Solve equation

i) $\frac{x}{2} \ -\ \frac{1}{5} \ =\ \frac{x}{3} \ +\ \frac{1}{4}$

And

ii) $\frac{n}{2} \ -\ \frac{3n}{4} \ +\ \frac{5n}{6} \ =\ 21$

i)

$ \begin{array}{l}
\frac{x}{2} \ -\ \frac{1}{5} \ =\ \frac{x}{3} \ +\ \frac{1}{4}\\
\\
\\
\\
\frac{x}{2} \ -\ \frac{x}{3} \ =\ \frac{1}{4} \ +\ \frac{1}{5}\\
\\
\\
\\
\frac{3x\ -\ 2x}{6} \ =\ \frac{5\ +\ 4}{20}\\
\\
\\
\\
\frac{x}{6} \ =\ \frac{9}{20}\\
\\
\\
\\
x\ =\ \frac{9\ \times \ 6}{20}\\
\\
\\
x\ =\ \frac{9\ \times \ 3}{10}\\
\\
\\
\mathbf{x\ =\ \frac{27}{10}}
\end{array}$

ii)

$ \begin{array}{l}
\frac{n}{2} \ -\ \frac{3n}{4} \ +\ \frac{5n}{6} \ =\ 21\\
\\
\\
\\
\frac{n( 6) \ -\ 3n( 3) \ +\ 5n( 2)}{12} \ =\ 21\\
\\
\\
\\
\frac{6n\ -\ 9n\ +\ 10n}{12} \ =\ 21\\
\\
\\
\\
\frac{7n}{12} \ =\ 21\\
\\
\\
\\
n\ =\ 21\ \times \ \frac{12}{7}\\
\\
\\
\\
n\ =\ 3\ \times \ 12\\
\\
\\
\mathbf{n\ =\ 36}
\end{array}$

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Updated on: 10-Oct-2022

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