How do you subtract $\frac{6}{x + 2} - \frac{2 x + 3}{x}$ $6/(x+2)-(2x+3)/x$ ?

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How do you subtract $\frac{6}{x + 2} - \frac{2 x + 3}{x}$ $6/(x+2)-(2x+3)/x$ ?

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Answer 1 · 2017-01-14T10:25:47

$- \frac{2 x^{2} + x + 6}{x (x + 2)}$ $-(2x^2+x+6)/(x(x+2))$

Explanation:

Before subtracting we require the fractions to have a $common denominator$ $color(blue)"common denominator"$

This is achieved as follows.

$(\frac{6}{x + 2} \times \frac{x}{x}) - (\frac{2 x + 3}{x} \times \frac{(x + 2)}{(x + 2)})$ $(6/(x+2)xxx/x)-((2x+3)/(x)xx((x+2))/((x+2)))$

That is, multiply the numerator/denominator of the first fraction by the denominator of the second fraction and multiply the numerator/denominator of the second fraction by the denominator of the first fraction.

$= \frac{6 x}{x (x + 2)} - \frac{(2 x + 3) (x + 2)}{x (x + 2)}$ $=(6x)/(x(x+2))-((2x+3)(x+2))/(x(x+2))$

We now have a common denominator and can subtract the terms on the numerator.

$= \frac{6 x - (2 x^{2} + 7 x + 6)}{x (x + 2)}$ $=(6x-(2x^2+7x+6))/(x(x+2))$

$= \frac{6 x - 2 x^{2} - 7 x - 6}{x (x + 2)} = \frac{- 2 x^{2} - x - 6}{x (x + 2)}$ $=(6x-2x^2-7x-6)/(x(x+2))=(-2x^2-x-6)/(x(x+2))$

Take out a common factor of - 1 on the numerator.

$= - \frac{2 x^{2} + x + 6}{x (x + 2)}$ $=-(2x^2+x+6)/(x(x+2))$

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How do you subtract $\frac{6}{x + 2} - \frac{2 x + 3}{x}$ $6/(x+2)-(2x+3)/x$ ?

1 Answer

Explanation:

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How do you subtract 6/(x+2)-(2x+3)/x6x+2−2x+3x6/(x+2)-(2x+3)/x?

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How do you subtract $\frac{6}{x + 2} - \frac{2 x + 3}{x}$ $6/(x+2)-(2x+3)/x$ ?