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what is the simplified form of 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? a.the quantity of 2x plus 11, all over the quantity of 3x plus 8 b.the quantity of 3x squared plus 24x plus 43, all over the quantity of 3x plus 8 c. the quantity of 3x squared plus 24x plus 43, all over the quantity of 2x squared plus 11x plus 15 d. the quantity of 2x plus 11, all over the quantity of 2x squared plus 11x plus 15

Respuesta :

nobillionaireNobley
[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.