### Video Transcript

If πΉπ½ is equal to six π₯ plus eight π¦, πΉπ is equal to three π₯ plus five π¦, πΊπ» equals 42, and πΊπ equals 24, what values of π₯ and π¦ make parallelogram πΉπΊπ»π½ a rectangle?

So weβve been given a lot of information in the question about different lengths within the diagram. So letβs begin by transferring this information onto the diagram itself. Weβre told that πΉπ½ is equal to six π₯ plus eight π¦. And the opposite side, πΊπ», is equal to 42. Weβre also given some information about parts of the diagonals of this parallelogram, which is that πΉπ is equal to three π₯ plus five π¦ and πΊπ is equal to 24.

Now weβre asked what values of π₯ and π¦ make this parallelogram into a rectangle. So we need to consider what properties a rectangle has that arenβt true of parallelograms in general. Now there are a couple of properties. But seeing as weβve been given the lengths of part of the diagonals of this parallelogram, the ones that are relevant here are these: firstly, that the diagonals of a rectangle are equal in length; secondly, they bisect each other, which means they cut each other in half.

This means then that all four of the line segments going from a vertex of the rectangle to the centre point π are equal in length. Specifically taking the two for which weβve been given expressions or values, we can write down the equation three π₯ plus five π¦ is equal to 24. And so we have an equation connecting the values of π₯ and π¦. However, this is only one equation with two unknowns. And so this isnβt sufficient to be able to solve it.

Letβs think about what other information weβve been given. Remember in any parallelogram, the lengths of opposite sides are equal. Therefore, the sides πΉπ½ and πΊπ» are equal in length. So we can write down another equation: six π₯ plus eight π¦ is equal to 42. Now we have a pair of simultaneous equations in π₯ and π¦, which we can solve in order to determine the values that make the first equation true and therefore make the parallelogram a rectangle.

In this second equation, all of the coefficients are even numbers, which means the whole equation can be divided by two, giving the simpler equation three π₯ plus four π¦ is equal to 21. Weβll now look at solving this simpler equation, which Iβve labelled equation two, along with the first equation, equation one. We first noticed that both equations have the same coefficient of π₯, itβs three, which means that if I subtract equation two from equation one, this will eliminate the π₯ terms.

So if I go ahead and do this, then as intended the π₯ terms will cancel out. Three π₯ minus three π₯ give no π₯s. Five π¦ minus four π¦ gives one π¦, which we just write as π¦. And 24 minus 21 gives three. So we now have the equation π¦ is equal to three. And in fact we found the value of π¦. In order to find the value of π₯, we can substitute this value of π¦ into any of the equations. Iβm going to choose to substitute into the equation labelled one.

Making this substitution gives three π₯ plus five multiplied by three is equal to 24. Five times three is 15. So we have three π₯ plus 15 is equal to 24. Next, I subtract 15 from each side of the equation, giving three π₯ is equal to nine. The final step is to divide both sides of the equation by three, giving π₯ is equal to three. So we found the values of π₯ and π¦ that make the parallelogram πΉπΊπ»π½ a rectangle. Theyβre both equal to three.

Remember, the key facts we used in this question, which are properties of rectangles but not all parallelograms, are these: the diagonals of a rectangle equal in length and they bisect each other.