What is (2x+6)/(x^2-3x)-(x+5)/(x^2-4x+3) ?

The solution to (2x+6)/(x^2-3x)-(x+5)/(x^2-4x+3) is (x+2)/(x(x-1))

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simplify (2x+6)/(x^2-3x)-(x+5)/(x^2-4x+3)

simplify

\frac{2 x + 6}{x ^{2} - 3 x} - \frac{x + 5}{x ^{2} - 4 x + 3}

\frac{x + 2}{x ( x - 1 )}

Solution steps

\frac{2 x + 6}{x ^{2} - 3 x} - \frac{x + 5}{x ^{2} - 4 x + 3}

Least Common Multiplier of

x^{2} - 3 x, x^{2} - 4 x + 3 : x (x - 3) (x - 1)

Adjust Fractions based on the LCM

= \frac{( 2 x + 6 ) ( x - 1 )}{x ( x - 3 ) ( x - 1 )} - \frac{( x + 5 ) x}{x ( x - 3 ) ( x - 1 )}

Apply the fraction rule:

\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}

= \frac{( 2 x + 6 ) ( x - 1 ) - ( x + 5 ) x}{x ( x - 3 ) ( x - 1 )}

Simplify

(2 x + 6) (x - 1) - (x + 5) x : x^{2} - x - 6

= \frac{x ^{2} - x - 6}{x ( x - 3 ) ( x - 1 )}

Cancel

\frac{x ^{2} - x - 6}{x ( x - 3 ) ( x - 1 )} : \frac{x + 2}{x ( x - 1 )}

= \frac{x + 2}{x ( x - 1 )}

8 x - 3 y - 9 x + 5 y - 2 x + y

(2 x - 3) 2

lo g_{6} (42)

(4 x^{4} - 6 y^{3})^{2}

ln (1^{2})

What is (2x+6)/(x^2-3x)-(x+5)/(x^2-4x+3) ?
The solution to (2x+6)/(x^2-3x)-(x+5)/(x^2-4x+3) is (x+2)/(x(x-1))