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Expert-verified Found in: Page 226 ### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903 # Alejandra and Kyle both simplified$\frac{2{a}^{2}b}{{\left(-2a{b}^{3}\right)}^{-2}}$ . Who is correct? Explain your reasoning. Alejandra is correct.

See the step by step solution

## Step 1. Write down the given information.

The given expression is $\frac{2{a}^{2}b}{{\left(-2a{b}^{3}\right)}^{-2}}$ which is solved by both Alejandra and Kyle. ## Step 2. Explanation.

The given expression $\frac{2{a}^{2}b}{{\left(-2a{b}^{3}\right)}^{-2}}$ is solved correctly by Alejandra and Kyle is not correct, because in first step Kyle wrote$\frac{2{a}^{2}b}{{\left(-2a{b}^{3}\right)}^{-2}}=\frac{2{a}^{2}b}{{\left(-2\right)}^{-2}a{\left({b}^{3}\right)}^{-2}}$ instead of$\frac{2{a}^{2}b}{{\left(-2a{b}^{3}\right)}^{-2}}=\frac{2{a}^{2}b}{{\left(-2\right)}^{-2}{\left(a\right)}^{-2}{\left({b}^{3}\right)}^{-2}}$ . Therefore, Kyle is incorrect because there should be${\left(a\right)}^{-2}$ in place of$\left(a\right)$ by using the concept of exponents of power property.

Hence, Alejandra is correct.

## Step 3. Conclusion.

Alejandra is correct. ### Want to see more solutions like these? 