Having problems solving this math equation been a while since I've been in high school appreciate the help?

Question

Whom can answer this equation:

{2X + Y + Z = 3

1/2X - 3Y - 2Z =1

7X + 3/2Z =10

Equation is all inside

So who can help.

Answer 1

Matrices are the easiest way to do it.

Just sub the coefficients for X, Y and Z into a 3 by 3 matrix

| 2     1     1 |

| 0.5  -3    -2|

| 7    0    1.5|

Find the inverse of that matrix (1 over the matrix) then times that new matrix by the matrix you get from your answers:  

|  3  |    |18/11   6/11   -4/11   |

|  1  | x |59/11   16/11  -18/11 |

| 10 |    |-84/11  -28/11  26/11 |

You get the answer:

| 20/11  |

| 13/11  |

| -20/11 |

Therfore X = 20/11, Y = 13/11, Z = -20/11

PS. My attempt of matrices on yahoo answers didn't turn out all that well cause they took all my spaces away which made them actually readable.

Answer 2

z=3-2x-y ... (0)

x/2-3y-2(3-2x-y)=1

=> x-6y-4(3-2x-y)=2

=> x-6y-12+8x+4y=2

=> 9x-2y=14 ... (1)

7x+3(3-2x-y)/2=10

=> 14x+3(3-2x-y)=20

=> 14x+9-6x-3y=20

=> 8x-3y=11 ... (2)

3*(1)-2*(2): 11x=20

x=20/11

substitute into (1) to get y and then into (0) to get z.

Answer 3

1st equation:

2x + y + z = 3

z = 3 - 2x - y

2nd equation (substitute z):

1/2x - 3y - 2(3 - 2x - y) = 1

1/2x - 3y - 6 + 4x + 2y = 1

y = 4 1/2x - 7

y = 9/2x - 7

1st equation (substitute y):

z = 3 - 2x - (9/2x - 7)

z = 3 - 4/2x - 9/2x + 7

z = 10 - 13/2x

3rd equation (substitute z):

7x + 3/2(10 - 13/2x) = 10

28/4x + 15 - 39/4x = 10

11/4x = 5

x = 20/11

1st equation (substitute x):

z = 10 - 13/2(20/11)

z = 110/11 - 130/11

z = - 20/11

2nd equation (substitute x):

y = 9/2(20/11) - 7

y = 90/11 - 77/11

y = 13/11

Answer: x = 20/11, y = 13/11, z = - 20/11

Proof (1st equation):

2(20/11) + 13/11 - 20/11 = 3

40/11 - 7/11 = 3

33/11 = 3

3 = 3

Proof (2nd equation):

1/2(20/11) - 3(13/11) - 2(- 20/11) = 1

10/11 - 39/11 + 40/11 = 1

11/11 = 1

1 = 1

Proof (3rd equation):

7(20/11) + 3/2(- 20/11) = 10

140/11 - 30/11 = 10

110/11 = 10

10 = 10

Answer 4

2X + Y + Z = 3 ----eq1

1/2X - 3Y - 2Z =1---eq2

7X + 3/2Z =10-------eq3

from eq1

Z= 3-Y-2X---eq4

multiply eq2 by 2

X - 6Y -4Z =2

substitute eq 4 in eq 2

X - 6Y -4(3 - Y -2X)=2

X - 6Y -12 +4Y +8X =2

9X - 2Y =14----eq5

multiply eq3 by2

14X +3Z =20

substitute eq4 in eq3

14X + 3(3 - 2X- Y)=20

14X + 9 -6X -3Y =20

8X - 3Y =11----eq6

solve eq5 and eq6 similtaneously

repeat similar procedure to find Z(replacing X or Y in this case)

Answer 5

multiply the first equation by 3 to eliminate one of the term then subtract  it to the second equation

3  ( 2x   + y  +  z  = 3)

    6x  + 3y  + 3z = 9

+  1/2x- 3y - 2z = 1

----------------------------------

  5.5x + z = 10

now multiply this equation by 7 and the last one by 5.5 the  you add them together

(5.5x + z = 10)7

(7X + 3/2Z =10)5.5

    38.5x +  7x      =  70

  - 38.5x +  8.25x  = 55

________________________

           1.25/ -1.25x=15/-1.25

                        x= 12

5.5(12) + z =10

66 -66 +z = 10 -66

           z = -56

2(12) + y + (-56) = 3

24  +  y  + (-56) = 3

-32 + 32 +y = 3 + 32

              y = 35

2(12) + 35 + (-56) = 3        

24+ 35 + (-56) = 3

                3=3

Answer 6

9 1/2 - 2y - 1/2z = 14

sorry i dont rmember the rest, but i hope this helps