Solve the system of equation.
x+y+z=6
3x-2y=4z=9
x-y-z=0
________
To solve the system of equations, we can use the method of substitution or the method of elimination. Let's use the method of substitution.
First, let's solve one equation for one variable in terms of the other variables and substitute it into the other equations.
From the third equation, we have:
x - y - z = 0
Rearranging this equation, we get:
x = y + z
Now, substitute this expression for x in the other equations.
Substituting x = y + z in the first equation, we get:
(y + z) + y + z = 6
2y + 2z = 6
Dividing both sides by 2, we have:
y + z = 3
Substituting x = y + z in the second equation, we get:
3(y + z) - 2y + 4z = 9
3y + 3z - 2y + 4z = 9
y + 7z = 9
Now, we have a system of two equations with two variables:
y + z = 3 ---(1)
y + 7z = 9 ---(2)
To solve this system, we can now use the method of substitution again, or solve it using the method of elimination.
Let's solve it using the method of elimination. Multiply equation (1) by 7 to get rid of y:
7(y + z) = 7 * 3
7y + 7z = 21 ---(3)
Now, subtract equation (3) from equation (2):
(y + 7z) - (7y + 7z) = 9 - 21
y + 7z - 7y - 7z = -12y
-6y = -12
y = -12 / -6
y = 2
Substituting this value of y back into equation (1), we have:
2 + z = 3
z = 3 - 2
z = 1
Now, substitute the values of y and z back into x = y + z:
x = 2 + 1
x = 3
Therefore, the solution to the system of equations is:
x = 3, y = 2, z = 1