Respuesta :
[tex] \frac{x+6}{2x+5}+ \frac{x+5}{x+3} \\ = \frac{(x+6)(x+3)+(x+5)(2x+5)}{2 x^{2} +11x+15} \\ = \frac{x^{2} +9x+18+2 x^{2} +15x+25}{2 x^{2} +11x+15} \\ = \frac{3 x^{2} +24x+43}{2 x^{2} +11x+15} [/tex]
Answer: Option 'C' is correct.
Step-by-step explanation:
Since we have given that
The quantity of x plus 6, all over the quantity of 2x plus 5 the quantity of x plus 5, all over the quantity of x plus 3.
[tex]\frac{x+6}{2x+5}+\frac{x+5}{x+3}\\[/tex]
Now, we need to simplify the above expression:
1) Taking the L.C.M. of denominator:
[tex]\frac{x+6}{2x+5}+\frac{x+5}{x+3}\\\\=\frac{(x+6)(x+3)+(x+5)(2x+5)}{(2x+5)(x+3)}\\[/tex]
2) Solving the brackets of the numerator and denominator:
[tex]\frac{x+6}{2x+5}+\frac{x+5}{x+3}\\\\=\frac{(x+6)(x+3)+(x+5)(2x+5)}{(2x+5)(x+3)}\\\\=\frac{x^2+18+9x+2x^2+25+15x}{2x^2+15+11x}\\\\=\frac{3x^2+43+24x}{2x^2+11x+15}[/tex]
Hence, Option 'C' is correct.