Question
Solve the linear equation $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$.
✓
Answer
$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
It is a linear equation since it involves linear expressions only.
$\therefore$ $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$ ... [Transposing $\frac{x}{3}$ to $L.H.S.$ and $ - \frac{1}{5}$ to $R.H.S]$
$\therefore$ $\frac{{3x - 2x}}{6} = \frac{{5 + 4}}{{20}}$
$\therefore$ $\frac{x}{6} = \frac{9}{{20}}$
$\therefore$ x = $\frac{9}{{20}} \times 6$ ... [Multiplying both sides by $6]$
$\therefore$ x = $\frac{{27}}{{10}}$
this is the required solution.
It is a linear equation since it involves linear expressions only.
$\therefore$ $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$ ... [Transposing $\frac{x}{3}$ to $L.H.S.$ and $ - \frac{1}{5}$ to $R.H.S]$
$\therefore$ $\frac{{3x - 2x}}{6} = \frac{{5 + 4}}{{20}}$
$\therefore$ $\frac{x}{6} = \frac{9}{{20}}$
$\therefore$ x = $\frac{9}{{20}} \times 6$ ... [Multiplying both sides by $6]$
$\therefore$ x = $\frac{{27}}{{10}}$
this is the required solution.
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→$x^2+ 12x - 45$