Exercise 17 Page 171

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

y

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

y

in the third equation; they will add to be

0 .

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 2 y + 3 z = 12 2 x - y - 2 z = 5 2 x + 2 y - z = 4 (I) (II) (III)

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

z - terms .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 2 y + 3 z = 12 2 x - y - 2 z = 5 2 x + 2 y - z = 4 (I) (II) (III)

$(I):$ Add $(III)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 2 y + 3 z + (2 x + 2 y - z) = 12 + (4) 2 x - y - 2 z = 5 2 x + 2 y - z = 4

$(I):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 2 y + 3 z + 2 x + 2 y - z = 12 + 4 2 x - y - 2 z = 5 2 x + 2 y - z = 4

$(I):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 2 x - y - 2 z = 5 2 x + 2 y - z = 4

Having eliminated the

y -

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

y

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

y

from the second equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 2 x - y - 2 z = 5 2 x + 2 y - z = 4

$(II):$ $LHS \cdot 2 = RHS \cdot 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 4 x - 2 y - 4 z = 10 2 x + 2 y - z = 4

$(II):$ Add $(III)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 4 x - 2 y - 4 z + (2 x + 2 y - z) = 10 + (4) 2 x + 2 y - z = 4

$(II):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 4 x - 2 y - 4 z + 2 x + 2 y - z = 10 + 4 2 x + 2 y - z = 4

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 2 z = 16 6 x - 5 z = 14 2 x + 2 y - z = 4

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