Which expression is equivalent to the one below?
[tex]\frac{\left(x^2 y\right)\left(x^4 y^3\right)}{x y^2}[/tex]
Answer (2)
The expression x y 2 ( x 2 y ) ( x 4 y 3 ) simplifies to x 5 y 2 . This is obtained by multiplying the terms in the numerator and then simplifying the fraction. The equivalent expression is therefore x 5 y 2 .
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Multiply the terms in the numerator: ( x 2 y ) ( x 4 y 3 ) = x 6 y 4 .
Rewrite the expression: x y 2 x 6 y 4 .
Simplify the expression by dividing common factors: x y 2 x 6 y 4 = x 5 y 2 .
The equivalent expression is x y 2 x 6 y 4 .
Explanation
Understanding the Problem We are given the expression x y 2 ( x 2 y ) ( x 4 y 3 ) and asked to find an equivalent expression from the given options.
Simplifying the Numerator First, we simplify the numerator by multiplying the terms: ( x 2 y ) ( x 4 y 3 ) = x 2 + 4 y 1 + 3 = x 6 y 4 .
Rewriting the Expression Now we rewrite the expression as x y 2 x 6 y 4 .
Simplifying the Expression Next, we simplify the expression by dividing the numerator and denominator by common factors of x and y : x y 2 x 6 y 4 = x 6 − 1 y 4 − 2 = x 5 y 2 .
Finding the Equivalent Expression Comparing our simplified expression x 5 y 2 with the given options, we see that none of the options match exactly. However, let's examine the options more closely to see if any of them can be further simplified to x 5 y 2 .
Option 1: x y 2 x 6 y 3 = x 6 − 1 y 3 − 2 = x 5 y Option 2: x y 2 x 8 y 3 = x 8 − 1 y 3 − 2 = x 7 y Option 3: x y 2 x 6 y 4 = x 6 − 1 y 4 − 2 = x 5 y 2
Option 3 simplifies to x 5 y 2 , which is what we found.
Examples
When simplifying algebraic expressions, we often use the rules of exponents. These rules are also useful in real-world applications such as calculating areas and volumes. For example, if you have a rectangular prism with length x 6 , width y 4 , and height x y 2 1 , the volume of the prism can be calculated by multiplying these dimensions together: V = x 6 \tims y 4 \tims x y 2 1 = x 5 y 2 . This shows how simplifying algebraic expressions can help solve practical problems involving geometric shapes and their properties.
