# Question Video: Forming and Solving a System of Linear and Quadratic Equations with Two Unknowns Mathematics

A right-angled triangle has a hypotenuse of 35 cm and a perimeter of 84 cm. Find the lengths of the other two sides.

09:07

### Video Transcript

A right-angled triangle has a hypotenuse of 35 centimeters and a perimeter of 84 centimeters. Find the lengths of the other two sides.

So to help us answer this question, the first thing I’ve done is drawing a right-angled triangle. Next I can add some information from the question itself. So we’re told that the right-angled triangle has a hypotenuse of 35 centimeters. So therefore, I can add on the annotation 35 centimeters. And I know it’s gonna be 35 centimeters in this point because this is a hypotenuse as it’s the longest side and it’s opposite the right angle as I’ve shown here with the arrow. But the other thing I want to add is the lengths of the other two sides, the lengths of the shorter sides.

However, we don’t know these because it says that we want to find the lengths of the other two sides. So therefore what I’ve labelled them as are 𝑥 and 𝑦. And the final bit of information that I can add is that the perimeter is equal to 84 centimeters because again we’re told that in the question. So we’ve got our information down. Where do we start? What do we do first? Well the first thing we want to do is actually form some equations. And the first equation we can form is 𝑥 plus 𝑦 plus 35 is equal to 84. And that’s because we’re told the perimeter is 34, and the perimeter of a shape is the distance of all the way around the outside. So that means we have to add together the sides of our triangle in this case. So that would be 𝑥 plus 𝑦 plus 35.

Now what we do is subtract 35 from each side of the equation. And we do this so that our variables are on one side and then our numbers on the other side. And when we do this, we get 𝑥 plus 𝑦 is equal to 49. And that’s because if you subtract 35 and 35 it’s just zero. And that’s on the left-hand side, so we’re left with 𝑥 plus 𝑦. And if you subtract 35 from 84, you get 49. And this equation is gonna be our first equation, and I’ve labelled it equation one. Now to form the next equation, what we can use is the Pythagorean theorem. And we can use that because we know that this is a right-angled triangle. Now if we have a right angled triangle, we know the Pythagorean theorem will work.

And what the Pythagorean theorem tells us is that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the hypotenuse. What this means is it’s the lengths of the shorter sides squared then added together is equal to the length of the longest side or the hypotenuse squared. So therefore, in our right-angled triangle, the two shorter sides are 𝑥 and 𝑦. So we’ll get 𝑥 squared plus 𝑦 squared is equal to the hypotenuse squared, which is 35 squared, which gives us 𝑥 squared plus 𝑦 squared is equal to 1225 because 35 squared is 1225. And this is gonna form our second equation. So now what we have are two equations, and these are a pair of simultaneous equations. And the method we’re going to use to solve these is the substitution method.

To use the substitution method, what we want to do is rearrange one of our equations to give us an equation in 𝑥 or 𝑦 that can substitute into the other equation. An equation we can use to do that in this case is the first equation: 𝑥 plus 𝑦 is equal to 49. Now we can make either 𝑥 or 𝑦 the subject, it doesn’t matter. Well I’ve chosen to make 𝑥 the subject. And to do this, what I do is subtract 𝑦 from each side of the equation. So I get 𝑥 is equal to 49 minus 𝑦, and I’m gonna call this equation equation three. And the reason I label my equations is because it’s easier to spot what we’re doing with them in the next few steps.

So the first thing I’m gonna do is substitute equation three into equation two. When I say I’m gonna substitute into it, what that means is I’m actually going to put 49 minus 𝑦 where 𝑥 is in equation two. And the reason I’ve done that is to make it so that we only have one variable. So now we have an equation just in 𝑦. So we can solve to find 𝑦. So the first thing we do when we’re trying to solve this equation is expand the parentheses because 49 minus 𝑦 all squared is the same as 49 minus 𝑦 multiplied by 49 minus 𝑦. So to expand our parentheses, what we do is multiply each term in the first parentheses by each term in the second parentheses. So it started with 49 multiplied by 49, which gives us 2401. And then we have 49 multiplied by negative 𝑦, which gives us negative 49𝑦. And then we have another negative 𝑦 multiplied by 49, which gives us another negative 49𝑦. And then finally, we have negative 𝑦 multiplied by negative 𝑦. Well a negative multiplied by a negative is a positive, so we get positive 𝑦 squared.

And then the final stage is to simplify. And to do that, what we’ve done is collected like terms, because we have 2401 then minus 98𝑦—and that’s because we had negative 49 𝑦 minus 49 𝑦, which gives us negative 98 𝑦—then plus 𝑦 squared. So then if we put this back into our equation, we get 2401 minus 98 𝑦 plus 𝑦 squared plus 𝑦 squared is equal to 1225. So then the next step is to have zero on the right-hand side. And in order to do this, what we’re gonna do is actually collect our like terms and then subtract 1225 from each side of the equation. When we do that, we get two 𝑦 squared minus 98 𝑦 plus 1178 is equal to zero.

So now to simplify this, we can divide each side of the eqaution by two. And when we do that, we get 𝑦 squared minus 49 𝑦 plus 588 is equal to zero. Then we factor this. And when we factor this, we get 𝑦 minus 21 multiplied by 𝑦 minus 28 is equal to zero, remembering that if you’re gonna factor what you need to do is to find a pair of factors that when multiplied together they give you 588 but then when added together they give you negative 49 because that’s the coefficient of 𝑦. The way to do that with these two that we found here was to find a couple that were within 20 and play around with them. And we knew that they both had to be negative or positive because they were making positive 588. So therefore, you either have to have a negative multiplied by a negative to give us a positive or a positive multiplied by a positive.

So now to find out the values of 𝑦, what we need to do is set each of our parentheses to zero. And that’s because to have a result of zero, we have to have one of them is zero cause zero multiplied by anything gives us an answer of zero. So from the first parentheses, we get that 𝑦 is equal to 21. And that’s because if you add 21 to each side of the equation, we get 𝑦 is equal to 21. And then for the second parentheses, you’re gonna get 𝑦 is equal to 28 because you had 𝑦 minus 28 is equal to zero. But if you add 28 to each side of the equation, you’re gonna get 𝑦 is equal to 28. So now we’ve found our two 𝑦-values. So now we know the possible values of 𝑦, we can use these to find the possible values of 𝑥. And in order to do this, what we’re gonna do is substitute each of our values for 𝑦 into equation three because this tells us what 𝑥 is equal to cause it says that 𝑥 is equal to 49 minus 𝑦.

So we’re gonna start with 𝑦 equals 21. If we substitute that in instead of 𝑦 in equation three, we get 𝑥 is equal to 49 minus 21. So therefore, 𝑥 will be equal to 28. If we use the 𝑦-value of 28, then we’re gonna get 𝑥 is equal to 49 minus 28. So this is gonna give us an 𝑥-value of 21. So therefore, we can see that our 𝑦-values are either 21 or 28 and the 𝑥-values are also either 21 or 28. So this tells us that we’ve actually found the lengths of the other two sides because, therefore, the lengths of the other two sides must be equal to 21 centimeters and 28 centimeters. And we’ve got the final answer, but what we can do is a quick check because we can use the Pythagorean theorem to check to see if this works. So if we do 21 squared plus 28 squared, because these are the lengths of the shorter sides, then this is equal to 441 plus 784, which is equal to 1225. And this is the same as 35 squared. So we know the length of our sides are definitely 21 centimeters and 28 centimeters.