Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Multiplying and Dividing Real Numbers
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Exercise 52 Page 43

Consider how multiplying or dividing the given values could lead to the desired result.

Example Solution: \text{-}\dfrac {3yz}{2x}

Practice makes perfect

There are many expressions we can create with the given values of x, y, and z to obtain a result of 1. We have that x= -3, y= -2, and z= -1. For our expression, we will arbitrarily choose to create a fraction. Now, for example, we can let the numerator of the fraction be 3yz. 3yz= 3( - 2)( - 1) = 6 From here, we can think about what we might divide 6 by to end with 1. Since the result needs to be positive, the number we are looking for is positive. The result also is a whole number. Consider a few possibilities.

Possible Number Division Result
1 6 ÷ 1 6
2 6 ÷ 2 3
3 6 ÷ 3 2
6 6 ÷ 6 1

That means that the denominator of our fraction must be 6. Notice that if we multiply -2 by x we will obtain exactly 6. -2 * x =- 2( - 3) = 6 Knowing that, we can complete our expression. \begin{gathered} \dfrac {3yz}{\text{-}2x} \quad \Leftrightarrow \quad \text{-}\dfrac{3yz}{2x} \end{gathered} We can verify that this works by substituting the given values into our expression.

-3yz/2x
\text{-}\dfrac {3({\color{#0000FF}{\text{-} 2}})({\color{#A800DD}{\text{-}1}})}{2({\color{#009600}{\text{-} 3}})}
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Evaluate
\dfrac {3(\text{-} 2)(\text{-}1)}{\text{-}2(\text{-} 3)}
\dfrac {3\cdot 2\cdot 1}{2\cdot 3}
6/6
1

Since the obtained value is 1, the expression is correct. Keep in mind that this is just one possible answer out of infinitely many.