Exercise 3 Page 171

The given system consists of equations of planes. Notice that the coefficient of

z

in the first equation is the additive inverse of the coefficient of

z

in the second equation; they will add to be

0 .

Let's use the Elimination Method to find a solution to this system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 x + 3 y - 2 z = 1 - x - y + 2 z = 5 3 x + 2 y - 3 z = - 6 (I) (II) (III)

We can start by adding the second equation to the first equation to eliminate the

z - terms .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 x + 3 y - 2 z = 1 - x - y + 2 z = 5 3 x + 2 y - 3 z = - 6 (I) (II) (III)

$(I):$ Add $(II)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 x + 3 y - 2 z + (- x - y + 2 z) = 1 + 5 - x - y + 2 z = 5 3 x + 2 y - 3 z = - 6

$(I):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 x + 3 y - 2 z - x - y + 2 z = 1 + 5 - x - y + 2 z = 5 3 x + 2 y - 3 z = - 6

Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y = 6 - x - y + 2 z = 5 3 x + 2 y - 3 z = - 6

Having eliminated the

z -

variable from the first equation, we can continue by creating additive inverse coefficients for

z

in the second and third equations. Then, we can add or subtract these equations to eliminate

z

from the second equation.