Question

use the substitution to solve the following equation 4x-y-3z=-3 x-2y-2z=-3 x+y+3z=-2 a. -1 2 -1 b. 1 -2 1 c. -1 -2 1 d.1 -2 -1

Answer 1

1) 4x - y -3z = -3 for -1 2 -1
Now LHS of equation = 4x - y - 3z
= 4(-1) - (2) - 3(-1)
= -4 -2 +3
= -6 + 3 = -3 = RHS

Hence verified that LHS = RHS

Answer 2

2) x - 2y - 2z = -3 for 1 -2 1
LHS = x - 2y - 2z
= (1) - 2(-2) - 2(1)
= 1 + 4 - 2
= 5 - 2 = 3

here LHS not equal to RHS (pls check if u have given values correct)


u can do the rest similarly..........I hope it is clear now??

Answer 3

a. solutions are correct