Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by the Substitution Method - Exercise Set - Page 300: 24

{(-$\frac{44}{3}$,-$\frac{7}{3}$)}

x=8y+4 Plug this value for x into the other equation. x=5y-3 8y+4=5y-3 8y+4-4=5y-3-4 8y=5y-7 8y-5y=5y-5y-7 3y=-7 y=-$\frac{7}{3}$ Plug this value for y into the first equation to find x. x=8(-$\frac{7}{3}$)+4 x=-$\frac{56}{3}$+4 x=-$\frac{56}{3}$+$\frac{12}{3}$ x=-$\frac{44}{3}$ Therefore the solution is {(-$\frac{44}{3}$,-$\frac{7}{3}$)}