# Fractional Algebriac Equation help!!

• Sep 16th 2012, 01:26 PM
Lotrnerd
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
• Sep 16th 2012, 01:31 PM
skeeter
Re: Fractional Algebriac Equation help!!
Quote:

Originally Posted by Lotrnerd
Hi, when I try to solve the equation below, I come up with 36, as seen below. I multiply each term by to in order to get ride of the fractions. However, I get an answer of 36, but the 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...

a + a/2 + a/4 + 3a/4 = 180

4a + 2a + a + 3a = 720

fixed ... now finish it
• Sep 16th 2012, 01:35 PM
Lotrnerd
Re: Fractional Algebriac Equation help!!
Thanks, so is it a rule that I need to solve within the parentheses, before I decide to multiply each term by a number?
• Sep 16th 2012, 01:38 PM
skeeter
Re: Fractional Algebriac Equation help!!
Quote:

Originally Posted by Lotrnerd
Thanks, so is it a rule that I need to solve within the parentheses, before I decide to multiply each term by a number?

using the order of operations helps ...