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An animal shelter has a total of three hundred and fifty animals comprised of cats, dogs, and rabbits.
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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?
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Letβs first begin by letting π₯ be the number of cats, π¦ be the number of dogs, and π§ be the number of rabbits.
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So it says this animal shelter has a total of three hundred and fifty animals.
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So the number of cats plus the number of dogs plus the number of rabbits, π₯ plus π¦ plus π§, should equal three hundred and fifty.
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Next it says the number of rabbits, π§, 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.
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And lastly, there are twenty more cats than dogs.
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So if there are more cats, then the number of cats would be the number of dogs plus twenty, because thereβs twenty more cats than dogs.
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So now we need to somehow take these three equations and solve for π₯, π¦, and π§.
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Notice the first, second, and third equation all have one common variable.
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They all have an π₯.
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So it says that π§ is equal to one half π₯ minus five, so we could plug that info into our first equation.
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And then we could keep the very first π₯ from the first equation.
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And then if we could somehow replace this π¦ with something in terms of π₯, then that first equation would be completely in terms of π₯ and we could solve for π₯.
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So our last equation, π₯ equals π¦ plus twenty, letβs solve it for π¦ and then it will be in terms of π₯.
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So if we would subtract twenty from both sides, we will get that π¦ is equal to π₯ minus twenty.
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So taking our first equation will keep the first π₯.
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And then instead of π¦, we are going to replace it with the π₯ minus twenty.
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Next, instead of π§, we will replace it with one half π₯ minus five, and then finally bring down the three hundred and fifty.
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Now we can solve for π₯ by combining like terms: π₯ plus π₯ plus one half π₯ is two and a half π₯.
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So letβs just go ahead and write it as two point five π₯.
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And then we have negative twenty and negative five, which is negative twenty five.
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So now letβs add twenty five to both sides of the equation.
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So we have two point five π₯ equals three hundred and seventy five.
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Now we divide both sides by two point five.
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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.
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So if π₯ is one hundred and fifty, there are one hundred and fifty cats.
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Now that we know π₯, we can plug that in to find π¦ and π§.
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So we can solve for π¦, because π¦ is equal to π₯ minus twenty.
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And replacing π₯ with one hundred and fifty, one hundred and fifty minus twenty means π¦ is one hundred and thirty.
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So there are one hundred and thirty dogs.
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And then lastly, π§ is equal to one half π₯ minus five, so we plug in one hundred and fifty.
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And half of one hundred and fifty is seventy five.
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And seventy five minus five means π§ is equal to seventy.
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So there are seventy rabbits.
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So again there are seventy rabbits, one hundred and fifty cats, and one hundred and thirty dogs.