Answered

Aimee and Theo both found the product of $\left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right)$.

Their work is shown below.

Aimee: $\left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right)=\left(\frac{1}{5}\right)\left(\frac{2 x^3}{5 y^4}\right)=\frac{2 x^3}{25 y^4}$

Theo: $\left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right)=\frac{8 x^5 y^3}{25 x^2 y^7}=\frac{8 x^3}{25 y^4}$



Which statements are true about Aimee's and Theo's work? Check all that apply.

  • Aimee simplified before she multiplied.
  • Theo's product is correct.
  • Aimee divided out the factors 4 and $x^2$ from $8 x^5$ and $20 x^2$.
  • Theo multiplied the numerators correctly but did not multiply the denominators correctly.
  • Multiplying rational expressions in different ways always results in different products.

Verified Answer :

  • Both Aimee's and Theo's methods correctly lead to the same product \(\frac{2 x^3}{25 y^4}\).
  • Aimee simplifies the expression by dividing out common factors like \(4\) and \(x^2\) before completing the multiplication.
  • Theo correctly multiplies the entire numerators and denominators and then simplifies the result.
  • The correct statements are:
  • \(\checkmark\) Aimee simplified before she multiplied.
  • \(\checkmark\) Theo's product is correct.
  • \(\checkmark\) Aimee divided out the factors \(4\) and \(x^2\) from \(8 x^5\) and \(20 x^2\).

Explanation

  1. Independent Calculation of the Product

We are evaluating the product of two rational expressions:
\[P = \left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right)\]
To calculate this product independently, we multiply the numerators together and the denominators together:
\[\text{Numerator} = y^3 \cdot 8 x^5 = 8 x^5 y^3\]
\[\text{Denominator} = 20 x^2 \cdot 5 y^7 = 100 x^2 y^7\]
So the product is:
\[P = \frac{8 x^5 y^3}{100 x^2 y^7}\]
We simplify this expression by dividing the coefficients and applying the laws of exponents \(\frac{x^a}{x^b} = x^{a-b}\):
\[\frac{8}{100} = \frac{2}{25}\]
\[\frac{x^5}{x^2} = x^{5-2} = x^3\]
\[\frac{y^3}{y^7} = y^{3-7} = y^{-4} = \frac{1}{y^4}\]
Combining these, we get:
\[P = \frac{2 x^3}{25 y^4}\]

  1. Analysis of Aimee's Work

Aimee's method: \(\left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right) = \left(\frac{1}{5}\right)\left(\frac{2 x^3}{5 y^4}\right) = \frac{2 x^3}{25 y^4}\).
Aimee simplified by grouping parts of the expressions before multiplying them fully. Specifically, she looked at:
\[\frac{8 x^5}{20 x^2} = \frac{2 x^3}{5}\]
This step involves dividing the coefficients \(8\) and \(20\) by their greatest common divisor \(4\), and dividing \(x^5\) and \(x^2\) by \(x^2\).
Then she looked at:
\[\frac{y^3}{5 y^7} = \frac{1}{5 y^4}\]
Multiplying these simplified parts gives \(\left(\frac{2 x^3}{5}\right)\left(\frac{1}{5 y^4}\right) = \frac{2 x^3}{25 y^4}\). This confirms that Aimee simplified before she multiplied, and in her simplification, she indeed divided out \(4\) and \(x^2\).

  1. Analysis of Theo's Work and Conclusion

Theo's method: \(\left(\frac{y^3}{20 x^2}\right)\left(\frac{8 x^5}{5 y^7}\right) = \frac{8 x^5 y^3}{100 x^2 y^7} = \frac{2 x^3}{25 y^4}\).
Theo followed the standard algorithmic approach by multiplying the full numerators and then the full denominators before simplifying. His multiplication is correct (\(8 \cdot 1 = 8\), \(20 \cdot 5 = 100\)). His final result \(\frac{2 x^3}{25 y^4}\) is also correct. Thus, the statement that Theo multiplied the denominators incorrectly is false. Furthermore, since both Aimee and Theo achieved the same correct result using different methods, the statement that different methods result in different products is false.

Examples

Understanding how to simplify rational expressions is like managing inventory. If you have to calculate the total cost of multiple shipments, you can either calculate the individual total cost of each item and sum them up (Aimee's method) or calculate the total quantity and total unit price of each category and then compute the final sum (Theo's method). Both approaches lead to the same final result, helping you maintain consistency and accuracy in your calculations, much like simplifying algebraic fractions.

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