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QuestionEvaluate x2+y3y\frac{x^{2}+y}{3 y} for x=−3x=-3 and y=−2y=-2. Choose A or B: A. x2+y3y=□\frac{x^{2}+y}{3 y}=\square or B. Undefined.

Studdy Solution

STEP 1

Assumptions1. The expression to evaluate is x+y3y\frac{x^{}+y}{3 y} . The given values are x=−3x=-3 and y=−y=-

STEP 2

We need to substitute the given values of xx and yy into the expression.x2+yy=(−)2+(−2)(−2)\frac{x^{2}+y}{ y} = \frac{(-)^{2}+(-2)}{ (-2)}

STEP 3

Calculate the numerator and the denominator separately.
umerator (−3)2+(−2)=9−2=7(-3)^{2}+(-2) =9 -2 =7
Denominator 3(−2)=−63 (-2) = -6
So, the expression becomes7−6\frac{7}{-6}

STEP 4

implify the expression. The negative sign can be placed in the numerator or the denominator.
7−6=−76\frac{7}{-6} = -\frac{7}{6}The expression x2+y3y\frac{x^{2}+y}{3 y} evaluates to −76-\frac{7}{6} when x=−3x=-3 and y=−2y=-2.
So, the correct choice is A. x2+y3y=−76\frac{x^{2}+y}{3 y}=-\frac{7}{6}

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