<span title="2x-y-z=6,3x+2y+3z=3,4x+y-2z=3?">2x-y-z=6,3x+2y+3z=3,4x+y-...</span>

Question

2x-y-z=6,3x+2y+3z=3,4x+y-...

Answer 1

2x - y - z = 6

3x + 2y +3z = 3

4x +y - 2z = 3

Adding Equation 1 to 3 you get

6x - 3z = 9

Doubling equatin 3 yields

8x + 2y - 4z = 6

taking this and substracting equation number 2 yiedls

5x - 7z =  3

we also know that

6x - 3z = 9 from above so multiplying the first by 3 and the second by 7 yields the follwing two

15x - 21z = 9

42x - 21z = 63

substracting the second from the first yields

27x = 54

x = 54/27 = 2

10 - 7z = 3

7 = 7z so z = 1

y = - 3

Answer 2

2x-y-z=6

4x-2y-2z=12 ; 4x-2z = 12+2y

4x+y-2z=3 ; 4x-2z = 3+y

3+y = 12+2y

y=-3

2x-z=3 ; z=2x-3

3x+3z=9

3x+3(2x-3)=9

9x=18

x=2

4+3-z=6

z=1

Answer: x=2, y=-3, z=1

Answer 3

(1)  2x - y - z = 6

(2)  3x + 2y + 3z = 3

(3)  4x + y - 2z = 3

So, rearranging (1) for x gives:

x = 1/2 * (6+y+z)

Substituting into (2) &amp; (3)

(2&#039;)  3 * 1/2 * (6+y+z) + 2y + 3z = 3

==&gt;  18 + 3y + 3z + 4y + 6z = 6

==&gt; 7y + 9z = -12

(3&#039;) 4 * 1/2 * (6+y+z)  + y - 2z = 3

==&gt; 12 + 2y + 2z + y - 2z = 3

==&gt;  3y = -9

==&gt; y = -3

Now from (2&#039;)

  9z = -12 - 7y = -12 + 21 = 9

==&gt; z = 1

And from (1)

  2x - 1 + 3 = 6

==&gt; x = 2