Make x the subject of p(x - q) = r(x + s) Step by Step

Step by step

The question

Question: make x the subject of p(x − q) = r(x + s).

p = r

The subject is the letter on its own: we want x = ...

Here x is in two places, so we must bring them together.

1 Open the brackets

What is in front of a bracket multiplies every part inside.

p times every part of (x − q).

r times every part of (x + s).

px − pq = rx + rs

2 Get the x terms on one side

rx is on the right.

To remove it, we take away rx.

We do it on both sides, so they stay equal.

px − pq − rx = rx + rs ~~− rx~~

Careful!

A term that moves to the other side changes its sign.

rx becomes −rx.

Careful: not px − pq + rx = rs ✗

rx and −rx make zero.

Cross them out.

Now we have px − pq − rx = rs.

px − pq − rx = rs

−pq has no x.

To remove it from the left, we add pq.

We do it on both sides, so they stay equal.

px ~~− pq~~ − rx ~~+ pq~~ = rs + pq

−pq and pq make zero.

Cross them out.

Now both x terms are on the left: px − rx = rs + pq.

px − rx = rs + pq

3 Take x out

x is in every part.

So we take it out once, and write what is left in a bracket.

ab + ac = a(b + c)

So x times (p − r) is rs + pq.

x(p − r) = rs + pq

4 Remove the (p − r)

(p − r)x means (p − r) times x.

We want just x.

To remove times (p − r), we divide by (p − r).

We do it on both sides.

~~(p − r)~~x~~(p − r)~~ = rs + pq(p − r)

(p − r) divided by (p − r) is 1.

So just x is left.

So x = (rs + pq) over (p − r).

x = rs + pq(p − r)

5 Check

Let's check with numbers.

We pick p = 3, q = 4, r = 2, s = 5.

Our answer gives x = 22.

x = (2)(5) + (3)(4)3 − 2 = 10 + 121 = 221 = 22

Now we put all the numbers into the start.

First the left side.

Left: (3)(22) − (2)(22) = 66 − 44 = 22

Now the right side.

Right: (2)(5) + (3)(4) = 10 + 12 = 22

Both sides are 22.

So our answer is right!

Both sides equal 22

Let's look back.

First: open the brackets.

Then: get the x terms on one side.

Then: take x out.

Then: remove the (p − r).

Last: check.

Answer