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A good example of this is the simple equation 3\({y}\) = 12. If we want to find out the value of \({y}\), we must divide both sides of the equation by 3. We know that this is the operation when we have three lots of \({y}\) on one side of the equation, and we want to find the value of one lot of \({y}\).

Dividing both sides gives \({y}\) = 4. This is the solution to the equation.

Similarly if we have 5\({z}\) = 30, we divide both sides of the equation by 5 to give \({z}\) = 6.

If we had 8\({d}\) = 20, we would divide both sides of the equation by 8 to give \({d}\) = 2.5.

If we had 9\({s}\) = 108 we would divide both sides by 9 to give \({s}\) = 12.

We can also use this method to solve equations like \(\frac{j}{4}=~12\). This time we only have one quarter of \({j}\), as we want to just have "\({j}=\)" we must multiply both sides of the equation by 4. This would allow us to obtain the result \({j}~=~48\)

If we had \(\frac{k}{3}={7}\) we multiply both sides by 3 to obtain \({k}\) = 21.

If we had \(\frac{z}{8}={3.5}\) we multiply both sides by 8 to obtain \({z}\) = 28.

If we had \(\frac{b}{2.5}={10}\) we multiply both sides by 2.5 to obtain \({b}\) = 25.

What if we had, for example, \({-z}~=~{2}\)? We first have to realise the "\({-z}\)" is actually (–1) × \({z}\), so we would need to divide both sides of the equation by –1 leaving \({z}\) = –2.

Similarly if we had –3\({P}\) = –6 we would divide both sides of the equation by –3 leaving \({P}\) = 2.

If we had \(\frac{–R}{6}={3.2}\) we would need to multiply by –6 giving \({R}\) = –19.2.