Review question

# Can we solve the simultaneous equations $x + y + \sqrt{xy} = 39$ and $x^2 + y^2 + xy = 741$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5682

## Solution

By substituting $x + y = S$ and $xy = P^2$, or otherwise, solve the simultaneous equations \begin{align*} x + y + \sqrt{\vphantom{1}xy} &= 39, \\ x^2 + y^2 + xy &= 741. \end{align*}

The expression $\sqrt{\vphantom{1}xy}$ denotes the positive square root of $xy$.

Using the suggested substitution, the first equation becomes $\begin{equation} S+P=39.\label{eq:1} \end{equation}$ Also, $S^2=x^2+y^2+2xy$ so the second equation becomes \begin{align} S^2-P^2&=741 \notag \\ (S-P)(S+P)&=741 \notag \\ S-P&=\frac{741}{39}=19. \label{eq:2} \end{align}

Solving $\eqref{eq:1}$ and $\eqref{eq:2}$ simultaneously, we obtain $S=29$ and $P=10$.

Thus $x + y = 29$ and $xy = 100$.

Combining these we have \begin{align*} y&=29-x \\ x(29-x)&=100\\ x^2-29x+100&=0 \\ (x-25)(x-4)&=0\\ x&=25\text{ or }4 \end{align*}

The original equations are symmetrical in $x$ and $y$ so the values of $x$ and $y$ are $25$ and $4$ in either order.