What is expand (x+1)/(2x)+(2x+3)/(3x) ?

The solution to expand (x+1)/(2x)+(2x+3)/(3x) is 7/6+3/(2x)

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expand (x+1)/(2x)+(2x+3)/(3x)

expand

\frac{x + 1}{2 x} + \frac{2 x + 3}{3 x}

\frac{7}{6} + \frac{3}{2 x}

Solution steps

\frac{x + 1}{2 x} + \frac{2 x + 3}{3 x}

Least Common Multiplier of

2 x, 3 x : 6 x

Adjust Fractions based on the LCM

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

Since the denominators are equal, combine the fractions:

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

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

Expand

(x + 1) \cdot 3 + (2 x + 3) \cdot 2 : 7 x + 9

= \frac{7 x + 9}{6 x}

Apply the fraction rule:

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

\frac{7 x + 9}{6 x} = \frac{7 x}{6 x} + \frac{9}{6 x}

= \frac{7 x}{6 x} + \frac{9}{6 x}

Cancel

\frac{7 x}{6 x} : \frac{7}{6}

= \frac{7}{6} + \frac{9}{6 x}

Cancel

\frac{9}{6 x} : \frac{3}{2 x}

= \frac{7}{6} + \frac{3}{2 x}

- \frac{8}{n ^{3}} [\frac{n ( n + 1 ) ( 2 n + 1 )}{4}]

x^{4} + (x^{5} + 2) \frac{d y}{d x}

(- 1 - i) e^{t} (cos (t) + i sin (t))

\frac{( 3 - 2 x )}{9}

sin (\frac{2}{3} (x + 6 π))

What is expand (x+1)/(2x)+(2x+3)/(3x) ?
The solution to expand (x+1)/(2x)+(2x+3)/(3x) is 7/6+3/(2x)