# Thread: Fractional Algebriac Equation help!!

1. ## 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

2. ## Re: Fractional Algebriac Equation help!!

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

3. ## 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?

4. ## Re: Fractional Algebriac Equation help!!

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 ...