\(2x+2y=20 \\ 5x - 3y=15 \\ \mbox{Using the first equation.}\\ 2x=20-2y \\ x=10-y \\ \mbox{now we plug that value of x into the second equation} \\ 5(10-y) - 3y = 15 \\ 50-5y-3y=15 \\ -8y = -35 \\ y = \dfrac {35}{8} \\ x=10-y = 10-\dfrac {35}{8}\\ x=\dfrac {45}{8}\)
