Question

X+y-z=1
2x-2y+3z=8
2x-y+2z=9

Answer 1

Hmmm. I get (0,11,10)

x+y-z=1
2x-2y+3z=8
2x-y+2z=9
subtract #3 and 2*#1 from #2 and we have
-4y+5z = 6
-y+z = -1
Now use z = y-1 to get
-4y+5(y-1) = 6
-y+y+8 = 8
So now we have
y = 11
z=10
x = z-y+1 = 0

Answer 2

We can solve the system of equations using the method of substitution or elimination.

Using substitution:
From the first equation, we can express x in terms of y and z:
x = 1 - y + z

Substituting this expression for x in the second and third equations, we get:
2(1 - y + z) - 2y + 3z = 8
2(1 - y + z) - y + 2z = 9

Simplifying these equations:
2 - 2y + 2z - 2y + 3z = 8
2 - 2y + 2z - y + 2z = 9

Combine like terms:
4z - 4y = 6
3z - 3y = 7

Now, we can solve this system of linear equations using the method of elimination.

Multiplying the second equation by 4:
4(3z - 3y) = 4(7)
12z - 12y = 28

Now we have two equations:
4z - 4y = 6
12z - 12y = 28

We can subtract the first equation from the second to eliminate y:
(12z - 12y) - (4z - 4y) = 28 - 6
8z = 22
z = 22/8
z = 2.75

Substituting the value of z back into the first equation:
4(2.75) - 4y = 6
11 - 4y = 6
-4y = 6 - 11
-4y = -5
y = (-5)/(-4)
y = 1.25

Finally, substituting the values of y and z back into the first equation:
x + 1.25 - 2.75 = 1
x - 1.5 = 1
x = 2.5

The solution to the system of equations is x = 2.5, y = 1.25, and z = 2.75.

Answer 3

Apologies for the error in my previous response. You are correct. The solution to the system of equations is indeed x = 0, y = 11, and z = 10. Thank you for catching that mistake.