### Video Transcript

In the given figure, find the values of π₯ and π¦.

Letβs look at the diagram carefully. We have three parallel lines as indicated by the blue arrows along their length. We then have two transversals cutting through these three parallel lines. We have the lengths of various different line segments in the diagram given in terms of the variables π₯ and π¦ whose values weβre looking to calculate. The key fact that weβre going to need within this question is this: parallel lines cut off transversals proportionally. Or, in other words, for this diagram, the ratio between the line segments π½πΎ and πΎπΏ in the top of the diagram is the same as the ratio between the line segments ππ and ππ at the bottom of the diagram.

Letβs substitute in the expressions we have for each of these line segments, in terms of π₯ and π¦. This gives six π₯ minus 20 over four π₯ minus eight is equal to five π¦ minus 25 over three π¦ minus seven. Now you may be thinking about how to solve this equation in order to find the values of π₯ and π¦. But the problem is, we have only one equation with two unknowns. And it isnβt possible to solve it. Letβs think about what other information weβve been given.

Looking carefully at the diagram, we can see the presence of these lines in the bottom section, which tells us that the line segments ππ and ππ are equal in length. This means that the ratio between these two line segments, and therefore also the ratio between π½πΎ and πΎπΏ, must be equal to one. So in actual fact, we can separate this into an equation only involving π₯s and an equation only involving π¦s, both equal to one. And solve them separately in order to find the values of π₯ and π¦.

So we have the two equations: six π₯ minus 20 over four π₯ minus eight equals one and five π¦ minus 25 over three π¦ minus seven equals one. Letβs solve the equation on the left-hand side. First of all, Iβm going to multiply both sides of the equation by four π₯ minus eight, in order to bring this out of the denominator. This gives six π₯ minus 20 is equal to four π₯ minus eight. Now there are π₯s on both sides of this equation. So I do want to bring them on to the same side. But first, Iβm going to add 20 to both sides of the equation. This gives six π₯ is equal to four π₯ plus 12. Now Iβm going to bring all the π₯s on to the same side by subtracting four π₯ from both sides of the equation. And this gives two π₯ is equal to 12. The final step is to divide both sides of the equation by two. So we found the value of π₯. π₯ is equal to six.

The equation for π¦ can be solved in a very similar way. First, multiplying by three π¦ minus seven which gives five π¦ minus 25 is equal to three π¦ minus seven. You may want to pause the video at this point and check that you can solve the equation independently. The next step is to add 25 to both sides. So I have five π¦ is equal to three π¦ plus 18. Next, Iβm going to subtract three π¦ from both sides of the equation, giving two π¦ is equal to 18. And finally, Iβm going to divide both sides of the equation by two. And we have then the value of π¦ is nine.

So we have the solution to the problem, the values of π₯ and π¦: π₯ is equal to six, π¦ is equal to nine. Remember, the key fact we used in this question was that parallel lines cut off transversals proportionally. So the ratio between the line segments of the first transversal at the top of the diagram was the same as the ratio between the line segments of the second transversal at the bottom of the diagram. We also needed to look at the diagram carefully in order to see that weβve been told that the two line segments ππ and ππ are in fact the same length. And therefore, the ratio was one to one.