Question
Solve the equations and check your result: $\frac{{2x}}{3} + 1 = \frac{{7x}}{{15}} + 3$
✓
Answer
$\frac{{2x}}{3} + 1 = \frac{{7x}}{{15}} + 3$
$\frac{{2x}}{3} - \frac{{7x}}{{15}} = 3 - 1 ...$ [Transposing $\frac{{7x}}{{15}}$ to $L.H.S$. and $1$ to $R.H.S.]$
$\therefore \frac{{2x}}{3} - \frac{{7x}}{{15}} = 2$
$\therefore 15\left( {\frac{{2x}}{3} - \frac{{7x}}{{15}}} \right)$ = 2 $\times 15 ... $[Multiplying both sides by $15]$
$\therefore 10x – 7x = 30$
$\therefore 3x = 30$
$\therefore x = \frac{{30}}{3}$ ... [Dividing both sides by $3]$
$\therefore x = 10$ this is the required solution.
Verification,
$L.H.S. = \frac{{2x}}{3} + 1 = \frac{2}{3}(10) + 1 = \frac{{20 + 3}}{3} = \frac{{23}}{3}$
$R.H.S. = \frac{{7x}}{{15}} + 3 = \frac{7}{{15}}(10) + 3 = \frac{{70}}{{15}} + 3 = \frac{{70 \div 5}}{{15 \div 5}} + 3 = \frac{{14 + 9}}{3} = \frac{{23}}{3}$
Therefore, $L.H.S. = R.H.S.$
$\frac{{2x}}{3} - \frac{{7x}}{{15}} = 3 - 1 ...$ [Transposing $\frac{{7x}}{{15}}$ to $L.H.S$. and $1$ to $R.H.S.]$
$\therefore \frac{{2x}}{3} - \frac{{7x}}{{15}} = 2$
$\therefore 15\left( {\frac{{2x}}{3} - \frac{{7x}}{{15}}} \right)$ = 2 $\times 15 ... $[Multiplying both sides by $15]$
$\therefore 10x – 7x = 30$
$\therefore 3x = 30$
$\therefore x = \frac{{30}}{3}$ ... [Dividing both sides by $3]$
$\therefore x = 10$ this is the required solution.
Verification,
$L.H.S. = \frac{{2x}}{3} + 1 = \frac{2}{3}(10) + 1 = \frac{{20 + 3}}{3} = \frac{{23}}{3}$
$R.H.S. = \frac{{7x}}{{15}} + 3 = \frac{7}{{15}}(10) + 3 = \frac{{70}}{{15}} + 3 = \frac{{70 \div 5}}{{15 \div 5}} + 3 = \frac{{14 + 9}}{3} = \frac{{23}}{3}$
Therefore, $L.H.S. = R.H.S.$
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