Solve the following system of equations using substitution or elimination. 4x-2y+3z=1 x+3y-4z=-7
3x+y+2z=5
PLEASE HELP !!!

[1]   4x - 2y + 3z = 1

[2]   x + 3y - 4z = -7

[3]   3x + y + 2z = 5

Add -4 [2] to [1], and add -3 [2] to [3] to eliminate x :

-4 (x + 3y - 4z) + (4x - 2y + 3z) = -4 (-7) + 1

-4x - 12y + 16z + 4x - 2y + 3z = 28 + 1

[4]   -14y + 19z = 29

-3(x + 3y - 4z) + (3x + y + 2z) = -3 (-7) + 5

-3x - 9y + 12z + 3x + y + 2z = 21 + 5

-8y + 14z = 26

[5]   -4y + 7z = 13

Add 2 [4] to -7 [5] to eliminate y :

2 (-14y + 19z) - 7 (-4y + 7z) = 2 (29) - 7 (13)

-28y + 38z + 28y - 49z = 58 - 91

-11z = -33

z = 3

Plug this into either [4] or [5] to solve for y :

-4y + 7 (3) = 13

-4y + 21 = 13

-4y = -8

y = 2

Plug both of these solutions into either of [1], [2], or [3] to solve for x :

x + 3 (2) - 4 (3) = -7

x + 6 - 12 = -7

x = -1

answer: 19 is   2

step-by-step explanation: i just looked at your answer and wrote it down

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.