Some time ago, I was surprised to see the following question in a predicted GCSE paper:

Solve for $x$:

$ \frac{2x}{3x+2} \leq \frac{3}{4x+1}$

Give your answers to two decimal places (3 marks)

## Why surprised?

Surprised because the techniques you need to solve it correctly are Further Maths A-level, rather than GCSE. A similar - if anything, less involved - question cropped up in an FP2 paper in 2013 for seven marks.

It’s not too tough to find where the two expressions are *equal* - in my opinion, that would be a fair top-end GCSE question.

If you multiply both sides by $3x+2$ and $4x+1$, you get $2x(4x+1) = 3(3x+2)$.

Expand and simply, and that’s $8x^2 - 7x - 3 = 0$.

That doesn’t factorise, but can be solved with the formula: $x = \frac{7 \pm \sqrt{145}}{16}$, and the calcu…

### OW!

“$\sqrt{145} \approx 12 + \frac{1}{24}$, so you’re looking at a shade more than $\frac{19}{16}\approx 1.19$ and a shade less than $-\frac{5}{16}\approx -0.32$.”

“Thank you, sensei.”

“Don’t let it happen again.”

So that’s all well and good for the *equation*, but what about the *inequality*? Naively – and what I expect the setters of the paper expected to see – you might say “does the inequality hold *between* the solutions or *outside*?” When $x=0$, it holds, so you’d give the answer $-0.32 \leq x \leq 1.19$.

## Unfortunately, it’s the wrong answer

This answer doesn’t account for the fact that the expressions are both discontinuous.

When $x$ is a smidge smaller than $-\frac{2}{3}$, the left-hand expression is an enormous positive number (and greater than the right-hand expression, so the inequality doesn’t hold). A smidge larger, and it’s a huge negative, and the inequality does.

Similarly, the right-hand expression changes sign at $x=-\frac{1}{4}$.

Instead of two places where the truth-value of the inquality changes, we have four: $x=-\frac{2}{3}$ (from false to true); $x\approx-0.32$ (true to false); $x=-\frac{1}{4}$ (false to true); and $x\approx 1.19$ (true to false).

So, you might write down the answer $-\frac{2}{3} \leq x \leq -0.32$ or $-\frac{1}{4} \leq x \leq 1.19$.

## But that’s also wrong

At $x=-\frac{2}{3}$, the inequality *has* no truth value, because the left-hand expression is undefined. The same is true for $x = -\frac{1}{4}$.

So the *correct* answer to the question as stated is $-\frac{2}{3} \lt x \leq -0.32$ or $-\frac{1}{4} \lt x \leq 1.19$.

I predict confidently that nothing along those lines will appear on a GCSE any time soon, even at the top end.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.