Solve the equations: The sides of an equilateral triangle are $(x+3 y) cm ,(3 x+2 y-2) cm$ and $\left(4 x+\frac{y}{2}+1\right)$ cm . Find the length of each side. [Hint. $x+3 y=3 x+2 y-2=4 x+\frac{y}{2}+1$ The solution gives x = 3 and y = 4. Each side of the triangle = (x + 3y) cm = (3 + 12) cm = 15 сm.]
Solve the equations: $\frac{3}{x+y}+\frac{2}{x-y}=3, \frac{2}{x+y}+\frac{3}{x-y}=\frac{11}{3} \quad\left[\right.$ Hint. Put $\frac{1}{x+y}=u$ and $\left.\frac{1}{x-y}=v\right]$