Nolan
28

# Use substitution to solve the following system of equations. d + e - f = -3 e = f + d + 7 f = 2e - 6 a. d = 3, e = -4, f = 2 b. d = -4, e = 3, f = 2 c. d = -2, e = 4, f = -5 d. d = -5, e = 4, f = 2

$d+e-f=-3 \\ e=f+d+7 \\ f=2e-6 \\ \\ \hbox{substitute 2e-6 for f in the 2nd equation and solve for e:} \\ e=2e-6+d+7 \\ e=2e+d+1 \ \ \ |-2e \\ -e=d+1 \ \ \ |\times (-1) \\ e=-d-1 \\ \\ \hbox{substitute -d-1 for e in the 3rd equation and solve for f:} \\ f=2(-d-1)-6 \\ f=-2d-2-6 \\ f=-2d-8$ $\hbox{substitute -d-1 for e and -2d-8 for f in the 1st equation and solve for d:} \\ d+(-d-1)-(-2d-8)=-3 \\ d-d-1+2d+8=-3 \\ 2d+7=-3 \ \ \ |-7 \\ 2d=-10 \ \ \ |\div 2 \\ d=-5 \\ \\ e=-d-1=-(-5)-1=5-1=4 \\ \\ f=-2d-8=-2 \times (-5)-8=10-8=2 \\ \\ \boxed{d=-5} \\ \boxed{e=4} \\ \boxed{f=2} \\ \hbox{answer D}$