# Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 1 • Question 16

𝐴𝐵, 𝐶𝐷, and 𝐸𝐹 are straight lines. 𝐴𝐵 is parallel to 𝐶𝐷. Find the value of 𝑦.

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### Video Transcript

𝐴𝐵, 𝐶𝐷, and 𝐸𝐹 are straight lines. 𝐴𝐵 is parallel to 𝐶𝐷. Find the value of 𝑦.

We can see that 𝑦 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 𝐴𝐵 is parallel to the line 𝐶𝐷. So in order to answer this question, we’re going to need to use some facts about angles in parallel lines. Notice that the expressions we’ve been given for two other angles in this diagram are in terms of a second variable 𝑥. So we’re probably going to need to calculate the value of 𝑥 before we can calculate the value of 𝑦. I’m going to introduce the letters 𝐺 and 𝐻 onto the diagram to represent the points where the line 𝐸𝐹 crosses 𝐴𝐵 and 𝐶𝐷.

Let’s look first of all at the angles 𝐴𝐺𝐻 and 𝐶𝐻𝐸. And we notice that they’re enclosed within an 𝐹 shape. It’s backwards. But it’s still an 𝐹 shape. The proper term for angles which are enclosed within an 𝐹 shape is corresponding angles. So we can conclude that angle 𝐴𝐺𝐻 and 𝐶𝐻𝐸 are corresponding angles. Corresponding angles are equal to each other. So angle 𝐴𝐺𝐻 is equal to angle 𝐶𝐻𝐸. This means that we can add the expression of two 𝑥 plus 27 degrees to the top part of our diagram.

Now let’s look at angles 𝐴𝐺𝐹 and 𝐴𝐺𝐻. That’s 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 𝐴𝐺𝐹 plus angle 𝐴𝐺𝐻 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 𝑥 plus three plus two 𝑥 plus 27 equals 180.

We can simplify this equation by grouping like terms. Firstly, four 𝑥 plus two 𝑥 is six 𝑥 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, we’re left with six 𝑥 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 𝑥 divided by six is just 𝑥 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 we’ve found the value of 𝑥. But how would this help us with the question which was to find the value of 𝑦? There are two ways that we could now work out the value of 𝑦. Firstly, we could note that angle 𝐴𝐺𝐻 and 𝐵𝐺𝐻 are on a straight line, which means the sum of these two angles must be 180 degrees. As we know the value of 𝑥, we can work out the size of angle 𝐴𝐺𝐻 and then form an equation to find the value of 𝑦. Or we can use the fact that angles 𝐴𝐺𝐹 and 𝐵𝐺𝐻 are vertically opposite angles. They’re formed by the intersection of two straight lines. And we know that vertically opposite angles are equal. Let’s use this method.

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

To solve for 𝑦, we must first add two to each side, giving five 𝑦 equals 105, and then divide both sides of the equation by five to give 𝑦 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 𝑦. 𝑦 is equal to 21.