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

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Answer 1 · 2017-06-22T18:01:43

$(4x^2+43x-9)/((3x-1)(x+4))$

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

$(4x)/(3x-1)(x+4)/(x+4)+(3x-1)/(3x-1)9/(x+4)$

$(4x^2+16x)/((3x-1)(x+4))+(27x-9)/((3x-1)(x+4))$

$(4x^2+16x+27x-9)/((3x-1)(x+4))$

$(4x^2+43x-9)/((3x-1)(x+4))$

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

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