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

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

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?

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 ...