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Chapter 7: Problem 14

Multiply as indicated. $$\frac{9 y+21}{y^{2}-2 y} \cdot \frac{y-2}{3 y+7}$$

Short Answer

Expert verified
\[3y + 7\]

Step by step solution

01

Simplify each fraction

Before you can multiply, you need to simplify both fraction. In the first fraction, both terms in the numerator (9y and 21) are divisible by 3, and both in the denominator (y^2 and 2y) by y. In the second fraction, again both terms in the numerator (y and 2) and in the denominator (3y and 7) can be expressed as separate fractions. This results in \[ \frac{3(3y + 7)}{y(y - 2)} \cdot \frac{y - 2}{3(y + (7/3))} \]
02

Multiply the fractions

Next, simply multiply across both the numerators and the denominators: \[ \frac{(3(3y +7))(y -2)}{(y(y -2))(3(y + (7/3)))} \] Notice that we can treat fractions like \((y-2)\) and \((y + (7/3))\), etc. as individual terms.
03

Cancel Common Factors

Look for terms in the numerator and denominator that are the same, and cancel them out: \[(3(3y +7))(3(y + (7/3))) \div (y(y -2)(3(y + (7/3)))) \] This simplifies to \(3y+7\).
04

Simplify

Simplify the final result: \[3y + 7\]

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