solve5(2x-3) = 3(x+4)
\(5(2x-3)=3(x+4) \\ 10x-15=3x+12 \\ 10x-3x=12+15 \\ 7x=27 \\ x=\frac{27}{7}=3 \frac{6}{7}\)
\(5(2x-3) = 3(x+4)\\ 10x-15=3x+12\\ 7x=27\\ x=\dfrac{27}{7}\)
