How do you add $\frac{4 x}{3 x - 1} + \frac{9}{x + 4}$ $\frac { 4x } { 3x - 1} + \frac { 9} { x + 4}$ ?

Question

1193 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-22T18:01:43

$\frac{4 x^{2} + 43 x - 9}{(3 x - 1) (x + 4)}$ $(4x^2+43x-9)/((3x-1)(x+4))$

The denominators must be the same in order to add fractions.

$\frac{4 x}{3 x - 1} \frac{x + 4}{x + 4} + \frac{3 x - 1}{3 x - 1} \frac{9}{x + 4}$ $(4x)/(3x-1)(x+4)/(x+4)+(3x-1)/(3x-1)9/(x+4)$

$\frac{4 x^{2} + 16 x}{(3 x - 1) (x + 4)} + \frac{27 x - 9}{(3 x - 1) (x + 4)}$ $(4x^2+16x)/((3x-1)(x+4))+(27x-9)/((3x-1)(x+4))$

$\frac{4 x^{2} + 16 x + 27 x - 9}{(3 x - 1) (x + 4)}$ $(4x^2+16x+27x-9)/((3x-1)(x+4))$

$\frac{4 x^{2} + 43 x - 9}{(3 x - 1) (x + 4)}$ $(4x^2+43x-9)/((3x-1)(x+4))$

Because the numerator does not factor (at least not to anything nicely), this completes the addition.

How do you add 4x3x−1+9x+4\frac { 4x } { 3x - 1} + \frac { 9} { x + 4}?