easy simultaneous equation, please solve

Question

how would you solve the following simultaneous equation

$$\dfrac{1}{x}+\dfrac{1}{y}=3$$

$$\dfrac{1}{x}-\dfrac{1}{y}=1$$

I have tried multiplying out the fraction so I multiplied both equation by $xy$ but that still gives two unknowns.

Any help would be much appreciated.

Answer 1

$$1/x + 1/y = 3$$

$$1/x - 1/y = 1$$ if we add we get $$2/x=4\Rightarrow x=1/2$$ if we subtract we get $$2/y=2\Rightarrow y=1$$

Answer 2

Hint

If we write $a=1/x,b=1/y$ you have the system $$\begin{cases}a+b&=3\\a-b&=1\end{cases}$$ Solve for $a,b$ and get $x=1/a,y=1/b.$

Answer 3

$$\dfrac 1x+\dfrac 1y=3\iff y+x=3xy\\\dfrac 1x-\dfrac 1y=1\iff y-x=xy$$ From this you have $$2y=4xy\\2x=2xy$$ Thus $$(x,y)=(\dfrac 12,1)$$