1. 2x − 3 = 1
N S
3x = 4 T IO
U A
E Q −1 = 2 − 4x
A R
IN E
L 3x − 1 = 5 − 2 x

2. A linear equation in one variable is equivalent to
an equation of the form: ax + b = 0
If we have a linear equation we can “manipulate” it to get it
in this form. We just need to make sure that whatever we
do preserves the equality (keeps both sides =)
3x + 1 = 4 We can add or subtract the same thing
from both sides of the equation.
-4 -4
-3 0
Notice this is the equation
3x − 3 = 0 above where a = 3 and b = -3.
While this is in the general form for a linear equation, we
often want to find all values of x so that the equation is true.
You could probably do this one in your head and see that
when x = 1 we’d have a true statement 0 = 0

3. We will want to find those values of x that make the
equation true by isolating the x (this means get the x all by
itself on one side of the equal sign)
Since the x is in more than
2( x − 4 ) = −5( 2 x + 1)
one place and inside of
parenthesis the first thing
we’ll do is get rid of
parenthesis by distributing.
2 x − 8 = −10 x − 5 Now let’s get all constants
+8 (terms without x’s) on the right
+8
side. We’ll do this by adding 8
+3 to both sides.
2 x = −10 x + 3 We are ready to get all x terms on the
left side by adding 10x to both sides.
+ 10x + 10x
1 Now get the x by itself by
12 x = 3 x= getting rid of the 12. 12x means
12 12 4 12 times x so we get rid of it by
dividing both sides by 12.

4. Let’s check this answer by substituting it into the original
equation to see if we get a true statement.
2( − 4 ) = −5( )
1  1
x
4
?
2 x + 1 Distribute and multiply
 4
1 ? 1 
− 8 = −5 + 1 Distribute
2 2 
1 ? 5
−8 = − −5 Get a common denominator
2 2
1 16 ? 5 10 15 15
− =− − − =−
2 2 2 2 2 2
It checks!

5. y +1 3 y
Solve the following equation: − =
3 4 2
When we see an equation with fractions, it is generally
easiest to find the common denominator and multiply all
terms by this common denominator. This will give you an
equivalent “fraction free” equation to solve.
4 3 6 The common denominator is 12 so we’ll
12 y +1 3(12) y (12) multiply each term by 12.
− =
3 4 2 Cancel all denominators
4( y + 1) − 9 = 6 y Here is our “fraction free” equation to solve
4y + 4 − 9 = 6y Distribute
−5 = 2y Subtract 4y from both sides and combine like terms
5
− =y Divide both sides by 2
2

6. 2 3 1
Solve the following equation: − = 2
x −1 x +1 x −1
Let's get a "fraction free" equation by multiplying by the
common denominator. Factor any denominators that will factor
so you can determine the lowest common denominator.
2(x+1)(x-1)3(x+1)(x-1) 1(x+1)(x-1)
− = The common denominator is (x+1)(x-1)
x − 1 x + 1 ( x + 1)( x − 1) so multiply all terms by this
Cancel all denominators
2( x + 1) − 3( x − 1) = 1 Here is our “fraction free” equation to solve
Distribute
2 x + 2 − 3x + 3 = 1
Combine like terms
− x +5 =1 subtract 5 from both sides
− x = −4 divide both sides by -1
Make sure the answer would not cause you to divide by 0 in
x=4 the original equation. (Only 1 and -1 would cause this so we
are okay).