SOLUTION: Solve the equation: (7/(2x+1))-(8x/(2x-1))=-4 I'm really bad at the equations with fractions! :)

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Question 95951This question is from textbook Precalculus with limits
: Solve the equation: (7/(2x+1))-(8x/(2x-1))=-4
I'm really bad at the equations with fractions! :)
This question is from textbook Precalculus with limits

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given to solve:
.
%287%2F%282x%2B1%29%29-%288x%2F%282x-1%29%29=-4
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If you don't like fractions, in this case you can get rid of the fractions. How? Begin by noting
that %282x%2B1%29%282x-1%29 is a common denominator. Suppose that we multiply both sides of this
equation (all terms) by %282x%2B1%29%282x-1%29. In doing that multiplication you make the original
problem become:
.

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Next, where it applies you can cancel the denominator with the like term in the numerator as
follows:
.

.
After these cancellations you are left with:
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%28%282x+-1%29%2A7%29+-+%28%282x%2B1%29%2A8x%29+=+-4%2A%282x%2B1%29%282x-1%29
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Multiplying %282x+-+1%29%2A7 results in 14x+-+7
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Multiplying %282x+%2B+1%29%2A8x results in 16x%5E2+%2B+8x
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and finally,
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Multiplying -4%2A%282x%2B1%29%282x+-1%29 results in -16x%5E2+%2B+4
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Substitute these results into the equation and you have:
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14x+-+7+-+%2816x%5E2+%2B+8x%29+=+-16x%5E2+%2B+4
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On the left side you can remove the parentheses if you change the signs of the terms inside
the parentheses to get:
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14x+-+7+-+16x%5E2+-+8x+=+-16x%5E2+%2B+4
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Notice that you have -16x%5E2 on both sides. Therefore, if you add 16x%5E2 to
both sides these two terms disappear and you are left with:
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14x+-+7+-+8x+=+4
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On the left side you can combine the 14x and the -8x to get 6x to reduce
the equation to:
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6x+-+7+=+4
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Then get rid of the -7 on the left side by adding 7 to both sides to get:
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6x+=+11
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Finally, solve for x by dividing both sides by 6 to get:
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x+=+11%2F6
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This is the answer you are looking for ... x+=+11%2F6
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You can check this answer by substituting 11%2F6 for x in the original problem and
ensuring that the equation will balance.
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Hope this helps you to understand the problem and how you can get rid of the denominators in
the equation so that you are not working with fractions.
.