Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Exercise Set - Page 301: 36

x =$\frac{1}{2}$ y = 3

x = $\frac{5}{6}$y -2 12x - 5y = -9 substitude x to the second equation 12( $\frac{5}{6}$y - 2 ) - 5y = -9 10y - 24 - 5y = -9 5y = 15 y = 3 substitude y = 3 to first equation x = $\frac{5}{6}$(3) -2 x = $\frac{5}{2}$ - $\frac{4}{2}$ x = $\frac{1}{2}$