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how to solve the system

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2 Answers

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 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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