Answer:
The answer is [tex]\frac{4y}{y+3}[/tex] ⇒ the 4th answer
Step-by-step explanation:
* Lets talk about the product of two fraction
- If we have two fraction a/b and c/d, the product of them
will be ac/bd
- The there is any simplify can do between numerator and
denominator we must to make it
Ex: 2a²/5b × 15b/4a, we can simplify 2a² with 4a at first and
simplify 15b with 5b and then put the answer
∵ 2a²/4a = a/2 ⇒ 2 ÷ 4 = 1/2 and a² ÷ a = a
∵ 15b/5b = 3 ⇒ 15 ÷ 5 = 3 and b ÷ b = 1
∴ 2a²/5b × 15b/4a = a/1 × 3/2 = 3a/2
* Now lets solve the problem
∵ [tex]\frac{2y}{y-3}*\frac{4y-12}{2y+6}[/tex]
- We can simplify the second fraction at first
∵ [tex]\frac{4y-12}{2y+6}=\frac{4(y-3)}{2(y+3}=\frac{2(y-3)}{y+3}[/tex]
- At first we took 4 as a common factor from 4y - 12 ⇒ 4(y - 3)
and then simplify 4 with 2 in the denominator
- we can cancel the term x - 3 in the numerator of the second
fraction with the same term in the denominator of the first fraction
∴ [tex]\frac{2y}{1}*\frac{2(1)}{y+3}=\frac{4y}{y+3}[/tex]
* The answer is [tex]\frac{4y}{y+3}[/tex]
