1. system!

Hello to everyone,

1) λχ+2y=λ
8y+λy=2λ

and

2) 7 + 4 = 47
--- --- ---
x-y x+y 15

3 - 2 = 3
--- --- ---
x-y x+y 5

they are fractions.
I know the --- are fractions.

Thanks.

2. Hello,

1) λχ+2y=λ
8y+λy=2λ
What are the variables ?

2) 7 + 4 = 47
--- --- ---
x-y x+y 15
For this one...

I notice that 15=5*3
So if x-y=3 and x+y=5, you will see that it works... Perhaps there is another solution.

Else put the fractions on the same denominator and solve..

3. Hello, gdespina!

$2)\;\;\begin{array}{ccc}\dfrac{7}{x-y} + \dfrac{4}{x+y} &=&\cfrac{47}{15} \\ \\ \dfrac{3}{x-y} - \dfrac{2}{x+y} &=& \dfrac{3}{5} \end{array}$

Let: . $\begin{array}{ccc}u &=&x-y \\ v &=& x+y \end{array}\;\;{\color{red}(A)}$

And we have: . $\begin{array}{cccc}\dfrac{7}{u} + \dfrac{4}{v} &=& \dfrac{47}{15} & {\color{blue}[1]} \\ \\ \dfrac{3}{u} - \dfrac{2}{v} &=& \dfrac{3}{5} & {\color{blue}[2]} \end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[2]} by 2:} & \dfrac{6}{u} - \dfrac{4}{v} &=&\dfrac{6}{5} \\ \\ \text{Add {\color{blue}[1]}:} & \dfrac{7}{u} + \dfrac{4}{v} &=&\dfrac{47}{15} \end{array}$

And we have: . $\frac{13}{u} \:=\:\frac{13}{3}\quad\Rightarrow\quad\boxed{u \:=\:3}$

Substitute into [1]: . $\frac{7}{3} + \frac{4}{v} \:=\:\frac{47}{15} \quad\Rightarrow\quad \frac{4}{v}\:=\:\frac{4}{5}\quad\Rightarrow\quad\b oxed{v \:=\:5}$

Substitute into (A): . $\begin{array}{ccc}x - y &=& 3 \\ x + y &=& 5 \end{array}$
. . . . . . . . . . Add: . . . . $2x \:=\:8 \qquad\Rightarrow\qquad \boxed{\begin{array}{ccc} x &=&4 \\ y&=&1 \end{array}}$