SOLUTION: how do you solve for the unknown of x/3 -7/8 = 2/5(x + 15/4) any help would be greatly appreciated.

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Question 551233: how do you solve for the unknown of x/3 -7/8 = 2/5(x + 15/4) any help would be greatly appreciated.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given to solve for x:
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x%2F3+-7%2F8+=+%282%2F5%29%28x+%2B+15%2F4%29
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There are a number of different things we could do to make the problem a little easier. A lot of students have trouble with fractions, so let's begin by getting rid of the fractions. The right side of the equation may cause us a little trouble, so let's begin by doing the distributed multiplication on the right side. Do that by multiplying 2%2F5 times each of the two terms in the parentheses to get:
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x%2F3+-7%2F8+=+%282%2F5%29x+%2B+%282%2F5%29%2A%2815%2F4%29
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When you multiply out the last term on the right side [that is when you multiply %282%2F5%29%2A%2815%2F4%29] you get 30%2F20+ which is 3%2F2. So you can now write the equation as:
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x%2F3+-7%2F8+=+%282%2F5%29x+%2B+3%2F2
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Now to get rid of the fractions, we can multiply every term on both sides by the product of all the denominators. That is multiply every term by 3%2A8%2A5%2A2=240. (We could use 120, but using 240 does not cause a problem).
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When we multiply by 240 the equation becomes:
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%28240x%29%2F3+-%28240%2A7%29%2F8+=+%28%28240%2A2%29%2F5%29x+%2B+%28240%2A3%29%2F2
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Now in each of the four terms, let's divide the denominator into the 240 in the numerator to get:
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80x+-%2830%2A7%29+=+%2848%2A2%29x+%2B%28120%2A3%29
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Multiply out all the terms and you have:
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80x+-210+=+96x+%2B+360
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This is in a form that is more like an ordinary algebraic equation. Start by subtracting 96x from both sides:
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-16x+-+210+=+360
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Next add 210 to both sides:
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-16x+=+570
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Solve for x by dividing both sides by -16 to get:
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x+=+570%2F%28-16%29
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and since both 570 and -16 are divisible by 2 this can be reduced to:
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x+=+-%28285%2F8%29
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Or by dividing it out, you can get the decimal equivalent form:
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x+=+-35.625
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Hope this helps you to understand the problem a little better. Sometimes it helps to make the problem a little easier by getting rid of the fractions, and you can do that by multiplying all terms by a common denominator (the product of all denominators) and then getting rid of each denominator by dividing it into the common denominator.
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