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1 vote
Help please! Thanks!!

Help please! Thanks!!-example-1
User Granero
by
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1 Answer

5 votes

Answer: None of them

I think your teacher made a typo. See explanation below for details

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Step-by-step explanation:

With math problems like this one, it is often very tricky to find integer solutions. Luckily, we're given a list of choices. Let's plug them in one at a time and see what happens.

If y = 5, then

y^2 - 7x^2 = 11

5^2 - 7x^2 = 11

25 - 7x^2 = 11

-7x^2 = 11-25

-7x^2 = -14

x^2 = -14/(-7)

x^2 = 2

But x^2 = 2 does not have any integer solutions. The two solutions are irrational. So we can rule out choice F.

Repeat for y = 4 and you should find that x^2 = 5/7, which also does not have any integer solutions. Choice G can be ruled out.

Choices H through K are even worse. Not only are there no integer solutions, but there aren't any real number solutions either (there are complex or imaginary solutions though). Plugging in y = 3 leads to x^2 = -2/7 which has imaginary solutions due to applying the square root to a negative radicand. Similar situations happen with J and K as well.

In summary, none of these answer choices are valid. We need to have two integers x and y that satisfy the equation. I think your teacher made a typo somewhere.

User Davidpodhola
by
5.1k points