### Video Transcript

Given that π΄π΅ equals two π₯ plus
one centimeters, π΅πΆ equals π₯ plus three centimeters, π·πΈ equals three π₯ minus
one centimeters, πΈπΉ equals π¦ centimeters, πΊπ» equals 11 centimeters, and π»πΌ
equals 10 centimeters, find the values of π₯ and π¦.

Itβs always a good idea to start a
question like this involving a diagram by filling in any length information that
isnβt already on the diagram. We can then see more clearly that
we have six given lengths. Three of these involve the unknown
value of π₯ and one of them involves the unknown value of π¦.

So what else should we observe
about this figure? Well, we can see from the line
marking here that we have three parallel lines: line π΄πΊ, π΅π», and πΆπΌ. The other three lines on this
diagram are all transversals. Remember that a transversal is a
line that intersects two or more lines in the same plane at distinct points. The fact that we have these three
parallel lines and transversals means that we can apply the basic proportionality
theorem or Thalesβs theorem. This theorem states that if three
or more parallel lines intersect two transversals, then they cut off the
transversals proportionally.

In order to work out the missing
values using this theorem, itβs worth noticing that the line segments πΊπ» and π»πΌ
both have numerical values for their lengths. If we wanted to work out the value
of π₯ first, then we wouldnβt want to include the line segment πΈπΉ in the
proportionality statement because that involves the unknown of π¦. So letβs see how we can write the
line segment π΄π΅ and π΅πΆ in a proportionality statement with πΊπ» and π»πΌ.

There are a few different
proportion statements that we could write, but one example would be that π΄π΅ over
π΅πΆ is equal to πΊπ» over π»πΌ. We then simply need to fill in the
length information that we have. So we have two π₯ plus one over π₯
plus three is equal to 11 over 10. To begin simplifying this, we can
cross multiply. So we have 10 times two π₯ plus one
equals 11 times π₯ plus three. We now need to perform some
algebraic manipulation. On the left-hand side, we multiply
10 by each of the terms within the parentheses. So we have 20π₯ plus 10. In the same way, on the right-hand
side, we multiply 11 by each of the terms. So we have 11π₯ plus 33.

The next stage is to gather all the
terms involving π₯ on one side of the equation and all the numerical values on the
other side. We can do this in two steps or in
one step by subtracting 11π₯ and 10 from both sides of the equation giving us nine
π₯ is equal to 23. Dividing both sides by nine, we
have π₯ is equal to 23 over nine. We can leave this answer as a
fraction or work it out as a decimal. If we didnβt have a calculator, we
could establish that 23 over nine is equal to two and five-ninths. Five-ninths is the recurring
decimal of 0.5 recurring. And so if we wanted a decimal to
two decimal places, we can say that π₯ is approximately 2.56.

And so we found an answer for the
value of π₯. Letβs see if we can work out the
second part of this question to find the value of π¦. So the unknown value of π¦ occurs
in the line segment of πΈπΉ. And so it might be sensible to use
the line segments πΈπΉ, π·πΈ, πΊπ», and π»πΌ in a proportion statement. We could write that π·πΈ over πΈπΉ
is equal to πΊπ» over π»πΌ. When it comes to filling in the
length of the line segment π·πΈ, weβll need to substitute in the value that we found
for π₯. We can use the fractional value of
23 over nine or two and five-ninths rather than the decimal value. But if we are using the decimal
value, make sure that we use the long recurring decimal.

When we have substituted this value
in, we get that the length of the line segment π·πΈ is 20 over three
centimeters. So once we have filled the length
information into the given equation, we have 20 over three over π¦ is equal to 11
over 10. Cross multiplying, we have that 10
times 20 over three is equal to 11π¦. We can then simplify the left-hand
side. 10 times 20 over three is 200 over
three.

Finally, we divide both sides of
this equation by 11. We can give the value of π¦, either
as a fraction or a decimal. As a fraction, itβs six and two
over 33. As a decimal, we get the answer
that π¦ is equal to 6.06 recurring. Rounding to two decimal places, we
can say that π¦ is approximately 6.06. We have now completed the
question. We found both values π₯ and π¦. As decimal approximations to two
decimal places, we have π₯ equals 2.56 and π¦ equals 6.06.