Question

How do you solve the system of equations algebraically `x-3z=7, 2x+y-2z=11, -x-2y+9z=13`?

Answer 1

`x = 10`; `y = -7`; `z = 1`

Explanation:

Write out the three equations to line up the unknown values. This will give you a clue how to proceed.

Always be careful of the signs `(-) (+)` as they move across the `(=)`. Always write the sign with the value to which it refers. `(-3z)`

`x -3z = 7`

`2x +y -2z = 11`

`-x -2y +9z = 13`

Already we can see that it is easy to solve for `x` in the first equation:

`x -3z = 7 >> x = 3z +7`

Insert the value for `x` into the second equation;

`2(3z +7) +y -2z = 11`

`6z +14 + -2z = 11`

`4z +y = -3`

`y = -4z -3`

Substitute the values of `x` and `y` into the third equation:

`-(3z+7) -2(-4z-3) +9z = 13`

`-3z -7 +8z +6 +9z = 13`

`14z = 14`

`z = 1`

Substitute the value of `z` into the first equation:

`x -3(1) = 7`

`x = 10`

Substitute the values of `x` and `z` into the second equation:

`2(10) +y -2(1) = 11`

`20 +y -2 = 11`

`y = -20 +2 +11`

`y = -7`

Use either equation that includes all three unknowns to check:

`-x -2y +9z = 13`

`-(10) -2(-7) +9(1) = 13`

`13 = 13`