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Video: Writing and Solving a System of Linear Equations in Three Unknowns in a Real-World Context

Bethani Gasparine

An animal shelter has a total of 350 animals comprised of cats, dogs, and rabbits. If the number of rabbits is 5 less than half the number of cats, and there are 20 more cats than dogs, how many of each are at the shelter?

03:47

Video Transcript

An animal shelter has a total of three hundred and fifty animals comprised of cats, dogs, and rabbits. If the number of rabbits is five less than half the number of cats, and there are twenty more cats than dogs, how many of each are at the shelter?

Let’s first begin by letting π‘₯ be the number of cats, 𝑦 be the number of dogs, and 𝑧 be the number of rabbits.

So it says this animal shelter has a total of three hundred and fifty animals. So the number of cats plus the number of dogs plus the number of rabbits, π‘₯ plus 𝑦 plus 𝑧, should equal three hundred and fifty. Next it says the number of rabbits, 𝑧, is five less than half the number of cats, which would be one half π‘₯ minus five: five less than half a number of cats. And lastly, there are twenty more cats than dogs. So if there are more cats, then the number of cats would be the number of dogs plus twenty, because there’s twenty more cats than dogs. So now we need to somehow take these three equations and solve for π‘₯, 𝑦, and 𝑧.

Notice the first, second, and third equation all have one common variable. They all have an π‘₯. So it says that 𝑧 is equal to one half π‘₯ minus five, so we could plug that info into our first equation. And then we could keep the very first π‘₯ from the first equation. And then if we could somehow replace this 𝑦 with something in terms of π‘₯, then that first equation would be completely in terms of π‘₯ and we could solve for π‘₯.

So our last equation, π‘₯ equals 𝑦 plus twenty, let’s solve it for 𝑦 and then it will be in terms of π‘₯. So if we would subtract twenty from both sides, we will get that 𝑦 is equal to π‘₯ minus twenty. So taking our first equation will keep the first π‘₯. And then instead of 𝑦, we are going to replace it with the π‘₯ minus twenty. Next, instead of 𝑧, we will replace it with one half π‘₯ minus five, and then finally bring down the three hundred and fifty. Now we can solve for π‘₯ by combining like terms: π‘₯ plus π‘₯ plus one half π‘₯ is two and a half π‘₯. So let’s just go ahead and write it as two point five π‘₯. And then we have negative twenty and negative five, which is negative twenty five.

So now let’s add twenty five to both sides of the equation. So we have two point five π‘₯ equals three hundred and seventy five. Now we divide both sides by two point five. The two point five cancels on the left, and then three hundred and seventy five divided by two point five is one hundred and fifty. So if π‘₯ is one hundred and fifty, there are one hundred and fifty cats. Now that we know π‘₯, we can plug that in to find 𝑦 and 𝑧. So we can solve for 𝑦, because 𝑦 is equal to π‘₯ minus twenty. And replacing π‘₯ with one hundred and fifty, one hundred and fifty minus twenty means 𝑦 is one hundred and thirty. So there are one hundred and thirty dogs. And then lastly, 𝑧 is equal to one half π‘₯ minus five, so we plug in one hundred and fifty. And half of one hundred and fifty is seventy five. And seventy five minus five means 𝑧 is equal to seventy. So there are seventy rabbits. So again there are seventy rabbits, one hundred and fifty cats, and one hundred and thirty dogs.