Fraction equations: 1 / z - 1 / 2z - 1 / 5z = 10 / z + 1

[iPwn]

I used to be pretty dang good at math, at least I thought. I waited a year between high school and college and then I took a college algebra course last semester (which is like Algebra II from high school). Now I'm in college algebra 2 which will be my final math class (English major). Anyway it's the second week of classes and we're doing some preliminary stuff and I'm lost already..LOL

Here is the first problem I ran into that I can't figure out how to solve:

1 / z - 1 / 2z - 1 / 5z = 10 / z + 1

I honestly don't know how to approach that. It seems simple but I hate fractions. I would think you would need to find the LCD. I think that's 5z - 5 isn't it? But when I do that then I get:

(5z-5) - (5z-5) - (5z-5) = 50z + 50.

Then I get -10 = 40z + 50, -60= 40z.

The back of the book says the answer is 3 / 97 which obviously I wouldn't get.

Thanks!

 

[soroban]

Hello, iPwn!

\(\displaystyle \L\frac{1}{z}\,-\,\frac{1}{2z}\,-\,\frac{1}{5z} \:= \:\frac{10}{z\,+\,1}\)

Evidently, you've forgotten how to find an LCD.

Factor the denominators: \(\displaystyle \:\begin{array}{cccc}z & = & z \\ 2z & = & 2\cdot z \\ 5z & = & 5\cdot z \\ z\,+\,1 & = & z\,+\,1\end{array}\)

Then "collect" the different factors: \(\displaystyle \:2\cdot5\cdot z\cdot(z\,+\,1) \:=\:10z(z\,+\,1)\;\;\leftarrow\;\) LCD


Multiply through by the LCD and simplify:

\(\displaystyle \L10\not{z}(z+1)\,\cdot\,\frac{1}{\not{z}} \,- \,\not{10}^{^5}\not{z}(z+1)\,\cdot\,\frac{1}{\not{2}\not{z}} \,-\, \not{10}^{^2}\not{z}(z+1)\,\cdot\,\frac{1}{\not{5}\not{z}}\;=\;10z(\sout{z+1})\,\cdot\,\frac{10}{\sout{z+1}}\)

We have: \(\displaystyle \L\:10(z\,+\,1)\,-\,5(z\,+\,1)\,-\,2(z\,+\,1)\;=\;10\cdot10z\)

. . \(\displaystyle \L10z\,+\,10\,-\,5z\,-\,5\,-\,2z\,-\,2\;=\;100z\)

. . \(\displaystyle \L3\;=\;97z\;\;\Rightarrow\;\;\fbox{z\:=\:\frac{3}{97}}\)