Is it a typo or am I missing something? SAT math problem

[m_0]

I've been mulling over this problem for a while now, but I can't come up with a solid answer...

If x - y = 5, and x² + y² = 15, what is the value of xy?

The answer is -5. That would imply that one of the variables = -1 and the other = 5 (or vice versa). But that doesn't fit the equations. I also can't think of two squared numbers that would add up to 15.

 

[Deleted member 4993]

I've been mulling over this problem for a while now, but I can't come up with a solid answer...

If x - y = 5, and x² + y² = 15, what is the value of xy?

The answer is -5. That would imply that one of the variables = -1 and the other = 5 (or vice versa). But that doesn't fit the equations. I also can't think of two squared numbers that would add up to 15.

Hint:

\(\displaystyle (x-y)^2 \ = \ x^2 \ + \ y^2 \ - \ 2 * x * y\)

Why are you assuming that x and y has to be integers or even rational numbers!!!

 

[m_0]

Hint:

\(\displaystyle (x-y)^2 \ = \ x^2 \ + \ y^2 \ - \ 2 * x * y\)

Why are you assuming that x and y has to be integers or even rational numbers!!!

That's how the answer sheet explained it! The problem is, I can't figure out where you're getting:

\(\displaystyle (x-y)^2\)

from.

I factored \(\displaystyle x^2 - y^2\) to get: \(\displaystyle (x-y)(x+y)\)

 

[Mrspi]

That's how the answer sheet explained it! The problem is, I can't figure out where you're getting:

\(\displaystyle (x-y)^2\)

from.

I factored \(\displaystyle x^2 - y^2\) to get: \(\displaystyle (x-y)(x+y)\)

You're GIVEN a value for x - y: x - y = 5

You're also GIVEN that x² + y² = 15

You are expected to realize that if you SQUARE (x - y), you will get an expression that involves x² + y² AND 2xy, which should make it easy for you to substitute what's given and solve the resulting expression for the value of xy:

(x - y)² = (x -y)(x - y) = x² - 2xy + y²

(x - y)² = (x² + y²) - 2xy

Substitute the values that were given in the problem:

5² = 15 - 2xy

25 = 15 - 2xy
10 = -2xy
-5 = xy

 

[m_0]

Thank God they didn't do it that way! Imagine the work the kids with TI-84s would have to do for a single problem.

Big thanks to all of you for your help! Mrspi, your explanation makes perfect sense.

Subhotosh Khan and Mrspi have given you excellent answers. I think what may be making this problem hard for you is that you are overlooking that you are NOT asked to find x and y. As the previous answers have indicated, the simple way to answer the question ACTUALLY ASKED is to recognize that

\(\displaystyle (x - y)^2 = (x^2 + y^2) - 2xy\) and that you are given the values of (x - y) and (x^2 + y^2).

However, you can solve this the way you were thinking, which is to find values for x and y and then compute xy.

\(\displaystyle x - y = 5 \longrightarrow x = y + 5 \longrightarrow (y + 5)^2 + y^2 = 15\)

\(\displaystyle (y + 5)^2 + y^2 = y^2 + 10y + 25 + y^2 = 2y^2 + 10y + 25\)

\(\displaystyle 2y^2 + 10y + 25 = 15 \longrightarrow 2y^2 + 10y + 10 = 0 \longrightarrow\) \(\displaystyle y = \dfrac{-10 \pm \sqrt{10^2 - (4 * 2 * 10}}{4} = \dfrac{-10 + (\pm \sqrt{100 - 80})}{4} = \dfrac{-10 + 2(\pm \sqrt{5})}{4}\)

\(\displaystyle So\ x = 5 + \dfrac{-10 +2(\pm \sqrt{5})}{4} = \dfrac{20 - 10 + 2(\pm \sqrt{5})}{4} = \dfrac{10 + 2(\pm \sqrt{5})}{4}\)

\(\displaystyle So\ xy = \dfrac{10 +2(\pm \sqrt{5})}{4} * \dfrac{-10 +2(\pm \sqrt{5})}{4} = \) \(\displaystyle \dfrac{[10 * (-10)] + [(10 - 10)(2)(\pm \sqrt{5}\ )] + [2(\pm \sqrt{5})]^2}{16} =\)

\(\displaystyle \dfrac{-100 + [(0)(2)(\pm \sqrt{5})] + [2^2 * (\sqrt{5})^2]}{16} = \dfrac{-100 + (4 * 5)}{16} = \dfrac{-80}{16} = - 5\)

I think you will agree that this is an awful way to solve this problem. BUT it does tell you that there are numbers that work.

\(\displaystyle x - y = \dfrac{10 + 2(\pm \sqrt{5})}{4} - \dfrac{-10 + 2(\pm\ sqrt{5})}{4} =\) \(\displaystyle \dfrac{10 - (-10) + 2(\pm \sqrt{5}) - 2(\pm \sqrt{5})}{4} =\dfrac{10+10}{4} = \dfrac{20}{4} = 5\)

\(\displaystyle x^2 = (\dfrac{10 + 2(\pm \sqrt{5})}{4})^2 = \dfrac{100 + 40(\pm \sqrt{5}) + 20}{16}\)

\(\displaystyle y^2 = (\dfrac{-10 + 2(\pm \sqrt{5})}{4})^2 = \dfrac{100 - 40(\pm \sqrt{5}) + 20}{16}\)

\(\displaystyle x^2 + y^2 = \dfrac{100 + 40(\pm \sqrt{5}) + 20}{16} + \dfrac{100 - 40(\pm \sqrt{5}) + 20}{16} =\) \(\displaystyle \dfrac{100+20+100+20}{16} = \dfrac{240}{16} = 15\)

Aren't you glad they did NOT ask you to find x and y. Moral: look for the easy way.

 

[Deleted member 4993]

A very similar problem that I have seen is:

Given:

x * y = 7

x + y = 35

Calculate:

\(\displaystyle \dfrac{1}{x} \ + \ \dfrac{1}{y} \ = \ ??\)

 

[Deleted member 4993]

Please show your work. Did you read "Read before posting"?

I only respond to threats written with blood (red).......

 

[marcochimio]

Sum of reciprocals solution

(Pardon my elementary editing, as this is my first time on the site.)

The common denominator for 1/x+1/y is xy. Given this, the expression can be re-written as y/xy+x/xy or (y+x)/xy .
Given that 1) y + x = 35 and 2) xy = 7, we can substitute these values into the above rearranged expression, to get 35/7 , which is, of course equal to 5.


Best regards,
Marco