Question

3. Solve:

2x + y + z = 1
x – y + 4z = 0
x + 2y – 2z = 3

Select the appropriate response

A) x=-3 y=5 z=2
B) x=2 y=-1 z=3
C) x=2 y=-1 z=3

Answer 1

Answer: A

Explanation:

Take 2 equations that make one of their letters disappear, and add them up:

2x + y + z = 1

x - y + 4z = 0

------------------------

3x + 5z = 1

Do the same with another 2 equations in which the letter, in this case y, can be removed. If you can't, take 2 of 3 equations and equal the value of the letter to make it eliminable.

x - y + 4z = 0

x + 2y - 2z = 3

Since we can't eliminate y, we have multiply as necessary to make it eliminable:

2 (x - y + 4z = 0)

= 2x - 2y + 8z = 0

add all up:

2x - 2y + 8z = 0

x + 2y - 2z = 3

-----------------------------

3x + 6z = 3

Now we've gone from a 3-variable equation to a 2-variable equation.

3x + 5z = 1

3x + 6z = 3

We can solve again by elimination; to get rid of z, for example, we cross multiply. the upper equation by 6 and the lower equation by 5. However, we have to make one of them negative in order to make them eliminable.

6 (3x + 5z = 1)

-5 (3x + 6z = 3)

----------------------------

18x + 30z = 6

-15x - 30z = -15

---------------------------

3x = - 9

Solving for x;

x =
`\frac {-9}{3}`

x = - 3

After finding one variable, we can use our 2-variable equations to find the next variable:

3x + 5z = 1

3 (-3) + 5z = 1

- 9 + 5z = 1

5z = 1 + 9

5z = 10

z =
`(10)/(5)`

z = 2

Having found these 2 variables, we can put them into one of our main 3-variable equations to find the last one:

2x + y + z = 1

2(-3) + y + 2 = 1

- 6 + y + 2 = 1

y - 4 = 1

y = 1 + 4

y = 5

And you've found all the variables in the equation; to prove if they're correct or not, you can replace them in any of the main equations and the result should be equal to each other:

x + 2y - 2z = 3

-3 + 2(5) - 2(2) = 3

- 3 + 10 - 4 = 3

10 - 7 = 3

3 = 3