Answer:
The expression simplifies to
.
Explanation:
The expression
![(2(x+3))/(x(x-1)) *(4x(x+2))/(10(x+3))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/alijhpk4bmf5dqtohp86hgmcmrpjzt2q8j.png)
can be rearranged and written as
![(8x(x+3)(x+2))/(10x(x-1)(x+3)).](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f02rpjs6cuuurv84755amru5yxsveld051.png)
In this form the
terms in the numerator and in the denominator cancel to give
![(8x(x+2))/(10x(x-1)).](https://img.qammunity.org/2020/formulas/mathematics/middle-school/stsfhy97da8e1cwq2glxkqbdcqzv7802sj.png)
The
are present both in the numerator and in the denominator, so they also cancel, and the fraction
simplifies to
, so finally our expression becomes:
![\boxed{(4(x+2))/(5(x-1))}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8tacg727iu9sqk69ze0pg91mwie9g5x9fb.png)
Which is our answer:)