## Fractional Algebriac Equation help!!

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Lotrnerd
Posts: 1
Joined: Sun Sep 16, 2012 9:06 pm
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### Fractional Algebriac Equation help!!

Hi, when I try to solve the equation below, I come up with 36, as seen below. I multiply each term by two in order to get rid of the fractions. The correct answer is 72. What am I doing wrong?

a+b+c+d = 180

b = a/2
c=b/2
d=3c

a + a/2 + (a/2)/2 + 3[(a/2)/2] = 180
which becomes...
a + a/2 + (a/2 * 1/2) + 3((a/2) * (1/2)) = 180
which becomes...
2a + a + a + 6a = 360
which becomes...
10a=360
which becomes...
a=36

Matt
Posts: 10
Joined: Mon May 18, 2009 3:30 pm
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### Re: Fractional Algebriac Equation help!!

Hi Lotrnerd,

Minor mistake here:

a + a/2 + (a/2)/2 + 3[(a/2)/2] = 180
a + a/2 + (a/2 * 1/2) + 3((a/2) * (1/2)) = 180
a + a/2 + a/4 + 3a/4 = 180

Here you have to multiply by 4 to clear the fractions:

4a + 2a + a + 3a = 720
10a = 720
a = 72