How do you add $\frac{3 x - 2}{x - 2} + \frac{1}{x^{2} - 4 x + 4}$ $\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}$ ?

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Answer 1 · 2014-12-11T06:38:03

$\frac{3 x - 2}{x - 2} + \frac{1}{x^{2} - 4 x + 4}$ ${3x-2}/{x-2}+1/{x^2-4x+4}$

by factoring out the denominator of the second quotient,

$= \frac{3 x - 2}{x - 2} + \frac{1}{{(x - 2)}^{2}}$ $={3x-2}/{x-2}+1/{(x-2)^2}$

by multiplying the numerator and the denominator of the first quotient by $(x - 2)$ $(x-2)$ ,

$= \frac{3 x^{2} - 8 x + 4}{{(x - 2)}^{2}} + \frac{1}{{(x - 2)}^{2}}$ $={3x^2-8x+4}/{(x-2)^2}+1/{(x-2)^2}$

by combining the two quotients together,

$= \frac{3 x^{2} - 8 x + 5}{{(x - 2)}^{2}}$ $={3x^2-8x+5}/{(x-2)^2}$

by factoring out the numerator,

$= \frac{(3 x - 5) (x - 1)}{{(x - 2)}^{2}}$ $={(3x-5)(x-1)}/{(x-2)^2}$

I hope that this was helpful.

How do you add \frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}3x−2x−2+1x2−4x+4\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}?