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

The solution to expand (x-3)/6+(3(x+2))/4 is (11x)/(12)+1

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expand (x-3)/6+(3(x+2))/4

expand

\frac{x - 3}{6} + \frac{3 ( x + 2 )}{4}

\frac{11 x}{12} + 1

Solution steps

\frac{x - 3}{6} + \frac{3 ( x + 2 )}{4}

Expand

3 (x + 2) : 3 x + 6

= \frac{x - 3}{6} + \frac{3 x + 6}{4}

Least Common Multiplier of

6, 4 : 12

Adjust Fractions based on the LCM

= \frac{( x - 3 ) \cdot 2}{12} + \frac{( 3 x + 6 ) \cdot 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{( x - 3 ) \cdot 2 + ( 3 x + 6 ) \cdot 3}{12}

Expand

(x - 3) \cdot 2 + (3 x + 6) \cdot 3 : 11 x + 12

= \frac{11 x + 12}{12}

Apply the fraction rule:

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

\frac{11 x + 12}{12} = \frac{11 x}{12} + \frac{12}{12}

= \frac{11 x}{12} + \frac{12}{12}

Apply rule

\frac{a}{a} = 1

\frac{12}{12} = 1

= \frac{11 x}{12} + 1

e^{y} d x + (x e^{y} + 2 y) d y

C_{4} (9 e^{- 3 x} x - 6 e^{- 3 x})

\frac{ln ( 2 n )}{ln ( 2 ( n + 1 ) )}

(cos (2 t) + 3 cos (t))^{2} (- 2 cos (2 t) + 3 cos (t))

\frac{2 ( 2 z - 5 )}{15}

What is expand (x-3)/6+(3(x+2))/4 ?
The solution to expand (x-3)/6+(3(x+2))/4 is (11x)/(12)+1