Solve the following system of equations.?

Question

x +2y-6=z
3y-2z=7
4 + 3x = 2y-5z

Answer 1

x + 2y -z = 6
0x +3y -2z = 7
3x-2y+5z=-4
now
| 1 2 -1 6 |
| 0 3 -2 7 |
| 3 -2 5 -4 |
solving matrix v get

| 1 0 0 7/4 |
| 0 1 0 3/2 |
| 0 0 1 -5/4 |

hence
x=7/4 y=3/2 z=-5/4

Answer 2

If there's a quadratic, then there'll maximum probable be 2 solutions. first of all, the addition technique could paintings completely in this methodology. upload the main dazzling factors and the left factors to get. y + 5x + 6x^2 - 13x - y = 3 - 5 word how the 'y's cancel out 6x^2 - 8x = -2 Divide the two factors by utilising 2 3x^2 - 4x = -a million 3x^2 - 4x + a million = 0 (3x - a million)(x-a million) = 0 x = a million ; a million/3 Now plug it into the less complicated equation to resolve for y y + 5x = 3 y = 3 - 5x y = 3 - 5(a million) y = 3 - 5(a million/3) y = -2 ; 4/3 so the suggestions are ( a million , -2 ) , ( a million/3 , 4/3)

Answer 3

Well, you are given
x+2y-6=z
3y-2z=7
4+3x=2y-5z

From that I got
z=x+2y-6
y=3.5-z
x=z-2y+6

And you subsitute x,y, and z into the three original equations or he ones you already have. Which, after about a year of calculations, you get
z=-y+3.5
y=2z+7
x=-1.5y+3.375

And then you do the rest, I’m not sure anymore

Answer 4

@ni has the correct answer. in addition to that method, you could always use linear algebra to solve it.

Write the matrix of coefficients:

[1, 2, -1]
[0, 3, -2] = A
[3, -2, 5]

Calculate A inverse as:

[11/8, -1, -1/8]
[-3/4, 1, 1/4 ] = A^-1
[-9/8, 1, 3/8 ]

Take your vector or solutions

[6 ]
[7 ] = y
[-4]

the x=A^(-1)*y

[x] [7/4 ]
[y] = [3/2 ]
[z] [-5/4]

Answer 5

x + 2y -z = 6
0x +3y -2z = 7
3x-2y+5z=-4

Then use the matrix to solve it.

Answer 6

x+2y-z=6..................(1)
3y-2z=7................(2)
3x-2y+5z=-4.............(3)
adding (1)&(2)
4x+4z=2....................(4)
adding(1)*-3&(2)*2
-3x-z=-4.................... (5)
(4)+(5)*4
-8x=-14 sox=7/4
substituting x in (4)z=-5/4
substituting z in (2)y=3/2
the solution set={7/4,3/2,-5/4}

Answer 7

x+2y-6=z
or, x+2y-z=6..........(i)

3y-2z=7..................(ii)

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

(i)+(iii)......
4x+4z=2
or, 2x+2z=1........(iv)

(iii)*3 + (ii)*2.....
9x+11z=2..........(v)

(v)*2 - (iv)*9.....
4z= -5
or, z= -5/4

so, x= 7/4 [from (iv)]

so, y = 3/2 [from (i)]

Answer 8

Wow. I used to be able to do these. That was a long time ago.

Answer 9

x=1.75
y=1.5
z=-1.25