Equations with fractions: 2nd Level
Back to Section 1
Let us now compare equations with fractions with adding fractions.
To add these fractions --
-- their LCM is abc.
2 a |
+ |
3 b |
+ |
4 c |
= |
2bc + 3ac + 4ab abc |
We have multiplied each numerator -- 2, 3, 4 -- by those factors of abc that are missing from its denominator.
Now, when we clear this equation,
2 a |
+ |
3 b |
= |
4 c |
|
2bc + 3ac |
= |
4ab
|
each term is the same as the numerator in the addition! Each numerator -- 2, 3, 4 -- is multiplied by those factors of the LCM that are missing from its denominator.
The terms of the cleared equation are the same as the numerators of the added fractions.
Example 1. Solve for x:
1 2x |
+ |
1 x − 1 |
= |
1 2(x − 1) |
|
Solution. The LCM is 2x(x −1). |
|
Here is the cleared equation: |
|
x − 1 + 2x |
= |
x |
|
Each numerator has been multiplied by those factors of the LCM that are missing from its denominator. The solution follows:
|
|
2x |
= |
1 |
|
x |
= |
1 2 |
Problem 9. Solve for x:
_9_ 3x − 5 |
+ |
_1_ x + 2 |
= |
4 x − 2 |
|
The LCM is
the product of the three denominators. Here is the cleared equation and its solution: |
|
9(x + 2)(x − 2) + (3x − 5)(x − 2) |
= |
4(3x − 5)(x + 2) |
|
9(x² − 4) + 3x² − 11x + 10 |
= |
4(3x² + x − 10) |
|
9x² − 36 + 3x² − 11x + 10 |
= |
12x² + 4x − 40 |
|
12x² + − 11x − 26 |
= |
12x² + 4x − 40 |
|
−11x − 4x |
= |
−40 + 26 |
|
−15x |
= |
−14 |
|
x |
= |
14 15 |
Problem 10. Solve for x:
1 x |
+ |
1 x − 1 |
= |
1 8x |
+ |
_1_ 8(x − 1) |
|
The LCM is
8x(x − 1). Here is the cleared equation and its solution: |
|
8(x − 1) + 8x |
= |
x − 1 + x |
|
8x − 8 + 8x |
= |
2x − 1 |
|
16x − 2x |
= |
−1 + 8 |
|
14x |
= |
7 |
|
x |
= |
1 2 |
Problem 11. Factor the denominators, clear of fractions, and solve for x:
_1_ x² − 2x |
− |
_8_ 3x² − 5x − 2 |
= |
_4_ 3x² + x
|
|
_1_ x(x − 2) |
− |
_8_ (3x + 1)(x − 2) |
= |
_4_ x(3x + 1)
|
The LCM is
x(x − 2)(3x + 1). Here is the cleared equation and its solution:
3x + 1 |
− 8x |
= |
4(x − 2) |
|
1 − 5x |
= |
4x − 8 |
|
−5x − 4x |
= |
−8 − 1 |
|
−9x |
= |
−9 |
|
x |
= |
1 |
Problem 12. Factor the denominators, clear of fractions, and solve for x:
x + 6 x² − 9 |
+ |
x − 9 x² − 4x + 3 |
= |
_2x − 1_ x² + 2 x − 3
|
|
__x + 6__ (x + 3)(x − 3) |
+ |
x − 9 (x − 1)(x − 3) |
= |
_2x − 1_ (x + 3)(x − 1)
|
The LCM is
(x + 3)(x − 3)(x − 1). Here is the cleared equation and its solution:
(x + 6)(x − 1) |
+ (x − 9)(x + 3) |
= |
(2x − 1)(x − 3) |
|
x² + 5x − 6 + x² − 6x − 27 |
= |
2x² − 7x + 3 |
|
2x² − x − 33 |
= |
2x² − 7x + 3 |
|
−x + 7x |
= |
3 + 33 |
|
6x |
= |
36 |
x |
= |
6 |
This is a simple fractional equation, which we saw in Lesson 9. It is
not necessary to clear of fractions. |
a b |
simply goes to the other side as its |
reciprocal.
Whatever multiplies on one side will divide on the other. And whatever divides on one side will multiply on the other.
We would like x to be in the numerator on the left. Imagine placing it there.
Then |
pq r |
stays on the right. And |
2s 3t |
goes to join them as its |
reciprocal!
Problem 13. Solve for x:
ab cd |
= |
mx npq |
|
mx npq |
= |
ab cd |
Exchange sides. |
|
x |
= |
npqab mcd |
Problem 14. Solve for x:
ab c |
= |
_st_ u(v + w)x |
|
x |
= |
__cst__ abu(v + w) |
In each of the following, solve for x.
Problem 15. |
A |
= |
½Bx |
|
|
2A |
= |
Bx |
|
|
x |
= |
2A B |
Problem 15. |
s |
= |
½(x + w)t |
|
|
(x + w)t |
= |
2s |
|
|
xt + wt |
= |
2s |
|
|
xt |
= |
2s − wt |
|
|
x |
= |
2s − wt t |
Problem 16. |
s |
= |
s − x at |
|
|
sat |
= |
s − x |
|
|
x |
= |
s − sat |
Problem 17. |
A |
= |
B( |
2x x − 2 |
) |
|
|
A(x − 2) |
= |
2Bx |
|
|
Ax − 2A |
= |
2Bx |
|
|
Ax − 2Bx |
= |
2A |
|
|
x(A − 2B) |
= |
2A |
|
|
x |
= |
2A A − 2B |
Example 4. Solving for the reciprocal.
While we could solve this by clearing of fractions, there is the more
  elegant method of solving for |
1 x |
, and then taking reciprocals. |
We have
|
1 3 |
+ |
1 x |
= |
1 2 |
|
This implies |
|
1 x |
= |
1 2 |
− |
1 3 |
|
|
= |
3 − 2 6 |
|
1 x |
= |
1 6 |
|
Therefore, on taking reciprocals, |
|
x |
= |
6 |
For, if two numbers are equal, then their reciprocals are also equal.(Except, if the number is 0.)
In each of the following, solve for x by first solving for its reciprocal.
Problem 18. |
1 r |
+ |
1 p |
= |
1 x |
|
|
p + r pr |
= |
1 x |
|
|
x |
= |
pr p + r |
Problem 19. |
1 a |
= |
1 x |
+ |
1 b |
|
|
1 a |
− |
1 b |
= |
1 x |
|
|
|
b − a ab |
= |
1 x |
|
|
x |
= |
ab b − a |
Next Lesson: Word problems
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