Review question

# Can we solve these two equations involving $x,y,1/x$ and $1/y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6694

## Solution

Solve the simultaneous equations $x+ \frac{1}{y} =2, \qquad 2y + \frac{1}{x}=3.$

We can obviously not have $x=0$ or $y=0$. Rearranging the first equation gives: $x=2-\frac{1}{y}=\frac{2y-1}{y}$ so that $\frac{1}{x}=\frac{y}{2y-1}.$

We can then substitute into the second equation to get: $2y+\frac{y}{2y-1}=3.$

Multiplying by $2y-1$ gives: $2y(2y-1)+y=3(2y-1)$ so expanding the brackets and rearranging, we get $4y^2-7y+3=0.$

This can now be factorised, giving: $(4y-3)(y-1)=0$ so $y=1$ or $y=\dfrac{3}{4}$.

Since $x=2-\dfrac{1}{y}$, we can substitute these $y$ values in to this to get the solutions:

$x=1,\ y=1\qquad\text{or}\qquad x=\frac{2}{3},\ y=\frac{3}{4}.$

Alternatively, we could have begun by rearranging the first equation to get $\frac{1}{y}=2-x$ so that $y=\frac{1}{2-x}$ and then substituted this into the second equation.

We could also have rearranged the second equation and then substituted this into the first one.