# Video: AQA GCSE Mathematics Higher Tier Pack 2 β’ Paper 3 β’ Question 4

Make π the subject of the equation π  = (3π‘)/(2π). Circle your answer. [A] π = (2π‘)/(3π ) [B] π = (3π‘)/(2π ) [C] π = (2π )/(3π‘) [D] π = (3π )/(2π‘)

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### Video Transcript

Make π the subject of the equation π  is equal to three π‘ over two π. Circle your answer. Is it π is equal to two π‘ over three π , π is equal to three π‘ over two π , π is equal to two π  over three π‘, or π is equal to three π  over two π‘?

It should be fairly obvious from the answers weβve been given that to make π the subject weβre looking for an equation which is π is equal to some other expression. Just like weβre solving equations, to achieve this, weβre going to need to perform a series of inverse, that means opposite, operations.

Letβs start with π  is equal to three π‘ over two π. At the moment, this fraction is making things look a little bit nasty. So weβre going to multiply both sides by two π. Now this isnβt the only first step that we can take, and weβll look at slightly different method in a moment. Multiplying both sides by two π gives us two π multiplied by π  on the left-hand side of this equation. Remember, we try to avoid using the multiplication symbol in algebra, so we just write two ππ . Three π‘ divided by two π and then multiply it by two π is simply three π‘. To get π to be the subject of this equation, we need to divide by two and π  since at the moment π is being multiplied by both two and π .

Two ππ  divided by two π  is simply π, and three π‘ divided by two π  is three π‘ over two π . And we can see that the correct answer is therefore π is equal to three π‘ over two π . Now you might have noticed that we started by multiplying by two π and then dividing by two π . Since multiplying by two and then dividing by two cancel each other out, we could have begun by multiplying both sides by π. Thatβs ππ  is equal to three π‘ over two. Then we would divide both sides by π  to once again give us π is equal to three π‘ over two π . Either method is absolutely fine. The answer is π is equal to three π‘ over two π .