how to solve the system
2x-3y=0
6y-6z=1
x+3z=1
If the system is dependent or inconsistant, indicate this. Show your work.
how to solve the system
2x-3y=0
6y-6z=1
x+3z=1
If the system is dependent or inconsistant, indicate this. Show your work.
Let's rewrite each equation with all three variables:
2x - 3y + 0z = 0 (Eqn. 1)
0x + 6y - 6z = 1 (2)
x + 0y + 3z = 1 (3)
I notice I can eliminate y by multiplying the first equation by 2 and adding all equations
So now we have:
4x - 6y + 0z = 0 (1)
0x + 6y - 6z = 1 (2)
x + 0y + 3z = 1 (3)
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5x + 0y - 3z = 2 or 5x - 3z = 2
also x + 3z = 1
Add these together and we have 6x = 3, so x = 1/2
Now substitute into (1): 2(1/2) - 3y = 0, so 1-3y = 0 and y = 1/3
From (3) we know that (1/2) + 3z = 1, so 3z = 1/2 and z = 1/6
1.) Write 2 of the variables in terms of the 3rd
Eq. 1 2x-3y=0
Eq. 2 6y-6z=1
Eq. 3 x+3z=1
Eq. 3 becomes:
x = 1-3z
Eq. 2 becomes:
6y = 1+ 6z
y = 1/6 + z
2.) Use Eq. 1 to solve for z
2(1 - 3z) - 3(1/6 + z) = 0
2 - 6z - 3/6 - 3z = 0
-9z + 3/2 = 0
-9z = -3/2
z = 1/6
3.) Then, solve for x and y using the value of z
x = 1-3 (1/6)
x = 1/2
y = 1/6 + (1/6)
y = 1/3
So you have: x = 1/2; y=1/3; z=1/6
Since you were able to solve for values of the three variables, the system is independent and consistent.
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