Fractional Algebriac Equation help!!

[lotrnerd]

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 is 72. What am I doing wrong? Do I need to reduce within the parentheses before I multiple across each term?

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

 

[Deleted member 4993]

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 is 72. What am I doing wrong? Do I need to reduce within the parentheses before I multiple across each term?

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 ............ correct upto here

a + a * 1/2 + a * 1/4 + a * 3/4 = 180

You need to multiply everything by 4 (not 2), to get

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

and continue.....


which becomes...
2a + a + a + 6a = 360
which becomes...
10a=360
which becomes...
a=36

.

 

[JeffM]

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 is 72. What am I doing wrong? Do I need to reduce within the parentheses before I multiple across each term?

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 How in the world do you multiply 2(a/2)(1/2) = (2/2)(a/2) = 1(a/2) = a when it equals a/2?
Similarly, how do you multiply 2 * 3(a/2)(1/2) = 3(2/2)(a/2) = 3 * 1 * (a/2) = 6a when it equals 3a/2?
180= a + (a/2) + (a/2)/2 + 3[(a/2)/2] = a + (a/2) + (a/4) + (3a/4) leads to 720 = 4a + 2a + a + 3a
which becomes...
10a=360 No it becomes 10a = 720
which becomes...
a=36 No it becomes a = 72

You had the right idea, but made a careless error.

 

[soroban]

Hello, lotrnerd!

You could have made it easier . . .

\(\displaystyle a+b+c+d \:=\: 180,\;\;b = \dfrac{a}{2},\;\;c=\dfrac{b}{2},\;\;d=3c\)
\(\displaystyle \text{Solve for }a,b,c,d.\)

\(\displaystyle \text{We have: }\;\begin{Bmatrix} b &=& \frac{a}{2} \\ c &=& \frac{b}{2} &=& \frac{a}{4} \\ d &=& 3c &=& \frac{3a}{4} \end{Bmatrix}\)

\(\displaystyle \text{Substitute: }\;a + \dfrac{a}{2} + \dfrac{a}{4} + \dfrac{3a}{4} \:=\:180\)

\(\displaystyle \text{Multiply by 4: }\;4a + 2a + a + 3a \:=\:720 \quad\Rightarrow\quad 10a \:=\:720\)

\(\displaystyle \text{Therefore: }\:\begin{Bmatrix}a &=& 72 \\ b &=& 36 \\ c &=& 18 \\ d &=& 54 \end{Bmatrix}\)