Answer:
[tex] \frac { 4y } { y + 3 } [/tex]
Step-by-step explanation:
We are given the following terms and we are to find their product:
[tex] \frac { 2y } { y - 3 } \times \frac { 4y - 12 } { 2y + 6 } [/tex]
[tex] \frac { 8y^2 - 24y } { 2y^2 - 6y + 6y - 18 } [/tex]
Taking the common terms out and cancelling them to get:
[tex] \frac { 2 (4y^2 - 12y ) } { 2 (y^2 - 9 ) } [/tex]
[tex] \frac { 4y ( y - 3 ) } { ( y + 3 ) ( y - 3 ) } [/tex]
[tex] \frac { 4y } { y + 3 } [/tex]
