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

Question

Eli Assouline · Accepted Answer

#{3x-2}/{x-2}+1/{x^2-4x+4}#

by factoring out the denominator of the second quotient,

#={3x-2}/{x-2}+1/{(x-2)^2}#

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

#={3x^2-8x+4}/{(x-2)^2}+1/{(x-2)^2}#

by combining the two quotients together,

#={3x^2-8x+5}/{(x-2)^2}#

by factoring out the numerator,

#={(3x-5)(x-1)}/{(x-2)^2}#

I hope that this was helpful.

Ethan Adkins · Answer

undefined To add the given expressions, $\frac{3x-2}{x-2}+\frac{1}{x^2-4x+4}$, we first need to find a common denominator. The denominator of the first fraction is $x-2$, and the denominator of the second fraction is $(x-2)^2$.

To make the denominators the same, we can rewrite the second fraction as $\frac{1}{(x-2)(x-2)}$.

Now that the denominators are the same, we can combine the fractions by adding their numerators and keeping the common denominator.

The numerator of the first fraction is $3x-2$, and the numerator of the second fraction is $1$.

Adding the numerators, we get $3x-2+1=3x-1$.

Therefore, the sum of the two fractions is $\frac{3x-1}{x-2}$.

Isaiah Anthony · Answer

undefined To add the given expressions, first find a common denominator, which is $ (x-2)(x-2) $. Then, combine the numerators and simplify the resulting expression. The simplified expression is $ \frac{3x^2 - 5x + 6}{(x - 2)^2} $.