Question

final exam - requires respondus lockdown browse
started: dec 10 at 8:58pm
quiz instructions
question 37
write the rational expression in lowest terms.
\\(\frac{y^2 - 5y - 14}{y^2 - 7y - 18}\\)
\\(\bigcirc\\) \\(\frac{y^2 - 5y - 14}{y^2 - 7y - 18}\\)
\\(\bigcirc\\) \\(\frac{y - 7}{y - 9}\\)
\\(\bigcirc\\) \\(\frac{y - 9}{y + 7}\\)
\\(\bigcirc\\) \\(\frac{y - 7}{y^2 - 7y - 18}\\)

Explanation:

Step1: Factor Numerator and Denominator

Factor \( y^2 - 5y - 14 \): Find two numbers that multiply to \(-14\) and add to \(-5\). The numbers are \(-7\) and \(2\), so \( y^2 - 5y - 14=(y - 7)(y + 2) \).
Factor \( y^2 - 7y - 18 \): Find two numbers that multiply to \(-18\) and add to \(-7\). The numbers are \(-9\) and \(2\), so \( y^2 - 7y - 18=(y - 9)(y + 2) \).

Step2: Simplify the Rational Expression

Substitute the factored forms into the expression:
\( \frac{(y - 7)(y + 2)}{(y - 9)(y + 2)} \).
Cancel the common factor \((y + 2)\) (assuming \( y
eq - 2 \) to avoid division by zero):
\( \frac{y - 7}{y - 9} \).

Answer:

\(\frac{y - 7}{y - 9}\) (corresponding to the option with this expression, e.g., the second option in the list)