What is the gradient of 2y = 10x?

It is 5, not 10. The gradient is only the number in front of x once the equation is in the form y = mx + c, and 2y = 10x is not — y is multiplied by 2. Divide both sides by 2 to get y = 5x, and now the coefficient of x is 5, so the gradient is 5. Reading 10 straight off the original equation reports the gradient of 2y, not of y.

Is the gradient of y = 10x equal to 10?

Yes. y = 10x is already in the form y = mx + c, with m = 10 and c = 0, so the gradient genuinely is 10. The fix is not to halve every coefficient — it is to rearrange to y = mx + c only when the equation is not already in that form. When y stands alone on the left, the number in front of x is the gradient as written.

Why do you have to rearrange to y = mx + c before reading the gradient?

Because the rule that the gradient is the coefficient of x only holds when y is by itself, equal to mx + c. If y is multiplied or divided by a number, the coefficient of x belongs to that multiple of y, not to y. For 2y = 10x the 10 describes how 2y changes with x; dividing by 2 to make it y = 5x reveals that y itself rises by 5 for each step in x, so the gradient is 5.

GCSE Maths Foundation · AQA · Linear graphs

Gradient of a straight line: rearrange to y = mx + c first, so 2y = 10x has gradient 5

Asked for the gradient of $2y = 10x$ , students read the number in front of x straight off the equation and answer 10. But that rule — gradient is the coefficient of x — only works once the equation is in the form $y = mx + c$ , and here y is multiplied by 2. Rearrange first: dividing by 2 gives $y = 5x$ , so the gradient is 5.

The thirty-second fix: get y on its own as y = mx + c, then the number in front of x is the gradient. Do not blindly halve every equation — $y = 10x$ is already in that form, so its gradient really is 10. Rearrange only when y is not yet alone.

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How to spot it in your own work

You read the coefficient of x straight off an equation that was not in $y = mx + c$ form, e.g. calling $2y = 10x$ gradient 10 instead of 5.
You forgot to divide through so that y stands alone, leaving y multiplied (or divided) by a number before reading off m.
You over-corrected and halved an equation that was already $y = mx + c$ , calling $y = 10x$ gradient 5 when it is genuinely 10.
You never rearranged to confirm the form, so you could not tell whether the number in front of x belonged to y or to a multiple of y.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (non-calculator paper, find the gradient):

A straight line has equation

$2y = 10x$

Write down the gradient of the line.

The misconception is to spot the 10 in front of x and write it down as the gradient. But $2y = 10x$ is not yet $y = mx + c$ — the y has a 2 attached — so the 10 describes $2y$ , not y.

Divide both sides by 2 to get y on its own: $y = 5x$ . Now the coefficient of x is 5, so the gradient is 5.

Why students fall for this

“The gradient is the number in front of x” is a true rule, but it comes with a condition that is easy to drop: the equation must read $y = mx + c$ , with y by itself. When students meet $2y = 10x$ they apply the rule to whatever is written, ignoring that the 10 sits against $2y$ , not y. The 10 tells you how $2y$ changes with x — and since $2y$ rises twice as fast as y, the real gradient of y is half of that.

Rearranging settles it. Divide every term by 2: $2y = 10x$ becomes $y = 5x$ . Now y stands alone, the coefficient of x is 5, and y rises by 5 for each step of 1 in x. The line passes through the origin and climbs steeply but not as steeply as a gradient of 10 would.

The trap cuts both ways, so a second discipline matters: do not assume every equation needs halving. $y = 10x$ is already in $y = mx + c$ form, so its gradient is genuinely 10. Rearrange when y is not yet alone; leave it when it already is.

The fix: Rearrange to y = mx + c, then the coefficient of x is the gradient

Check the form. The gradient is the number in front of x only when the equation reads $y = mx + c$ , with y by itself. $2y = 10x$ is not in that form.

Get y on its own. Divide every term by the number attached to y. Here divide by 2: $2y = 10x$ becomes $y = 5x$ .

Read off the gradient. Now the coefficient of x is 5, so the gradient is $5$ — not the 10 you would have read before rearranging.

Don't over-correct. If y is already alone, as in $y = 10x$ , there is nothing to rearrange — the gradient is 10. Halve only when you actually divided y by a number.

Worked example

Find the gradient of $2y = 10x$ . The trap is to read 10 straight off; the fix is to rearrange to $y = mx + c$ first.

Spot that y is not alone. The left side is $2y$ , not y, so the equation is not yet in $y = mx + c$ form and the 10 cannot be read off as the gradient.
Divide every term by 2.
$\frac{2y}{2} = \frac{10x}{2} \quad\Longrightarrow\quad y = 5x$
Read off the coefficient of x. Now y stands alone, so the gradient is the number in front of x.
$\text{gradient} = 5$
Sanity-check against over-correcting. Had the equation been $y = 10x$ — already in $y = mx + c$ form — the gradient would genuinely be 10. You halve only because you divided y by 2 here.

So the gradient of $2y = 10x$ is $5$ , found by rearranging to $y = 5x$ — not the 10 that comes from reading the coefficient before y is on its own.

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Common questions

What is the gradient of $2y = 10x$ ?: It is $5$ , not 10. The gradient is only the number in front of x once the equation is in the form $y = mx + c$ , and $2y = 10x$ is not — y is multiplied by 2. Divide both sides by 2 to get $y = 5x$ , and now the coefficient of x is 5, so the gradient is 5. Reading 10 straight off the original equation reports the gradient of $2y$ , not of y.
Is the gradient of $y = 10x$ equal to 10?: Yes. $y = 10x$ is already in the form $y = mx + c$ , with $m = 10$ and $c = 0$ , so the gradient genuinely is 10. The fix is not to halve every coefficient — it is to rearrange to $y = mx + c$ only when the equation is not already in that form. When y stands alone on the left, the number in front of x is the gradient as written.
Why do you have to rearrange to $y = mx + c$ before reading the gradient?: Because the rule that the gradient is the coefficient of x only holds when y is by itself, equal to $mx + c$ . If y is multiplied or divided by a number, the coefficient of x belongs to that multiple of y, not to y. For $2y = 10x$ the 10 describes how $2y$ changes with x; dividing by 2 to make it $y = 5x$ reveals that y itself rises by 5 for each step in x, so the gradient is 5.

Related misconceptions

Reading and finding interceptsThe neighbouring linear graphs skill: the y-intercept is where x = 0 and the x-intercept is where y = 0, so y = 3x + 8 crosses the y-axis at (0, 8), not (8, 0).

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