Question

unit: rational expressions
progress:
the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer.
multiply. state the product in simplest form.
\\(\frac{27y}{y^2 - 1} \cdot \frac{y^2 + y}{36y^2}\\)
\\(\frac{3}{4y - 4}\\)\\(\frac{-3}{4}\\)\\(\frac{12}{y - 1}\\)\\(\frac{3y}{4y^2 - 4y}\\)

Explanation:

Step1: Factor numerators and denominators

First, factor the expressions:

• \(y^2 - 1=(y - 1)(y + 1)\) (difference of squares)
• \(y^2 + y=y(y + 1)\) (factor out \(y\))
• \(36y^2 = 4\times9y^2\)
• \(27y = 3\times9y\)

So the original expression \(\frac{27y}{y^2 - 1}\cdot\frac{y^2 + y}{36y^2}\) becomes \(\frac{27y}{(y - 1)(y + 1)}\cdot\frac{y(y + 1)}{36y^2}\)

Step2: Multiply numerators and denominators

Multiply the numerators together and the denominators together:
Numerator: \(27y\times y(y + 1)=27y^2(y + 1)\)
Denominator: \((y - 1)(y + 1)\times36y^2 = 36y^2(y - 1)(y + 1)\)

So the expression is \(\frac{27y^2(y + 1)}{36y^2(y - 1)(y + 1)}\)

Step3: Cancel common factors

Cancel out the common factors:

• \(y^2\) cancels out from numerator and denominator.
• \((y + 1)\) cancels out from numerator and denominator.
• Simplify \(\frac{27}{36}=\frac{3}{4}\) (divide numerator and denominator by 9)

After canceling, we get \(\frac{3}{4(y - 1)}\)? Wait, no, wait. Wait, let's re - check. Wait, \(27y\times y(y + 1)=27y^2(y + 1)\), \(36y^2(y - 1)(y + 1)\). Then cancel \(y^2\) and \((y + 1)\), we have \(\frac{27}{36(y - 1)}=\frac{3}{4(y - 1)}\)? But that's not one of the options. Wait, maybe I made a mistake in factoring. Wait, the options have \(\frac{3y}{4y^2 - 4y}\). Let's try again.

Wait, original expression: \(\frac{27y}{y^2 - 1}\cdot\frac{y^2 + y}{36y^2}\)

Factor \(y^2-1=(y - 1)(y + 1)\), \(y^2 + y=y(y + 1)\)

So \(\frac{27y}{(y - 1)(y + 1)}\times\frac{y(y + 1)}{36y^2}\)

Multiply numerators: \(27y\times y(y + 1)=27y^2(y + 1)\)

Multiply denominators: \((y - 1)(y + 1)\times36y^2=36y^2(y - 1)(y + 1)\)

Now, cancel \(y^2\) and \((y + 1)\):

\(\frac{27}{36(y - 1)}=\frac{3}{4(y - 1)}\)? No, that's not right. Wait, maybe I messed up the multiplication. Wait, no, wait the numerator is \(27y\times(y^2 + y)=27y(y^2 + y)=27y^3+27y^2\), denominator is \((y^2 - 1)\times36y^2=(y - 1)(y + 1)\times36y^2\). Wait, no, the original problem is \(\frac{27y}{y^2 - 1}\cdot\frac{y^2 + y}{36y^2}\), so it's \(\frac{27y(y^2 + y)}{(y^2 - 1)36y^2}\)

Factor \(y^2 + y=y(y + 1)\), \(y^2-1=(y - 1)(y + 1)\)

So \(\frac{27y\times y(y + 1)}{(y - 1)(y + 1)\times36y^2}=\frac{27y^2(y + 1)}{36y^2(y - 1)(y + 1)}\)

Cancel \(y^2\) and \((y + 1)\), we get \(\frac{27}{36(y - 1)}=\frac{3}{4(y - 1)}\). But the option is \(\frac{3y}{4y^2 - 4y}\). Let's simplify \(\frac{3y}{4y^2 - 4y}=\frac{3y}{4y(y - 1)}=\frac{3}{4(y - 1)}\) (when \(y
eq0\)). Ah! So \(\frac{3y}{4y^2 - 4y}=\frac{3y}{4y(y - 1)}=\frac{3}{4(y - 1)}\) (after canceling \(y\) when \(y
eq0\)). So that's the same as our result.

Let's check the option \(\frac{3y}{4y^2 - 4y}\): factor denominator \(4y^2-4y = 4y(y - 1)\), so \(\frac{3y}{4y(y - 1)}=\frac{3}{4(y - 1)}\) (for \(y
eq0\)), which matches our simplified form.

Answer:

\(\frac{3y}{4y^2 - 4y}\) (the last option, the one with \(\frac{3y}{4y^2-4y}\))