SOLUTION: can you please help me in solving for x;y;z x+y-2z=5 .........equation 1 x+z=4.......equation 2 y-z =6........equation 3

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Question 965464: can you please help me in solving for x;y;z
x+y-2z=5 .........equation 1
x+z=4.......equation 2
y-z =6........equation 3

Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
3 equations need to be solved simultaneously.

the equations are:

x + y - 2z = 5 (e1)
x + z = 4 (e2)
y - z = 6 (e3

subtract e2 from e1 to get:

x + y - 2z = 5 minus:
x + 0 + z = 4 becomes:
0 + 6 - 3z = 1

the result of the operation is y - 3z = 1 (e4).

take e3 and e4 and solve them simultaneously for either y or z.

the equations are:

y - z = 6 (e3)
y - 3z = 1 (e4)

subtract e4 from e3 to get:

2z = 5

solve for z to get z = 2.5

in e3, if z = 2.5, then y - z = 6 becomes y - 2.5 = 6 which becomes y = 8.5.

in e1, if z = 2.5 and y = 8.5, then x + y - 2z = 5 becomes x + 8.5 - 2 * 2.5 = 5 becomes x + 3.5 = 5 which becomes x = 1.5

if all was done correctly, your solution should be:

x = 1.5
y = 8.5
z = 2.5

let's see if that's true.
it has to be true in all 3 equations.

the original 3 equations are:

x + y - 2z = 5 (e1)
x + z = 4 (e2)
y - z = 6 (e3)

when x = 1.5 and y = 8.5 and z = 2.5, these equations become:

1.5 + 8.5 - (2 * 2.5) = 5 (e1) which becomes 5 = 5 which is true.
1.5 + 2.5 = 4 (e2) which becomes 4 = 4 which is true.
8.5 - 2.5 = 6 (e3) which becomes 6 = 6 which is true.

all 3 equations are true, so the solution is good.

x = 1.5
y = 8.5
z = 2.5

Answer by Edwin McCravy(20054) (Show Source):
You can put this solution on YOUR website!

x+y-2z=5 .........equation 1 x+z=4.......equation 2 y-z =6........equation 3 Rule for special cases of systems, that applies here: When one of the equations has no term in one of the letters, eliminate that letter from the other two. There are two ways we can go here. Equation 3 has no term in x so we would eliminate x from equations 1 and 2. Then we'd have two equations in only y and z. OR Equation 2 has no term in y so we would eliminate y from equations 1 and 3. Then we'd have two equations in only x and z. I'll choose the second way: x + y - 2z = 5 .........equation 1 y - z = 6........equation 3 Multiply equation 3 through by -1 so that y will become -y and will cancel the +y in the first equation: x + y - 2z = 5 .........equation 1 -y + z = -6........equation 3 multiplied through by -1. ------------------------------------------------------------- x - z = -1........equation 4 found by adding term by term. Now when we put equation 4 together with equation 2 and we have this system: Adding the two equations just as they are gives Substituting in Multiply through by LCD = 2 Substituting in y - z = 6........equation 3 Multiply through by LCD = 2 Solution: , , and Edwin