Review question

# Can we solve these simultaneous equations that involve reciprocals? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8128

## Solution

Find the values of $x$ and $y$ which satisfy the simultaneous equations $2x + 3y = 1,$ $\frac{2}{x} + \frac{3}{y} = 1.$

Substitution works well as a method here. Elimination won’t work, as subtracting (a multiple of) one equation from the other will not cancel any terms. (This is because the terms in the second equation are not multiples of $x$ and $y$ as in the first equation, but instead multiples of $1/x$ and $1/y$.)

From the first equation, $y = \frac{1}{3} - \frac{2}{3} x.$ Multiplying the second equation by $xy$ gives $2y + 3x = xy,$ and substituting the value of $y$ from above into this gives $\frac{2}{3} - \frac{4}{3} x + 3x =x\left(\frac{1}{3} - \frac{2}{3} x\right).$ Therefore $\frac{2}{3} - \frac{4}{3} x + 3x = \frac{1}{3} x -\frac{2}{3} x^2.$ Rearranging this gives the quadratic $x^2 + 2x + 1 = 0.$ This can be rewritten as $(x + 1)^2 = 0,$ and therefore the only solution is $x = -1$. Substituting this into the rearranged first equation gives $y = \frac{1}{3} - \frac{2}{3} \times (-1) = 1.$

We can check our answers by substituting $x=-1$ and $y=1$ into both equations; the first gives $-2+3=1$ and the second also gives $-2+3=1$, so our solution correctly solves the simultaneous equations.

We might be able to see this solution to the equations quickly by inspection (just by looking at them!).

This doesn’t, however, prove that we have found all possible solutions—we need to work a little harder to do this.

Notice too that we can get one equation from the other by replacing $x$ with $\dfrac{1}{x}$ and $y$ with $\dfrac{1}{y}$. Thus if $(x,y)$ is a solution, so is $\left(\dfrac{1}{x}, \dfrac{1}{y}\right)$.