Video: Pack 3 β€’ Paper 1 β€’ Question 19

Pack 3 β€’ Paper 1 β€’ Question 19

02:56

Video Transcript

The ratio three 𝑦 to four 𝑦 minus four π‘₯ is equivalent to π‘˜ to two. Show that 𝑦 is equal to two π‘˜π‘₯ all over two π‘˜ minus three.

The question tells us that these two ratios are equivalent. This means that when we divide what’s on the left of each ratio by the expression on the right, we get a pair of equivalent expressions. These must give us the same value.

Note that because these ratios are equivalent, this would also work by performing the division the other way. Four 𝑦 minus four π‘₯ all over three 𝑦 is equal to two over π‘˜.

Let’s use the first equation we formed. We are asked to find 𝑦 in terms of π‘₯. So let’s make 𝑦 the subject. We can start by multiplying both sides of the equation by four 𝑦 minus four π‘₯. Remember, that’s the same as multiplying by four 𝑦 minus four π‘₯ all over one.

To multiply two fractions, we multiply the numerator by the numerator and then multiply the denominator by the denominator. Our equation becomes three 𝑦 is equal to π‘˜ multiplied by four 𝑦 minus four π‘₯ all over two. We can expand these brackets by multiplying each part inside the bracket by the letter π‘˜. This becomes four π‘˜π‘¦ minus four π‘˜π‘₯. And it’s still all over two.

Notice how both the numerator and the denominator have a common factor of two. We can, therefore, simplify this fraction slightly by dividing three by two. This gives us three 𝑦 is equal to two π‘˜π‘¦ minus two π‘˜π‘₯. We are aiming to make 𝑦 the subject of the equation. So we now need to gather all of the 𝑦s on the same side. We can do this by subtracting two π‘˜π‘¦ from both sides of the equation. This gives us three 𝑦 minus two π‘˜π‘¦ is equal to negative two π‘˜π‘₯.

Now this is the tricky part. We factorise the expression on the left-hand side of our equation. Each term has a common factor of 𝑦. So this goes on the outside of the bracket. This gives us 𝑦 multiplied by three minus two π‘˜ is equal to negative two π‘˜π‘₯. The inverse or the opposite of multiplying by three minus two π‘˜ is to divide both sides of the equation by three minus two π‘˜.

Now at this point, we’ve not quite got what we need. However, we can create an equivalent fraction by multiplying both the numerator and the denominator by negative one. This changes the numerator of our fraction to two π‘˜π‘₯. And the denominator becomes positive two π‘˜ minus three. 𝑦 is, therefore, equal to two π‘˜π‘₯ all over two π‘˜ minus three.

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