Mrs. Cho wrote the following problem on the board. Problem: ((1)/(x^(2))-(2)/(y))/(y-2x^(2)) Step 1: ((y)/(x^(2)y)-(2x^(2))/(x^(2)y))/(y-2x^(2)) Step 2: ((y-2x^(2))/(x^(2)y))/((y-2x^(2))/(1))x
The Correct Answer And Explanation is:
Let’s carefully evaluate the expression Mrs. Cho wrote and explain the steps in detail.
Given Expression:
1×2−2yy−2×2\frac{\frac{1}{x^2} – \frac{2}{y}}{y – 2x^2}
Step 1:
Mrs. Cho rewrote the numerator with a common denominator: yx2y−2x2x2yy−2×2\frac{\frac{y}{x^2y} – \frac{2x^2}{x^2y}}{y – 2x^2}
This is correct. The least common denominator (LCD) of the two terms in the numerator 1×2\frac{1}{x^2} and 2y\frac{2}{y} is x2yx^2y. So:
- 1×2=yx2y\frac{1}{x^2} = \frac{y}{x^2y}
- 2y=2x2x2y\frac{2}{y} = \frac{2x^2}{x^2y}
Now the numerator becomes: y−2x2x2y\frac{y – 2x^2}{x^2y}
So the entire expression becomes: y−2x2x2yy−2×2\frac{\frac{y – 2x^2}{x^2y}}{y – 2x^2}
Step 2:
Mrs. Cho wrote: (y−2x2x2y)÷(y−2×21)\left( \frac{y – 2x^2}{x^2y} \right) \div \left( \frac{y – 2x^2}{1} \right)
This is a correct transformation because dividing by a number is the same as multiplying by its reciprocal. So: y−2x2x2yy−2×21=y−2x2x2y⋅1y−2×2\frac{\frac{y – 2x^2}{x^2y}}{\frac{y – 2x^2}{1}} = \frac{y – 2x^2}{x^2y} \cdot \frac{1}{y – 2x^2}
Now cancel the common factor (y−2×2)(y – 2x^2) in the numerator and denominator: =1x2y= \frac{1}{x^2y}
✅ Final Answer:
1x2y\boxed{\frac{1}{x^2y}}
300-Word Explanation:
Mrs. Cho presents a rational algebraic expression involving division of fractions. The original problem is a complex fraction: 1×2−2yy−2×2\frac{\frac{1}{x^2} – \frac{2}{y}}{y – 2x^2}
To simplify, start by combining the two fractions in the numerator. The least common denominator (LCD) of x2x^2 and yy is x2yx^2y. Rewriting each term with this denominator:
- 1×2=yx2y\frac{1}{x^2} = \frac{y}{x^2y}
- 2y=2x2x2y\frac{2}{y} = \frac{2x^2}{x^2y}
Now subtract the two fractions: y−2x2x2y\frac{y – 2x^2}{x^2y}
So the entire expression becomes: y−2x2x2yy−2×2\frac{\frac{y – 2x^2}{x^2y}}{y – 2x^2}
This is now a division of two expressions. Since dividing by a number is the same as multiplying by its reciprocal, rewrite as: y−2x2x2y⋅1y−2×2\frac{y – 2x^2}{x^2y} \cdot \frac{1}{y – 2x^2}
At this point, the common factor y−2x2y – 2x^2 in the numerator and denominator cancels out. The result is: 1x2y\frac{1}{x^2y}
Thus, Mrs. Cho’s simplification is valid, and the final answer is 1x2y\boxed{\frac{1}{x^2y}}. This process demonstrates the importance of identifying common denominators and using algebraic rules for fraction operations such as multiplying by the reciprocal and canceling common terms.