Review question

# Can we match this equation to its graph? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7757

## Solution

A sketch of the curve with equation $x^2y^2(x+y)=1$ is drawn in which of the following diagrams?

Firstly, we can rule out options (a) and (b) because points which satisfy $x^2y^2(x+y)=1$ cannot have $x=0$ or $y=0$. Since graphs (a) and (b) both cross the axes, these cannot give the correct solution.

Now let’s rearrange the equation of the curve. We have $x+y=\frac{1}{x^2y^2}>0$ so points on the curve must lie only in regions where $x+y>0$, or $y>-x$. This rules out option (d).

Hence, by process of elimination, (c) is the correct answer.

Alternatively, can we have $x=y\:$? We can, but only where $x=y=2^{-1/5}>0$. Graphs (a) and (b) show no such point, while graph (d) seems to show $4$ of them. Hence the answer must be (c).