Given that \(2x+3y=5\)...(1)

and

\(5x-4y=2\)...(2).

From (1) \(\displaystyle{2}{x}={5}-{3}{y}\to{x}=\frac{{{5}-{3}{y}}}{{2}}\). Putting this value of x in (2) we get

\(5 \times \frac{5-3y}{2}-4y=2\)

\(\rightarrow 5 \times (5-3y)-8y=4 \rightarrow -23y=-21 \rightarrow y=\frac{21}{23}\)

Therefore from (1) we have \(x=\frac{5-3y}{2} =\frac{5-3(\frac{21}{23})}{2} =\frac{115-63}{46} =\frac{26}{23}\)