# Thread: all sorts of algebra questions

1. ## all sorts of algebra questions

1. simplify

2. If ,what does equal to?

3.

2. Originally Posted by masterwizard
1. simplify
$\frac{m}{m - n} + \frac{n}{n - m} = \frac{m}{m - n} - \frac{n}{m - n} = \frac{m - n}{m - n} = 1$

2. If ,what does equal to?
$a + b = 7$ and $a - b = 4$
$a = b + 4$ Set this value into the first equation.
$b + 4 + b = 7$
$b = 1,5$

Therefore $a = 5,5$

Now you'll be able to do $\frac{a^2}{b^2}$

3.
$\frac{1}{ab + b^2} + \frac{1}{ab + a^2} = \frac{1}{b(a + b)} + \frac{1}{a(a + b)} = \frac{a + b}{ab(a + b)} = \frac{1}{ab}$

3. Hello, masterwizard!

#2 could be a dangerous problem . . .

If . $\frac{a+b}{a-b} = \frac{7}{4}$, what does $\frac{a^2}{b^2}$ equal?

We should not assume that the fraction is already reduced to lowest terms.

If . $\frac{a+b}{a-b}$ has been reduced to $\frac{7}{4}$,

. . then we have: . $\begin{array}{c}a + b \:=\:7k \\ a - b \:=\:4k\end{array}$ . . . for some integer $k$.

Hence: . $\begin{array}{c}a \:=\:\frac{11}{2}k \\ b\:=\:\frac{3}{2}k\end{array}$

Therefore: . $\frac{a^2}{b^2}\;=\;\frac{\left(\frac{11}{2}k\righ t)^2}{\left(\frac{3}{2}k\right)^2} \;=\;\frac{\frac{121}{4}k^2}{\frac{9}{4}k^2} \;=\;\frac{121}{9}$

We arrive at the same answer, but if we are required to show our work,
. . janvdl's solution could lose some points.

4. Well, I don't understand this part.

5. Originally Posted by masterwizard
Well, I don't understand this part.
Look at what we are given:
$a + b = 7k$

$a - b = 4k$
Add those 2 equations together and you'll get:

$a + a + b - b = 7k + 4k$

$2a = 11k$

$a = \frac{11k}{2}$

Subtract those 2 equations from each other and you'll get:

$a - a + b + b = 7k - 4k$

$2b = 3k$

$b = \frac{3k}{2}$