What is the solution to the following system?

Question

Answer 1

There are multiple ways to solve this problem.

The answer is x=2,y=1, and z=0

I solved by using elimination. I took equation 3 and equation 1 to get

9y+6z=9

Then I multiplied every term in equation 2 by 3 and then I subtracted the new equation 2 from equation 3 to get

3y+6z=3

I subtracted second new equation from 9y+6z=9 and solved for y. Substitute y=1back into equations to get z and x.

Answer 2

x=2, y=1 and z=0

I used primarily elimination with one substitution.

To keep track I numbered the equations

Equation 1

3x + 3y + 6z = 9

Equation 2

x + 3y + 2z = 5

Equation 3

3x + 12y + 12z = 18

First multiplying Equation 2 by negative 3 and combining it with Equation 1 eliminates both x and z leaving a solution for y

Multiply Equation 2 by -3

-3(x + 3y + 2z = 5) gives

-3x - 9y - 6z = -15

Combine with Equation 1

3x + 3y + 6z = 9

-3x - 9y - 6z = -15 The x and z are eliminated

Leaving

-6y = -6

Divide both sides by -6

y = 1

Substitute this value for y Equations 1 and Equation 3 to give two equations in x and z

Equation 1

3x + 3y + 6z = 9

3x + 3(1) + 6z = 9

3x + 3 + 6z = 9

Subtract 3 from both sides of the equation

3x + 6z = 6 Your first equation in x and z

Do the same for Equation 3

Equation 3

3x + 12y + 12z = 18

3x + 12(1) + 12z = 18

3x + 12 + 12z = 18

Subtract 12 from both sides of the equation

3x + 12z = 6 Your second equation in x and z

Now we have two equations in two unknowns

3x + 6z = 6

3x + 12z = 6

We can easily eliminate x by multiplying either one (not both) of the equations by negative 1

-1(3x + 6z = 6) gives

-3x - 6z = -6

Combine this with our other equation in x and z we have

-3x - 6z = -6

3x + 12z = 6 The x is eliminated

This leaves

6z = 0

z= 0

Since we now have y and z we can plug those values into any one of the equations to find x, lets use

Equation 2

x + 3y - 2z = 5

x + 3(1) - 2(0) =5

x + 3 = 5

Subtract 3 from both sides of the equation to give

x = 2

Finally we have x = 2, y = 1, and z = 0

Check the values in all the original equations

Equation 1

3x + 3y + 6z = 9

3(2) + 3(1) + 6(0) = 9

6 + 3 + 0 = 9

9 = 9

Equation 2

x + 3y + 2z = 5

2 + 3(1) + 2(0) = 5

2 + 3 + 0 = 5

5 = 5

Equation 3

3x + 12y + 12z = 18

3(2) + 12(1) + 12(0) = 18

6 + 12 + 0 = 18