• Jul 26th 2010, 01:30 PM
Ingersoll
5/z+4 + 3/3z+12

I can figure out that the LCM is z+4. But the answer to the problem is 6/z+4. At some point I have an idea that I multiply the 3 in the second expression by 3 to get 9, and then multiply it by -1 to get -9. Then (or before that) I multiply 5 by 3 to get 15. That would give me the answer but I just can't seem to get there the right way without basically making up my own rules.

Help is appreciated more than words can say. I'm ready to pull my hair out over these problems.
• Jul 26th 2010, 01:34 PM
Quote:

Originally Posted by Ingersoll
5/z+4 + 3/3z+12

I can figure out that the LCM is z+4. But the answer to the problem is 6/z+4. At some point I have an idea that I multiply the 3 in the second expression by 3 to get 9, and then multiply it by -1 to get -9. Then (or before that) I multiply 5 by 3 to get 15. That would give me the answer but I just can't seem to get there the right way without basically making up my own rules.

Help is appreciated more than words can say. I'm ready to pull my hair out over these problems.

Hi Ingersoll,

$\displaystyle \frac{5}{z+4}+\frac{3}{3z+12}=\frac{5}{z+4}+\frac{ 3}{3(z+4)}=\frac{5}{z+4}+\frac{1}{z+4}$
• Jul 26th 2010, 01:59 PM
Ingersoll
Where does the 1 come from? The above seems very simple compared to what my textbook illustrates.
• Jul 26th 2010, 02:32 PM
Quote:

Originally Posted by Ingersoll
Where does the 1 come from? The above seems very simple compared to what my textbook illustrates.

How does your textbook do it?

The way I look at this is....

5 apples + 1 apple = 6 apples.
5 halves + 1 half = 6 halves
5 tenths + 1 tenth = 6 tenths

$\displaystyle 5\ \frac{1}{x}'s+1\ \frac{1}{x}=6\ \frac{1}{x}'s$

$\displaystyle 5\ \frac{1}{z+4}'s+1\ \frac{1}{z+4}'s=6 \frac{1}{z+4}'s$

though you're much likelier to have heard about it in terms of the "common denominator".

$\displaystyle \frac{5}{z+4}+\frac{3}{3(z+4)}=5\left(\frac{1}{z+4 }\right)+\frac{3}{3}\left(\frac{1}{z+4}\right)$

and $\displaystyle \frac{3}{3}=1$

therefore

$\displaystyle 5\left(\frac{1}{z+4}\right)+1\left(\frac{1}{z+4}\r ight)=(5+1)\left(\frac{1}{z+4}\right)=\frac{5+1}{z +4}$
• Jul 26th 2010, 03:06 PM
Ingersoll
Actually, the textbook doen't really cover it. It's only in the exercises secton but there's no comparable example. They're getting rid of this particular text at the end of this semester. Anyway, I think I've got the gist of it from what you've shown me. I'm still really shaky on these things though. Every problem seems different from the last.
• Jul 26th 2010, 03:52 PM