24
EQUATIONS
WITH FRACTIONS
TO SOLVE AN EQUATION with fractions, we transform it into an equation without fractions -- which we know how to solve. The technique is called clearing of fractions.
Example 1. Solve for x:
| x 3 |
+ | x − 2 5 |
= 6 |
Solution. Clear of fractions as follows:
Multiply both sides of the equation -- every term -- by the LCM of denominators. Every denominator will then cancel. We will then have an equation without fractions.
The LCM of 3 and 5 is 15. Therefore, multiply every term on both sides by 15:
| 15· | x 3 |
+ | 15· | x − 2 5 |
= 15· 6 |
Each denominator will now cancel into 15, and we have the following simple equation that has been "cleared" of fractions:
| 5x + 3(x − 2) | = | 90 |
| It is easily solved as follows: | ||
| 5x + 3x − 6 | = | 90 |
| 8x | = | 90 + 6 |
| x | = | 96 8 |
| = | 12. | |
We say "multiply" both sides of the equation, yet we take advantage of the fact that the order in which we multiply or divide does not matter (Lesson 1). Therefore we divide by each denominator first, and thus clear of fractions.
Example 2. Clear of fractions and solve for x:
| x 2 |
− | 5x 6 |
= | 1 9 |
Solution. The LCM of 2, 6, and 9 is 18. Multiply each term by 18 -- and cancel.
9x − 15x = 2.
It should not be necessary to actually write 18. The student should
| simply look at | x 2 |
, and see that 2 will go into 18 nine (9) times. That term |
therefore becomes 9x.
| Next, look at | 5x 6 |
, and see that 6 will to into 18 three (3) times. That |
term therefore becomes 3· −5x = −15x.
| Finally, look at | 1 9 |
, and see that 9 will to into 18 two (2) times. That |
term therefore becomes 2· 1 = 2.
Here is the cleared equation, followed by its solution:
| 9x − 15x | = | 2 | |
| −6x | = | 2 | |
| x | = | 2 −6 |
|
| x | = | − | 1 3 |
In the following problems, clear of fractions and solve for x:
To see each answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
| Problem 1. | x 2 |
− | x 5 |
= | 3 |
| The LCM is 10. Here is the cleared equation and its solution: | |||||
| 5x | − | 2x | = | 30 | |
| 3x | = | 30 | |||
| x | = | 10 | |||
| Problem 2. | x 6 |
= | 1 12 |
+ | x 8 |
| The LCM is 24. Here is the cleared equation and its solution: | |||||
| 4x | = | 2 + 3x | |||
| 4x − 3x | = | 2 | |||
| x | = | 2 | |||
| Problem 3. | x − 2 5 |
+ | x 3 |
= | x 2 |
| The LCM is 30. Here is the cleared equation and its solution: | |||||
| 6(x − 2) + 10x | = | 15x | |||
| 6x − 12 + 10x | = | 15x | |||
| 16x − 15x | = | 12 | |||
| x | = | 12 | |||
| Problem 4. | x − 1 4 |
= | x 7 |
| The LCM is 28. Here is the cleared equation and its solution: | |||
| 7(x − 1) | = | 4x | |
| 7x − 7 | = | 4x | |
| 7x − 4x | = | 7 | |
| 3x | = | 7 | |
| x | = |
7 3 |
|
We see, as in this Problem, that when a single fraction is equal to another single fraction, then the equation can be cleared by "cross-multiplying."
| If | ||||
| a b |
= | c d |
, | |
| then | ||||
| ad | = | bc. | ||
| Problem 5. | x − 3 3 |
= | x − 5 2 |
| Here is the cleared equation and its solution: | |||
| 2(x − 3) | = | 3(x − 5) | |
| 2x − 6 | = | 3x − 15 | |
| 2x − 3x | = | − 15 + 6 | |
| −x | = | −9 | |
| x | = | 9 | |
| Problem 6. | x − 3 x − 1 |
= | x + 1 x + 2 |
||
| Here is the cleared equation and its solution: | |||||
| (x − 3)(x + 2) | = | (x − 1)(x + 1) | |||
| x² −x − 6 | = | x² − 1 | |||
| −x | = | −1 + 6 | |||
| −x | = | 5 | |||
| x | = | −5 | |||
| Problem 7. | 2x − 3 9 |
+ | x + 1 2 |
= | x − 4 |
| The LCM is 18. Here is the cleared equation and its solution: | |||||
| 4x − 6 + 9x + 9 | = | 18x − 72 | |||
| 13x + 3 | = | 18x − 72 | |||
| 13x − 18x | = | − 72 − 3 | |||
| −5x | = | −75 | |||
| x | = | 15 | |||
| Problem 8. | 2 x |
− | 3 8x |
= | 1 4 |
| The LCM is 8x. Here is the cleared equation and its solution: | |||||
| 16 − 3 | = | 2x | |||
| 2x | = | 13 | |||
| x | = |
13 2 |
|||
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