\(\frac{2x}{3}+1 =\frac{ 7x}{15}+3\)
Transposing \(\frac{7x}{15}\) to L.H.S and \(1\) to R.H.S, we obtain
\(\frac{2x}{3}-\frac{7x}{15}\) = \(3-1\)
\(\frac{5 × 2x - 7x }{ 15}\) = 2
\(\frac{3x}{15} = 2\)
\(\frac{x}{5}\) = 2
Multiplying both sides by \(5\), we obtain
\(x = 10\)
L.H.S = \(\frac{2x}{3}+1\)
= \(\frac{2×10}{3}+1\)
= \(\frac{2 ×10+1×3}{3}\)
= \(\frac{23}{3}\)
R.H.S = \(\frac{7x}{15}+3\)
=\(\frac{7×10}{15}+3\)
= \(\frac{7 × 2}{3}+3\)
= \(\frac{14}{3}+3\)
= \(\frac{14+3 × 3}{3}\)
= \(\frac{23}{3}\)
L.H.S. = R.H.S.
Hence, the result obtained above is correct