Exercise 16 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 second equation; they will add to be

0 .

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

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

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

y - terms .

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

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

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

$(II):$ Remove parentheses

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

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x - y + z = 3 4 x + 3 z = 7 x + 2 y + z = 4

Having eliminated the

y -

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

y

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

y

from the first equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x - y + z = 3 4 x + 3 z = 7 x + 2 y + z = 4

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 6 x - 2 y + 2 z = 6 4 x + 3 z = 7 x + 2 y + z = 4

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 6 x - 2 y + 2 z + (x + 2 y + z) = 6 + (4) 4 x + 3 z = 7 x + 2 y + z = 4

$(I):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 6 x - 2 y + 2 z + x + 2 y + z = 6 + 4 4 x + 3 z = 7 x + 2 y + z = 4

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 7 x + 3 z = 10 4 x + 3 z = 7 x + 2 y + z = 4

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