Expert: Scott A Wilson - 3/21/2010

Question
Please help me. I have been out of school for 10 year and am now in college algebra.  I don't have a clue at how to even start these two problems. The problems are as follows
a/5=a/2-3/2                x/5-3/5+x/2-2/2=1/2

Any help would be appreciated. Thank you ever so much

Answer
The first thing that I'd do is get rid of the fractions.
On the 1st problem, there is a 2 and a 5 in the denominator, so use 10.
That is, 10(a/5) = 10(a/2) - 10(3/2) => 2a = 5a - 15.

Subtracting 5a from both sides and then dividing by -3 would give us the answer.
That is, 2a-5a = 5a-5a - 15 => -3a = -15.

The only thing left to do is to divide by -3, and that gives a=5.

For the 2nd problem, there is a 5 and a 2 in the denominators, so use 10 again.
That is, 10(x/5) - 10(3/5) + 10(x/2) - 10(2/2) = 10(1/2).
This reduces to 2x - 6 + 5x - 10 = 5.

The 2x and 5x combine, as well as the -6 and -10.  This gives 7x - 16 = 5.
Once this has been done, add 16 to both sides and get 7x = 21.

Divide both sides by 7 and get x = 21/7 = 3.