 4 = 3x + 1 + 4 Adding 4 to both sides
2x = 3x + 5 Simplifying
2x + 3x = 3x + 3x + 5 Adding 3x to both sides
5x = 5 Combining like terms and simplifying
Dividing both sides by 5
x = 1 Simplifying
Check:
The solution is 1.
c) 6x + 5  7x = 10  4x + 7
x + 5 = 17  4x Combining like terms on both sides
x + 5 + 4x = 17  4x + 4x Adding 4x to both sides
5 + 3x = 17 Simplifying. This is identical to Example 1
3x = 12 Subtracting 5 from both sides and simplifying
x = 4 Dividing both sides by 3 and simplifying.
Check:
The solution is 4.
d) 2  5(x + 5) = 3(x  2)  1
2  5x  25 = 3x  6  1 Using the distributive law. This is
now similar to part (c) above.
5x  23 = 3x  6 Combining like terms on both sides
Adding 7 and 5x to both sides. This isolates the xSimplifying
16 = 8x Simplifying
2 = x Dividing both sides by 8
The student can confirm that 2 checks and is the solution.
