Examkin

Question

Difficulty: MediumEvaluating Algebraic Expressions

If a=−2a = -2, b=−12b = -\frac{1}{2}, and c=3c = 3, what is the value of the algebraic expression a3b−2c2a−b−2\frac{a^3 b - 2c^2}{a - b^{-2}}?

  1. 73\frac{7}{3}Answer
  2. B
    113\frac{11}{3}
  3. C
    -7
  4. D
    163\frac{16}{3}
  5. E
    −73-\frac{7}{3}$

Answer

The correct answer is 73\frac{7}{3}.
Substituting a=−2a = -2, b=−12b = -\frac{1}{2}, and c=3c = 3 gives a numerator of (−2)3(−12)−2(3)2=(−8)(−12)−18=4−18=−14(-2)^3 \left(-\frac{1}{2}\right) - 2(3)^2 = (-8)\left(-\frac{1}{2}\right) - 18 = 4 - 18 = -14, and a denominator of −2−(−12)−2=−2−4=−6-2 - \left(-\frac{1}{2}\right)^{-2} = -2 - 4 = -6. Dividing −14-14 by −6-6 simplifies to 73\frac{7}{3}.

Step-by-Step Solution

1
Substitute a=−2a = -2, b=−12b = -\frac{1}{2}, and c=3c = 3 into the numerator a3b−2c2a^3 b - 2c^2.
Numerator =(−2)3⋅(−12)−2(3)2=(−8)⋅(−12)−2(9)=4−18=−14= (-2)^3 \cdot \left(-\frac{1}{2}\right) - 2(3)^2 = (-8) \cdot \left(-\frac{1}{2}\right) - 2(9) = 4 - 18 = -14.
Apply exponents first, then multiplication, and finally subtraction.
2
Substitute a=−2a = -2 and b=−12b = -\frac{1}{2} into the denominator a−b−2a - b^{-2}.
Denominator =−2−(−12)−2=−2−(−2)2=−2−4=−6= -2 - \left(-\frac{1}{2}\right)^{-2} = -2 - (-2)^2 = -2 - 4 = -6.
A negative exponent indicates the reciprocal of the base, so (−12)−2=(−2)2=4\left(-\frac{1}{2}\right)^{-2} = (-2)^2 = 4.
3
Divide the numerator by the denominator and simplify the fraction.
−14−6=146=73.\frac{-14}{-6} = \frac{14}{6} = \frac{7}{3}.
Dividing two negative numbers yields a positive quotient, which simplifies by dividing the numerator and denominator by 2.

Key Concept

Evaluating algebraic expressions with negative bases and negative integer exponents
Estimated Time:1m 30s
Rate this question

Topics

Evaluating Algebraic Expressions

More questions from this topic