Question

What is the simplified form of the quantity of x plus 5, all over the quantity of 3x plus 4 + the quantity of x plus 4, all over the quantity of x plus 3?

Answer 1

Answer: $\frac{4x^2+24x+31}{3x^2+13x+12}$

Step-by-step explanation:

Since the given expression ,

$\frac{x+5}{3x+4} + \frac{x+4}{x+3}$

=$\frac{(x+5)(x+3)+(x+4)(3x+4)}{(3x+4)(x+3)}$ ( by taking the LCM for solving the fraction)

=$\frac{x^2+5x+3x+15+3x^2+12x+4x+16}{3x^2+9x+4x+12}$ ( By multiplying)

=$\frac{x^2+8x+15+3x^2+16x+16}{3x^2+13x+12}$

=$\frac{4x^2+24x+31}{3x^2+13x+12}$ ( by adding the like terms)

Answer 2

$\frac{x + 5}{3x + 4} + \frac{x + 4}{x + 3} = \frac{(x + 5)(x + 3)}{(3x + 4)(x + 3)} + \frac{(x + 4)(3x + 4)}{(3x + 4)(x + 3)} = \frac{x(x + 3) + 5(x + 3)}{(3x(x + 3) + 4(x + 3)} + \frac{x(3x + 4) + 4(3x + 4)}{3x(x + 3) + 4(x + 3)} = \frac{x(x) + x(3) + 5(x) + 5(3)}{3x(x) + 3x(3) + 4(x) + 4(3)} + \frac{x(3x) + x(4) + 4(3x) + 4(4)}{3x(x) + 3x(3) + 4(x) + 4(3)} = \frac{x^{2} + 3x + 5x + 15}{3x^{2} + 9x + 4x + 12} + \frac{3x^{2} + 4x + 12x + 16}{3x^{2} + 9x + 4x + 12} = \frac{x^{2} + 8x + 15}{3x^{2} + 13x + 12} + \frac{3x^{2} + 16x + 16}{3x^{2} + 13x + 12} = \frac{(x^{2} + 8x + 15) + (3x^{2} + 16x + 16)}{3x^{2} + 13x + 12} = \frac{(x^{2} + 3x^{2}) + (8x + 16x) + (15 + 16)}{3x^{2} + 13x + 12} = \frac{4x^{2} + 24x + 31}{3x^{2} + 13x + 12}$