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Fraction Equations

Reference > Mathematics > Algebra > Algebraic Fractions
 

Whenever I'm teaching algebraic fractions (rational expressions) to an algebra class, this is my favorite section to teach. I tell my students that "this is where the magic happens."

They don't believe me. They don't think there's anything magical about fractions.

And they may be right, but it certainly is magical when you can make all the fractions disappear. And that's what we're going to do. Are you ready for the magic?

Example #1

Solve for x if
x
3
+
x
2
=
25
6

Solution #1

Just as we've always done, we're going to check to see if anything will factor and/or reduce. Nothing, in this case. So now we move on to our next step: since we have addition of fractions, we're going to find an LCD. Since the denominators are 3, 2, and 2·3, the LCD is 2·3 = 6.

Now, if these were just rational expressions, we would convert each of them to the common denominator. But they're not just rational expressions; they are part of an equation. And in an equation, we can multiply both sides of the equation by the same thing, and we won't change the meaning of the equation. So what do we multiply by? We multiply by the LCD. Watch what happens:

6(
x
3
+
x
2
) = 6(
25
6
)

2x + 3x = 25

5x = 25

x = 5

Did you see the magic happen? In one line we had three fractions, and in one simple step (multiplying both sides of the equation by the LCD) we made the fractions vanish. We didn't even have to wave a magic wand! And then we were back to the kinds of problems we were all solving in the first couple months of algebra class!

Of course, you'll want to plug that answer back into the original equation to make sure it works.

Example #2

Solve for x:
x + 1
4
-
x - 5
3
= 1

Solution #2

Let's rewrite the this with parentheses, just to remind ourselves that the fraction bar is grouping those binomials:

(x + 1)
4
-
(x - 5)
3
= 1

Nothing cancels or reduces, so we find a common denominator. LCD = 12. Now comes the magic step - we're going to multiply both sides of the equation by 12.

12(
(x + 1)
4
-
(x - 5)
3
) = 12(1)

3(x + 1) - 4(x - 5) = 12

3x + 3 - 4x + 20 = 12

-x + 23 = 12

x = 11

The process is no different even if there are variables in the denominator.

Example #3

Solve for x:
x
x + 6
+
1
4
=
x
4

Solution #3

x
(x + 6)
+
1
4
=
x
4

The LCD for these three fractions is 4(x + 6), so that's what we'll multiply both sides of the equation by:

4(x + 6)(
x
(x + 6
} +
1
4
) = 4(x + 6)(
x
4
)

4x + x + 6 = x2 + 6x

This equation is a quadratic, so we want to get everything to one side so we can factor and use the zero rule of multiplication:

x2 + x - 6 = 0
(x + 3)(x - 2) = 0

x = -3 or x = 2

Both of these answers work in the original equation.

With this, you've reached the end of our unit on Algebraic Fractions. Congratulations! Try the problems below for practice.

Questions

1.
Solve for x:
x
3
+
x
4
= 7
2.
Solve for x:
x + 1
2
-
x - 1
3
= 5
3.
Solve for x:
x/2
+
x
3
-
x
5
= 57
4.
Solve for x:
24
x
+
1
3
=
5
6
5.
Solve for x:
12
x
-
10
x + 1
= 1
6.
Solve for x:
1
x - 1
-
1
x + 1
=
1
40
7.
Solve for x:
x
x + 1
+
3
2
=
33
10
8.
Solve for x:
h + 1
hx + x
+ 1 =
3h - 3
2h - 2
9.
Solve for x:
1
x
+
1
x + 1
+
1
x - 1
=
11
3x
10.
Solve for x:
1
x3 - 27
+ 1 =
38
37
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