How do you solve the system of equations 4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-84x+2y+z=7,2x+2y+4z=4,x+3y2z=8?

1 Answer
Apr 17, 2017

x=47/11x=4711
y=(-51)/11y=5111
z=(-9)/11z=911

Explanation:

Given -

4x+2y+z=74x+2y+z=7----------------(1)
2x+2y+4z=-42x+2y+4z=4-------------(2)
x+3y-2z=-8x+3y2z=8----------------(3)

Take the first two equations and eliminate yy

4x+2y+z=74x+2y+z=7----------------(1)
2x+2y+4z=-42x+2y+4z=4-------------(2) -------- (1)-(2)(1)(2)
2x-3z=112x3z=11 -------------(4)

Take equations (1) and (3)
We have to elimiate yy term

4x+2y+z=74x+2y+z=7----------------(1) -------xx3×3
x+3y-2z=-8x+3y2z=8----------------(3) -------xx2×2

12x+6y+3z=2112x+6y+3z=21 ------------(5)
2x+6y-4z=-162x+6y4z=16-------------(6) ----- (5)-(6)(5)(6)
10x+7z=3710x+7z=37 ----------------(7)

Take equations (4) and (7)

2x-3z=112x3z=11 -------------(4) --------xx5×5
10x+7z=3710x+7z=37 ----------------(7)

10x-15z=5510x15z=55 ---------(8)
10x+7z=3710x+7z=37-----------(7) ------(8)-(7)(8)(7)
-22z=1822z=18
z=-18/22z=1822

Plugin >z=-18/22z=1822 in equation (7)

10x+7(-18/22)=3710x+7(1822)=37
10x-126/22=3710x12622=37
10x=37+126/22=(814+126)/22=940/22=470/1110x=37+12622=814+12622=94022=47011
x=470/11xx1/10=470/110=47/11x=47011×110=470110=4711
x=47/11x=4711

Plugin >z=-18/22z=1822 and >x=47/11x=4711 in equation (1)

4x+2y+z=74x+2y+z=7----------------(1)
4(47/11)+2y-18/22=74(4711)+2y1822=7
4(47/11)+2y-9/11=74(4711)+2y911=7
188/11+2y-9/11=718811+2y911=7
(188-9)/11+2y=7188911+2y=7
179/11+2y=717911+2y=7
2y=7-179/11=(77-179)/11=(-102)/112y=717911=7717911=10211
y=(-102)/11xx1/2=(-102)/22=y=10211×12=10222=
y=(-51)/11y=5111

x=47/11x=4711
y=(-51)/11y=5111
z=-18/22=(-9)/11z=1822=911