1. simplify
2. If ,what does equal to?
3.
$\displaystyle \frac{m}{m - n} + \frac{n}{n - m} = \frac{m}{m - n} - \frac{n}{m - n} = \frac{m - n}{m - n} = 1 $
$\displaystyle a + b = 7 $ and $\displaystyle a - b = 4 $2. If ,what does equal to?
$\displaystyle a = b + 4 $ Set this value into the first equation.
$\displaystyle b + 4 + b = 7 $
$\displaystyle b = 1,5 $
Therefore $\displaystyle a = 5,5 $
Now you'll be able to do $\displaystyle \frac{a^2}{b^2} $
$\displaystyle \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} $3.
Hello, masterwizard!
#2 could be a dangerous problem . . .
If .$\displaystyle \frac{a+b}{a-b} = \frac{7}{4}$, what does $\displaystyle \frac{a^2}{b^2}$ equal?
We should not assume that the fraction is already reduced to lowest terms.
If .$\displaystyle \frac{a+b}{a-b}$ has been reduced to $\displaystyle \frac{7}{4}$,
. . then we have: .$\displaystyle \begin{array}{c}a + b \:=\:7k \\ a - b \:=\:4k\end{array}$ . . . for some integer $\displaystyle k$.
Hence: .$\displaystyle \begin{array}{c}a \:=\:\frac{11}{2}k \\ b\:=\:\frac{3}{2}k\end{array}$
Therefore: .$\displaystyle \frac{a^2}{b^2}\;=\;\frac{\left(\frac{11}{2}k\righ t)^2}{\left(\frac{3}{2}k\right)^2} \;=\;\frac{\frac{121}{4}k^2}{\frac{9}{4}k^2} \;=\;\frac{121}{9}$
We arrive at the same answer, but if we are required to show our work,
. . janvdl's solution could lose some points.
Look at what we are given:
Add those 2 equations together and you'll get:$\displaystyle a + b = 7k $
$\displaystyle a - b = 4k $
$\displaystyle a + a + b - b = 7k + 4k $
$\displaystyle 2a = 11k $
$\displaystyle a = \frac{11k}{2} $
Subtract those 2 equations from each other and you'll get:
$\displaystyle a - a + b + b = 7k - 4k $
$\displaystyle 2b = 3k $
$\displaystyle b = \frac{3k}{2} $