If x and y are non-zero real numbers such that x \neq y, \frac{x^2 - y^2}{x - y} = 12, and… — GRE General Test

If $x$ and $y$ are non-zero real numbers such that $x \neq y$ , $\frac{x^2 - y^2}{x - y} = 12$ , and $\frac{x^2y - xy^2}{xy} = 4$ , what is the value of $x^2 + y^2$ ?

Answer: 80

Answer

The value of

x^2 + y^2

is 80.

Simplifying the first equation by factoring the difference of squares

\frac{(x-y)(x+y)}{x-y} = 12

gives

x + y = 12

. Simplifying the second equation by factoring out

xy

yields

\frac{xy(x-y)}{xy} = 4

, which gives

x - y = 4

. Solving this system yields

x = 8

and

y = 4

. Squaring and summing these values gives

8^2 + 4^2 = 64 + 16 = 80

. Alternatively, using the algebraic identity

\frac{(x+y)^2 + (x-y)^2}{2} = \frac{12^2 + 4^2}{2} = \frac{144 + 16}{2} = 80

directly yields the correct answer.

Step-by-Step Solution

Simplify the first given algebraic expression using the difference of squares identity.

x + y = 12

Factoring

x^2 - y^2

gives

(x - y)(x + y)

. Since

x \neq y

, we can cancel the non-zero common factor

(x - y)

from the numerator and denominator.

Simplify the second given algebraic expression by factoring out the greatest common factor.

x - y = 4

Factoring

xy

from

x^2y - xy^2

yields

xy(x - y)

. Since

x, y \neq 0

, we cancel

xy

from the numerator and denominator.

Solve for the individual values of

x

and

y

x = 8

and

y = 4

Adding

(x + y = 12)

and

(x - y = 4)

gives

2x = 16

, so

x = 8

. Subtracting the equations gives

2y = 8

, so

y = 4

Calculate the target expression

x^2 + y^2

x^2 + y^2 = 8^2 + 4^2 = 64 + 16 = 80

Key Concept

Factoring Algebraic Expressions (Difference of Squares and GCF Extraction)

Question

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