# Video: Forming and Solving a System of Linear Equations with Three Unknowns

The sum of the length and width of a cuboid is 24 cm. Its width plus its height is 19 cm and the sum of its height and length is 31 cm. Calculate the volume of the cuboid.

03:45

### Video Transcript

The sum of the length and width of a cuboid is 24 centimetres. It’s width plus its height is 19 centimetres and sum of its height and length is 31 centimetres. Calculate the volume of the cuboid.

The volume of a cuboid is equal to its length multiplied by its width multiplied by its height. So it’s quite clear we’re going to need to calculate each of these dimensions before we can find the volume of the cuboid. We begin by defining the length of our cuboid to be 𝑙 in centimetres. 𝑤 is its width, which is also in centimetres. And we’ll say ℎ is the height of the cuboid. Then the first piece of information we’re given in this question says that the sum of the length and the width of the cuboid is 24 centimetres. So we form an equation using this information. We said that 𝑙 plus 𝑤 is equal to 24. Next, we’re told that the width plus the height is 19. So we can say that 𝑤 plus ℎ must be equal to 19. And thirdly, we know that the sum of its height and length is 31 centimetres. So ℎ plus 𝑙 equals 31.

We formed a system of linear equation with three variables. Those are 𝑙, 𝑤, and ℎ. So we’re going to need to come up with a way to solve these three equations simultaneously. Remember, that means at the same time. Ideally, we’d like to come up with two equations with just two unknowns. So we’ll label our original three equations, equation one, two, and three. And then we notice equation one and two have 𝑤 in common. This means if we subtract one of these equations from the other, we will eliminate 𝑤. And we’ll actually end up with a second equation in terms of 𝑙 and ℎ. So let’s subtract equation two from equation one. 𝑙 plus 𝑤 minus 𝑤 plus ℎ becomes 𝑙 minus ℎ. And then we have 24 minus 19 which is five.

So we have a fourth equation. It’s 𝑙 minus ℎ equals five. We now notice that equation three and four have both ℎ’s and 𝑙’s in common. So we can solve these simultaneously to calculate the value of both 𝑙 and ℎ. You might have seen the acronym SSS when it comes to solving simultaneous equation with two variables. This stands for if the signs are the same, we subtract. Let’s have a look at our ℎ’s. We have the same number of ℎ’s in each equation, but their signs are different. So we’re going to add equation three and four. Notice though that we could have eliminated 𝑙 by subtracting the equations. We’re looking to eliminate ℎ though. When we add the two equations, ℎ plus 𝑙 plus 𝑙 minus ℎ gives us two 𝑙. 31 plus five is 36. So we find that our equation is equal to two 𝑙 equals 36.

We solve this equation for 𝑙 by dividing through by two. And we get 𝑙 equals 18. We can substitute this value of 𝑙 into equation three and form an equation purely in terms of ℎ. When we do that, it gives us ℎ plus 18 equals 31. And we solve for ℎ by subtracting 18 from both sides. That gives us ℎ is equal to 13. So we found 𝑙 and we found ℎ, but what about 𝑤? Let’s clear some space. If we substitute ℎ equals 13 into equation two, we’ll end up with an equation just in terms of 𝑤. That gives Us 𝑤 plus 13 equals 19. And we solve for 𝑤 by subtracting 13 from both sides. What gives us 𝑤 equal six.

Now, before we calculate the volume, it’s really sensible to use one of the equations we haven’t yet used to check whether what we’ve done is correct. We’re going to check in equation one. We’ll let 𝑙 be equal to 18 and 𝑤 be equal to six. When we do, 𝑙 plus 𝑤 becomes 18 plus six. And that is indeed equal to 24 as required. Now that we know the length, width, and height of our cuboid, we’re ready to calculate the volume. It’s 18 times six times 13 which is equal to 1404 cubic centimetres.