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Algebra

by BTGmoderatorRO » Thu Nov 02, 2017 12:35 pm
$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
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by [email protected] » Thu Nov 02, 2017 1:18 pm
Hi Roland2rule,

This question can be approached in a couple of different ways, Here's how you can TEST VALUES to get to the answer:

IF...
X=3 and Y=2
A=5 and B=1

So we're looking for an answer that equals (2)(3)(2) = 12 when A=5 and B=1. There's only one answer that matches...

Final Answer: A

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by EconomistGMATTutor » Thu Nov 02, 2017 3:04 pm
$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
Hi Roland2rule,
Lets take a look at your question,

We can use this problem using the formula:
$$\left(x+y\right)^2-\left(x-y\right)^2=4xy$$

Plugin the given values,
$$a^2-b^2=4xy$$
$$a^2-b^2=2\times2xy$$
$$\frac{a^2-b^2}{2}=2xy$$
$$2xy=\frac{a^2-b^2}{2}$$

Therefore, option A is correct.

Hope it helps.
I am available if you'd like any follow up.
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hi

by Scott@TargetTestPrep » Sun Dec 15, 2019 6:43 pm
BTGmoderatorRO wrote:$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
Squaring both equations, we have:

(x + y)^2 = a^2

x^2 + y^2 + 2xy = a^2

and

(x - y)^2 = b^2

x^2 + y^2 - 2xy = b^2

Subtracting the second equation from the first, we have:

4xy = a^2 - b^2

2xy = (a^2 - b^2)/2

Answer: A

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by SampathKp » Tue Dec 17, 2019 7:27 pm
BTGmoderatorRO wrote:$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
(x+y)^2 = x^2+y^2+2xy
(x-y)^2 = x^2+y^2-2xy

a^2=x^2+y^2+2xy is equation 1
b^2 =x^2+Y^2-2xy is equation 2

Subtracting equation 2 from equation 1 will give

a^2-b^2 = 4xy

2xy = (a^2-b^2)/2
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