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Solving Equations
1.
2.
Clear the fractions.
3
5

4
LCD  2x x  1
a. Multiply through by the LCD. If the LCD contains a
2x  1 2x
variable, note which values will make the original
32x x  1 52x x  1
fractions undefined. These values will be excluded

 42x x  1
2x  1
2x
from the solution set.
b. Continue solving depending on the type of equation
3x  5x  1  8xx  1
you have left.
Decide which of the following types of equations you are solving.
Type
A.
DO THIS
LINEAR
ax  b  c
a  0
2
(No x term)
x  0,1
EXAMPLE
a. Use the distributive property to remove
parenthesis and combine like terms.
b. Use the addition property of equality to get
the variable terms on one side of the
equation and the numbers on the other.
c. Then use the multiplication property of
equality to make the coefficient of the
variable equal to 1.
48  3t   20  8t
32  12t  20  8t
32  12t  12t  20  8t  12t
32  20  4t
32  20  20  4t  20
12  4t
12
4t

4
4
t  3
Set equation = 0.
B.
QUADRATIC
ax
2
 bx  c  0
a  0
2
(Contains an x term)
Try to factor first.
a. Factor if you can. Look for GCF.
b. Set each factor equal to 0.
c. Solve those equations.
2x
2x
 5x  3
2x  1  0
x  3  0
2
 5x  3  0
2x  1
1
x
2
x  3
2x  1x  3  0
Complete the square.
2
a. Be sure the coefficient of the x term is 1.
b. Rewrite the equation so that both terms
containing variables are on one side of
the equal sign and the constant is on the
other.
c. Take half of the coefficient of x and square
it.
d. Add this amount to both sides of the
equation.
e. It is now an equation like Type E.
(Continued)
vz
95
2
2x
x
x
x
2
 4 x  18  0
2
 2x  9  0
2
 2x  9
2
 2x  1  9  1
x  12
 10
1

 2  2


2
  12  1
QUADRATIC
(Continued)
Use the quadratic formula
x 
C.
ABSOLUTE VALUE
3x  4  5  2
b 
b  4ac
2a
a. Isolate the absolute value. There will
generally be two cases. Set the quantity
inside the absolute value equal to  the
value on the other side of the equal sign.
b. If the absolute value equals 0, there will be
only one solution. If the absolute value
equals a negative number, there will be no
solutions.
D.
a. Isolate the radical.
b. Use the power rule. Raise both sides of the
equation to the same power as the index of
the radical.
c. This will leave either a linear or a quadratic
equation. Follow directions above.
d. Check all solutions.
RADICAL
2x  3  x
E.
PERFECT SQUARE
x  12
F.
 10
QUADRATIC IN FORM.
23x  12  53x  1  2  0
3x
2
2
a. Isolate the perfect square.
b. Use the square root property. Take the
square root of both sides.
c. Remember, there will be a positive and
negative square root.
d. Simplify the solution.
Use substitution.
a. Let another variable equal the repeated
quantity.
b. Solve the simpler equation.
c. Set the repeated quantity equal to those
solutions.
d. Solve for the original variables.
x
 5x  2  0
 5  5 2  432
23
2
x  1 or 3
3x  4  5  2
3x  4  2  5
3x  4  7
3 x  11
11
x
3
3x  4  7
3x  3
3
x
3
x  1
2x  3  x
2x  3  x
x
2
x  12
y
2
 2y  6  y  2
y 2  2y  6  y  22
2
 2x  3  0
2
2
y  2y  6  y  4y  4
 2y  6  4 y  4
1
y 
3
 10
x  1  
x  1
10
10
23x  12  53x  1  2  0
Let a  3x  1
2a
2
 5a  2  0
2a  1a  2  0
a  
vz
95
a  3 b  5 c  2
1
, 2
2
3x  1  
1
2
1
x 
6
3x 
1
2
3x  1   2
3x   1
x  
1
3