# Question Video: Using the Properties of Parallel Lines and Solving Linear Equations to Determine the Value of Two Unknowns Mathematics • 11th Grade

Given that 𝐴𝐵 = (2𝑥 + 1) cm, 𝐵𝐶 = (𝑥 + 3) cm, 𝐷𝐸 = (3𝑥 − 1) cm, 𝐸𝐹 = 𝑦 cm, 𝐺𝐻 = 11 cm, and 𝐻𝐼 = 10 cm, find the values of 𝑥 and 𝑦.

05:58

### Video Transcript

Given that 𝐴𝐵 equals two 𝑥 plus one centimeters, 𝐵𝐶 equals 𝑥 plus three centimeters, 𝐷𝐸 equals three 𝑥 minus one centimeters, 𝐸𝐹 equals 𝑦 centimeters, 𝐺𝐻 equals 11 centimeters, and 𝐻𝐼 equals 10 centimeters, find the values of 𝑥 and 𝑦.

It’s always a good idea to start a question like this involving a diagram by filling in any length information that isn’t already on the diagram. We can then see more clearly that we have six given lengths. Three of these involve the unknown value of 𝑥 and one of them involves the unknown value of 𝑦.

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 𝐴𝐺, 𝐵𝐻, and 𝐶𝐼. 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 Thales’s 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, it’s worth noticing that the line segments 𝐺𝐻 and 𝐻𝐼 both have numerical values for their lengths. If we wanted to work out the value of 𝑥 first, then we wouldn’t want to include the line segment 𝐸𝐹 in the proportionality statement because that involves the unknown of 𝑦. So let’s see how we can write the line segment 𝐴𝐵 and 𝐵𝐶 in a proportionality statement with 𝐺𝐻 and 𝐻𝐼.

There are a few different proportion statements that we could write, but one example would be that 𝐴𝐵 over 𝐵𝐶 is equal to 𝐺𝐻 over 𝐻𝐼. We then simply need to fill in the length information that we have. So we have two 𝑥 plus one over 𝑥 plus three is equal to 11 over 10. To begin simplifying this, we can cross multiply. So we have 10 times two 𝑥 plus one equals 11 times 𝑥 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 20𝑥 plus 10. In the same way, on the right-hand side, we multiply 11 by each of the terms. So we have 11𝑥 plus 33.

The next stage is to gather all the terms involving 𝑥 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 11𝑥 and 10 from both sides of the equation giving us nine 𝑥 is equal to 23. Dividing both sides by nine, we have 𝑥 is equal to 23 over nine. We can leave this answer as a fraction or work it out as a decimal. If we didn’t 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 𝑥 is approximately 2.56.

And so we found an answer for the value of 𝑥. Let’s see if we can work out the second part of this question to find the value of 𝑦. So the unknown value of 𝑦 occurs in the line segment of 𝐸𝐹. And so it might be sensible to use the line segments 𝐸𝐹, 𝐷𝐸, 𝐺𝐻, and 𝐻𝐼 in a proportion statement. We could write that 𝐷𝐸 over 𝐸𝐹 is equal to 𝐺𝐻 over 𝐻𝐼. When it comes to filling in the length of the line segment 𝐷𝐸, we’ll need to substitute in the value that we found for 𝑥. 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 𝐷𝐸 is 20 over three centimeters. So once we have filled the length information into the given equation, we have 20 over three over 𝑦 is equal to 11 over 10. Cross multiplying, we have that 10 times 20 over three is equal to 11𝑦. 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 𝑦, either as a fraction or a decimal. As a fraction, it’s six and two over 33. As a decimal, we get the answer that 𝑦 is equal to 6.06 recurring. Rounding to two decimal places, we can say that 𝑦 is approximately 6.06. We have now completed the question. We found both values 𝑥 and 𝑦. As decimal approximations to two decimal places, we have 𝑥 equals 2.56 and 𝑦 equals 6.06.