Answer: Third Option is correct.
Step-by-step explanation:
Since we have given that

Now, we will simplify it, step by step:
First we take 3 as common factor :
![3[(1)/(4x+3)+(7)/(8x^2-14x-15)]](https://img.qammunity.org/2018/formulas/mathematics/high-school/cime0avw4qvlga0yihqj3mwiqbx32r6fv6.png)
Now, we will do the method " Splitting the middle term" we get,
![3[(1)/(4x+3)+(7)/(8x^2-20x+6x-15)]\\\\3[(1)/(4x+3)+(7)/(4x(2x-5)+3(2x-5))]\\\\3[(1)/(4x+3)+(7)/((4x+3)(2x-5))]\\\\3[(2x-5+7)/((2x-5)(4x+3))]\\\\=3[(2x+2)/((2x-5)(4x+3))]\\\\=(6(x+1))/((2x-5)(4x+3))](https://img.qammunity.org/2018/formulas/mathematics/high-school/z2f93mbvy0mnh6ihg6waagpla199zlar7r.png)
Hence, 6 times the quantity x plus 1 end quantity over the quantity 2x minus 5 end quantity times 4x plus 3.
Therefore, Third Option is correct.