### Video Transcript

The ratio three π¦ to four π¦ minus
four π₯ is equivalent to π to two. Show that π¦ is equal to two ππ₯
all over two π minus three.

The question tells us that these
two ratios are equivalent. This means that when we divide
whatβs 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 π¦ minus four π₯ all over
three π¦ is equal to two over π.

Letβs use the first equation we
formed. We are asked to find π¦ in terms of
π₯. So letβs make π¦ the subject. We can start by multiplying both
sides of the equation by four π¦ minus four π₯. Remember, thatβs the same as
multiplying by four π¦ 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 π¦ is
equal to π multiplied by four π¦ 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 ππ¦ minus four
ππ₯. And itβs 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 π¦ is equal to
two ππ¦ minus two ππ₯. We are aiming to make π¦ the
subject of the equation. So we now need to gather all of the
π¦s on the same side. We can do this by subtracting two
ππ¦ from both sides of the equation. This gives us three π¦ minus two
ππ¦ is equal to negative two ππ₯.

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
π¦. So this goes on the outside of the
bracket. This gives us π¦ multiplied by
three minus two π is equal to negative two ππ₯. 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, weβve 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 ππ₯. And the denominator becomes
positive two π minus three. π¦ is, therefore, equal to two ππ₯
all over two π minus three.