This is an A* grade extension GCSE maths question on rearranging equations.

The question is asking us to find an equation that would represent x. In the question this x appears on the denominator (bottom half) and numerator (upper half). So the first thing we have to do is multiply out by the denominators to get rid of them. Lets do this one by one.

This gives:

y(3x-1) = 2(1+x)

To simplify the equation, we expand (multiply out) the bracket on the right hand side (RHS). Which gives us:

y (3x - 1) = 2 + 2x

Now we do the same thing to the left hand side (LHS):

3yx - y = 2 + 2x

Now, since we want x by itself, it is better to group x all together. So, we want to have the everything with x on the LHS and everything without an x on the RHS.

To do this, we'll have to add y to both sides and take away 2x on both sides to the above equation.

3yx - y + y - 2x = 2 + 2x + y - 2x

As you can see, the positive and negatives will cancel and so this will give us:

3yx - 2x = y + 2

Excellent. Now we've made everything a lot simpler and makes it easy to answer the question. As you can see, x on the LHS is the common factor. So if we take out the x on the LHS like:

x(3y - 2) = y + 2

we can see that to get x by itself, we will then need to divide both sides by 3y - 2 like so:

x(3y - 2)/(3y - 2) = y + 2/(3y - 2)

As you can see the (3y - 2) on the LHS will cancel and so this will give us:

x = (y + 2)/(3y - 2)

And there you have it!

4x^2+5x+2=10

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