# Solving Algebraic Fraction Equations and Proportions

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```Solving an equation involving fractions.
315
x 2 x
315
2x
x 2 x

Variable in denominator

6  x  10
6
6
x 4
Restriction: x  0
Since our answer
is not 0 we are ok
We are going to
multiply through by
the LCD to get rid of
the fractions
LCD  2x
We multiply the top by
what isn’t in the bottom –
the bottoms cancel out
We have to be careful
that x doesn’t make
the denominator 0
Solve.
2
x
 2
1
x  2 x  2x
2
x

 1 Restriction: x  2
x  2 x(x  2)
or x  0
 2

x
x(x  2) 

 1 LCD x(x+2)
 x  2 x(x  2) 
2x  x  x(x  2)
2
x  x  2x
x
x
2
0x x
0  x(x  1)
x  0 or x 1
Objective - To use proportions to solve
problems.
Proportion - A statement of equality between
two ratios.
a
c

b
d
extremes
Solve for x.
1) 3 x

5 20
60  5x
5
5
12 x
means
ad  bc
2) 5
30

7 x
5x  210
5
5
x 42
Solve the proportions below.
2x 5
1)

3 30
60x  15
60 60
1
x
4
2x 1 5
2)

4
3
3(2x  1)  20
6x  3  20
3  3
6x  17
6
6
17
5
x
or 2
6
6
```