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.

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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)

_____________

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

**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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