Video: GCSE Mathematics Foundation Tier Pack 5 β€’ Paper 2 β€’ Question 18

GCSE Mathematics Foundation Tier Pack 5 β€’ Paper 2 β€’ Question 18

02:32

Video Transcript

Consider the equation 𝑦 equals seven π‘₯ minus five.

Part a) Find the value of 𝑦 when π‘₯ equals four.

So we’ve been given an equation for 𝑦 in terms of π‘₯ and asked to work out the value of 𝑦 when π‘₯ is four. To do this, we need to substitute the value of four where π‘₯ appears in this equation. So when π‘₯ is equal to four, we have 𝑦 is equal to seven π‘₯, which is now seven multiplied by four, minus five. Seven multiplied by four is 28, and 28 minus five is 23. The value of 𝑦 when π‘₯ equals four is 23.

Part b) Make π‘₯ the subject of the equation.

To make π‘₯ the subject of the equation means we need to write the equation in the form π‘₯ equals something. You’ll also see this being referred to as rearranging the equation. π‘₯ is currently on the right of this equation. And as we have a positive number of π‘₯s, we’ll keep them on the right. We need to perform operations to both sides of this equation so that we’re just left with π‘₯ on its own on the right.

Now at the moment, we have a negative five on the right of the equation. So the first step to leaving π‘₯ on its own is going to be to add five. But whatever we do to one side of the equation we must make sure we do to the other in order to keep the equation balanced. So we’re adding five to both sides.

On the left, we now have 𝑦 plus five. And on the right, the negative five has canceled out with the positive five, and we’re just left with seven π‘₯. The final step in rearranging this equation to make π‘₯ the subject is to divide both sides of the equation by seven, as we currently have seven π‘₯ and we want just π‘₯ or one π‘₯ left on the right.

On the left of the equation then, we’ll now have 𝑦 plus five all over seven. And on the right, we’re just left with π‘₯, so we’ve completed our rearrangement and made π‘₯ the subject. Our answer then is π‘₯ equals 𝑦 plus five all over seven.

It would also be acceptable to separate this fraction out into the sum of two separate fractions, where they both have a denominator of seven. So π‘₯ equals 𝑦 over seven plus five over seven or five-sevenths would be another acceptable answer.

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