### Video Transcript

π΄π΅, πΆπ·, and πΈπΉ are straight
lines. π΄π΅ is parallel to πΆπ·. Find the value of π¦.

We can see that π¦ has been used in
an expression for the size of one of the angles in this diagram. We were told in the question that
the line π΄π΅ is parallel to the line πΆπ·. So in order to answer this
question, weβre going to need to use some facts about angles in parallel lines. Notice that the expressions weβve
been given for two other angles in this diagram are in terms of a second variable
π₯. So weβre probably going to need to
calculate the value of π₯ before we can calculate the value of π¦. Iβm going to introduce the letters
πΊ and π» onto the diagram to represent the points where the line πΈπΉ crosses π΄π΅
and πΆπ·.

Letβs look first of all at the
angles π΄πΊπ» and πΆπ»πΈ. And we notice that theyβre enclosed
within an πΉ shape. Itβs backwards. But itβs still an πΉ shape. The proper term for angles which
are enclosed within an πΉ shape is corresponding angles. So we can conclude that angle
π΄πΊπ» and πΆπ»πΈ are corresponding angles. Corresponding angles are equal to
each other. So angle π΄πΊπ» is equal to angle
πΆπ»πΈ. This means that we can add the
expression of two π₯ plus 27 degrees to the top part of our diagram.

Now letβs look at angles π΄πΊπΉ and
π΄πΊπ». Thatβs the angle I have marked in
pink and the angle I have marked in orange. These two angles lie on a straight
line. And we know that the sum of angles
on a straight line is 180 degrees. So angle π΄πΊπΉ plus angle π΄πΊπ»
equals 180 degrees. We can form an equation by adding
the expressions for these two angles and setting it equal to 180. We have four π₯ plus three plus two
π₯ plus 27 equals 180.

We can simplify this equation by
grouping like terms. Firstly, four π₯ plus two π₯ is six
π₯ and three plus 27 is 30. We are now in a position to be able
to solve this equation. The first step is to subtract 30
from each side of the equation. On the left-hand side, weβre left
with six π₯ and on the right-hand side, 180 minus 30 is 150.

Next, we need to divide both sides
of this equation by six. On the left-hand side six π₯
divided by six is just π₯ and on the right-hand side, 150 divided by six is 25. You can work this out with a bit of
logic if you remember that four lots of 25 are 100. So six lots of 25 are 150 or you
can use a short division method. There are no sixes is in one. So we carry the one into the next
column. There are two sixes in 15 as two
sixes are 12 with a remainder of three. And there are five sixes in 30 with
no remainder, giving our answer of 25.

So weβve found the value of π₯. But how would this help us with the
question which was to find the value of π¦? There are two ways that we could
now work out the value of π¦. Firstly, we could note that angle
π΄πΊπ» and π΅πΊπ» are on a straight line, which means the sum of these two angles
must be 180 degrees. As we know the value of π₯, we can
work out the size of angle π΄πΊπ» and then form an equation to find the value of
π¦. Or we can use the fact that angles
π΄πΊπΉ and π΅πΊπ» are vertically opposite angles. Theyβre formed by the intersection
of two straight lines. And we know that vertically
opposite angles are equal. Letβs use this method.

Angle π΄πΊπΉ is four π₯ plus three
degrees. And we know the value of π₯ is
25. So we can work out the size of this
angle by substituting 25 for π₯. Four multiplied by 25 is 100 and
adding three gives 103. So angle π΄πΊπΉ is 103 degrees. But as vertically opposite angles
are equal, angle π΅πΊπ» is also equal to 103 degrees. So we can form an equation: five π¦
minus two equals 103.

To solve for π¦, we must first add
two to each side, giving five π¦ equals 105, and then divide both sides of the
equation by five to give π¦ equals 21. You can see that 105 divided by
five is 21, either using a short division method or remembering that five times 20
is 100. So five times 21 will be 105.

We found the value of π¦. π¦ is equal to 21.