Solving Equations With Algebraic Fractions

This article explores solving equations with algebraic fractions. It covers understanding how to solve equations invo...

This article explores solving equations with algebraic fractions. It covers understanding how to solve equations involving fractions, working with denominators with either constants or linear factors. Before attempting this chapter you must have prior knowledge of expanding brackets and factorising quadratics.

Simple fraction equations

Let us look at simple equations first. Suppose we had to solve;
Example: Solve the following algebraic fraction.
2x + 5/3 = 17
Answer
2x + 5× 3/3 = 17 × 3
2x + 5 = 51
2x = 46
x = 23
Explanation The problem is very simple that you may think. First we have to get rid of the fraction by multiplying by 3.
2x + 5× 3/3 = 17 × 3
The above gives;
2x + 5 = 51
Then we solve as usual by subtracting 5 from both sides to get;
2x = 46
Then we divide both sides by 2 to get;
x = 23

Multiplying up a linear factor

This following example shows equations with linear factors on top and bottom for example suppose we had to solve;
Example: Solve the following algebraic fraction by finding the value of x.
5x + 1/x – 1 = 8
Answer:
5x + 1/x – 1(x – 1) = 8(x – 1)
5x + 1/x – 1(x – 1) = 8(x – 1)
5x + 1 = 8(x – 1)
5x + 1 = 8x – 8
1 = 3x – 8
x = 3
Explanation: Again here we have to get rid of the fraction first we need to undo the divide operation by multiplying both sides by (x – 1)
5x + 1/x – 1(x – 1) = 8(x – 1)
5x + 1/x – 1(x – 1) = 8(x – 1)
That would give;
5x + 1 = 8(x – 1)
Now we can expand the brackets and solve;
5x + 1 = 8x – 8
1 = 3x – 8
x = 3

Multiplying 2 linear factors

The following shows an example of equations with linear factors
Example: Find the value of x in the following expression.
Answer:
x + 1/x + 3 = 2x – 1/3x – 1
x + 1/x + 3(x + 3) = (2x – 1)(x + 3)/3x – 1
x + 1 = (2x – 1)(x + 3)/3x – 1
(x + 1)(3x – 1) =(2x – 1)(x + 3)/(3x – 1)(3x – 1)
(3x – 1)(x + 1) = (2x – 1)(x + 3)
3x2 + 2x – 1 = 2x2 – 5x – 3
x2 – 3x + 2 =
(x – 2)(x – 1) = 0
x – 2 = 0
x = 2
x – 1 = 0
x = 1
x = 2 or x = 1
Explanation:
x + 1/x + 3 = 2x – 1/3x – 1
Again to simplify we have to get rid og the fractions or undo the division operation by multiplying; First we multiply everything by (x + 3)
x + 1/x + 3(x + 3) = (2x – 1)(x + 3)/3x – 1
…to get;
x + 1 = (2x – 1)(x + 3)/3x – 1
Next we multiply again (3x – 1)
(x + 1)(3x – 1) =(2x – 1)(x + 3)/(3x – 1)(3x – 1)
…the above gives us;
(3x – 1)(x + 1) = (2x – 1)(x + 3)
Now we can solve normally by first expanding the brackets and rearranging.
3x2 + 2x – 1 = 2x2 – 5x – 3
x2 – 3x + 2 =
Now we can factorise and solve;
(x – 2)(x – 1) = 0
…so now we have.
x – 2 = 0
x = 2
…and…
x – 1 = 0
x = 1
So the solutions to the equations are;
x = 2 or x = 1

Equations with 2 fractions on the same side

Now we shall explore equations with 2 fractions on the same side.
Example: Find the value of x in the following algebraic expression.
Answer:
5x – 3/2 + x + 7/3 = 15
5x – 3/2 × 2 + x + 7/3 × 2 = 15 × 2
(5x – 3) + 2(x + 7)/3 = 30
(5x – 3)×3 + 2(x + 7)/3 × 3 = 30 × 3
3(5x – 3) + 2(x + 7) = 90
5x – 9 + 2x + 14 = 90
17x – 5 = 90
17x = 85
x = 5
Explanation:
5x – 3/2 + x + 7/3 = 15
The above equation is slightly tricker than the other equations we have seen so far. First let us get rid of the divide by 2 fraction we do that by multiplying through by 2.
5x – 3/2 × 2 + x + 7/3 × 2 = 15 × 2
…that gives us;
(5x – 3) + 2(x + 7)/3 = 30
Remember that when you multiply something out in an equation all the terms must get multiplied as well. By multiplying above by two we get rid of the denominator. Now we have to undo the divide by 3 by multiplying through by 3.
(5x – 3)×3 + 2(x + 7)/3 × 3 = 30 × 3
…that gives use…
3(5x – 3) + 2(x + 7) = 90
We have managed to form a standard equation that we can expand and solve.
5x – 9 + 2x + 14 = 90
17x – 5 = 90
17x = 85
x = 5

Equations with 3 fractions

The following example shows equations with 3 fractions.
Example: Solve the following algebraic expression by finding the value of x.
2/x + 8 + 1/x – 2 = 1/3
Answer:
2 + x + 8/x – 2 = x + 8/3
2(x – 2) + x + 8 = (x + 8)(x – 2)/3
3 × 2(x – 2) + 3(x + 8) = (x + 8)(x – 2)
6x – 12 + 3x + 24 = x2 + 6x – 16
0 = x2 – 3x – 28
0 = (x – 7)(x + 4)
x = 7 or x = -4
Explanation:We have 3 fractions that we have to undo first. First we multiply through to get;
2 + x + 8/x – 2 = x + 8/3
We undo the second fraction by multiplying through by x-2 to get;
2(x – 2) + x + 8 = (x + 8)(x – 2)/3
Next we undo the division by 3 to get;
3 × 2(x – 2) + 3(x + 8) = (x + 8)(x – 2)
6x – 12 + 3x + 24 = x2 + 6x – 16
0 = x2 – 3x – 28
Now we can factorise and solve;
0 = (x – 7)(x + 4)
The solutions are;
x = 7 or x = -4
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