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505-555 Level|   Algebra|                        
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jlgdr
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x + y = a

\(x^2 + 2xy + y^2 = a^2\) ............. (1)

x - y = b

\(x^2 - 2xy + y^2 = b^2\) ................ (2)

(1) - (2)

\(4xy = a^2 - b^2\)

\(2xy = \frac{a^2-b^2}{2}\)

Answer = A
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jlgdr
If x+y=a and x-y=b, then 2xy=?

a) \(\frac{(a^2 - b^2)}{2}\)
b) \(\frac{(b^2 - a^2)}{2}\)
c) \(\frac{(a - b )}{2}\)
d) \(\frac{(ab)}{2}\)
e) \(\frac{(a^2 + b^2)}{2}\)

Squaring the first equation, we have:

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

Squaring the second equation, we have:

x^2 - 2xy + y^2 = 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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I assigned values to the variables

x = 3
y = 2

X + Y = A, 3 + 2 = 5 (A)

X - Y = B, 3 -2 = 1 (B)

So 2XY = 2(3)2 = 12, look for the answer choice that gives you 12. A is the only one.
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jlgdr
If x+y=a and x-y=b, then 2xy=?

a) \(\frac{(a^2 - b^2)}{2}\)
b) \(\frac{(b^2 - a^2)}{2}\)
c) \(\frac{(a - b )}{2}\)
d) \(\frac{(ab)}{2}\)
e) \(\frac{(a^2 + b^2)}{2}\)

Given: x+y=a and x-y=b

Asked: 2xy=?

x = (a+b)/2
y = (a-b)/2
2xy = \(\frac{(a^2-b^2)}{2}\)

IMO A
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jlgdr
If x + y = a and x - y = b, then 2xy = ?


A. \(\frac{(a^2 - b^2)}{2}\)

B. \(\frac{(b^2 - a^2)}{2}\)

C. \(\frac{(a - b )}{2}\)

D. \(\frac{(ab)}{2}\)

E. \(\frac{(a^2 + b^2)}{2}\)

STRATEGY: Upon reading any GMAT Problem Solving question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily test values of x and y.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we might try some algebraic manipulation, but it's not 100% apparent that this strategy will work. So I'll stick with the strategy I know will work...


APPROACH #1: Testing values that satisfy the given information
Since x + y = a and x - y = b, it could be the case that x = 3 and y = 2, in which case 2xy = 2(3)(2) = 12
So, we're looking for an answer choice that evaluates to 12

Since the answer choices are terms of a and b, we need to find the corresponding values of a and b when x = 3 and y = 2
So, we'll plug x = 3 and y = 2 into the equations x + y = a and x - y = b

Take: x + y = a
Substitute values to get: 3 + 2 = 5, which means a = 5

Take: x - y = a
Substitute values to get: 3 - 2 = 1, which means b = 1

Now plug a = 5 and b = 1 into the 5 answer choices to see which one(s) evaluate to 12...

A. \(\frac{(5^2 - 1^2)}{2} = \) 12. KEEP!

B. \(\frac{(1^2 - 5^2)}{2}=\) -12. Eliminate

C. \(\frac{(5 - 1 )}{2}=\) 2. Eliminate

D. \(\frac{(5)(1)}{2}=\) 2.5. Eliminate

E. \(\frac{(5^2 + 1^2)}{2}=\) 13. Eliminate

By the process of elimination, the correct answer is A.

APPROACH #2: Algebraic manipulation
Given:
x + y = a
x - y = b

Add the two equations to get: 2x = a + b

Similarly, we can take the two equations....
x + y = a
x - y = b

.... And subtract the bottom equation from the top equation to get: 2y = a - b

This means we can write: (2x)(2y) = (a + b)(a - b)
Expand and simplify both sides: 4xy = a² - b²
Divide both sides of the equation by 2 to get: 2xy = (a² - b²)/2

Answer: A
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BrentGMATPrepNow
A. \(\frac{(5^2 - 1^2)}{2} = \) 12. KEEP!

B. \(\frac{(1^2 - 5^2)}{2}=\) -12. Eliminate

C. \(\frac{(5 - 1 )}{2}=\) 2. Eliminate

D. \(\frac{(5)(1)}{2}=\) 2.5. Eliminate

E. \(\frac{(5^2 + 1^2)}{2}=\) 13. Eliminate

By the process of elimination, the correct answer is A.

So we test each answer choice just in case more than one of them give the same answer calculated before?
And then we pick another set of numbers to test both answer choices?
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criticall
So we test each answer choice just in case more than one of them give the same answer calculated before?
And then we pick another set of numbers to test both answer choices?

That's correct.
For example, if x = 0 and y = 0, then all of the answer choices would work.
So, we'd have to test more values.
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