Question

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)

Answer 1

\( 2\left( {x + \dfrac{3}{5}} \right) = 5 - \left( {\dfrac{{13}}{5} + x} \right)\\ \Leftrightarrow 2x + \dfrac{6}{5} = \dfrac{{12}}{5} - x\\ \Leftrightarrow 3x = \dfrac{6}{5}\\ \Leftrightarrow x = \dfrac{2}{5} \)

Answer 2

\( \dfrac{{10x + 3}}{{12}} = 1 + \dfrac{{6 + 8x}}{9}\\ \Leftrightarrow 3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\ \Leftrightarrow 30x + 9 = 36 + 24 + 32x\\ \Leftrightarrow - 2x = 51\\ \Leftrightarrow x = - \dfrac{{51}}{2} \)