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28 Mar 2022, 06:45
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For all x and y, \((x + 1)(y + 1) – x – y =\)
A. \(xy – x – y + 1\)
B. \(xy + 1\)
C. \(–x – y + 1\)
D. \(x^2 + y^2 – 1\)
E. \(1\)
A. \(xy – x – y + 1\)
B. \(xy + 1\)
C. \(–x – y + 1\)
D. \(x^2 + y^2 – 1\)
E. \(1\)
28 Mar 2022, 07:32
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For all x and y, \((x + 1)(y + 1) – x – y =\)
A. \(xy – x – y + 1\)
B. \(xy + 1\)
C. \(–x – y + 1\)
D. \(x^2 + y^2 – 1\)
E. \(1\)
A. \(xy – x – y + 1\)
B. \(xy + 1\)
C. \(–x – y + 1\)
D. \(x^2 + y^2 – 1\)
E. \(1\)
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 to test for equivalency.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also expand and simplify the given expression.
On test day, I would probably choose the latter approach, but I'm going to show both approaches.
APPROACH #1: Expand and simplify
Given:\((x + 1)(y + 1) – x – y\)
Expand: \((xy + x + y + 1) – x – y =\)
Simplify: \(xy + 1\)
Answer: B
APPROACH #2: Test for equivalency
Key concept: If two expressions are equivalent, they must evaluate to the same value for every possible value of x.
For example, since the expression 2x + 3x is equivalent to the expression 5x, the two expressions will evaluate to the same number for every value of x.
So, if x = 7, the expression 2x + 3x = 2(7) + 3(7) = 14 + 21 = 35, and the expression 5x = 5(7) = 35
Let's evaluate the given expression for \(x = 1\) and \(y = 1\).
We get: \((x + 1)(y + 1) – x – y =(1 + 1)(1 + 1) – 1 – 1 = 2\)
We will now evaluate each answer choice for \(x = 1\) and \(y = 1\) and eliminate those that don't evaluate to \(2\)
A. \((1)(1) – 1 – 1 + 1 = 0\). Doesn't evaluate to \(2\). ELIMINATE.
B. \(xy + 1 = (1)(1) + 1 = 2\). KEEP
C. \(–1 – 1 + 1 = -1\). Doesn't evaluate to \(2\). ELIMINATE.
D. \(1^2 + 1^2 – 1 = 1\). Doesn't evaluate to \(2\). ELIMINATE.
E. \(1\). Doesn't evaluate to \(2\). ELIMINATE.
Answer: B
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