Exercise 5 Page 29

We will apply the strategy from a previous exercise to solve the given linear system.

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

Our strategy consists of four steps.

Isolate either variable in any of the equations.
In the other equations, substitute the expression obtained in the first step for the variable. In this way we will get a system of two linear equations in two variables.
Solve the $2 \times 2$ system.
Substitute the obtained values for the corresponding variables in the equation obtained in the first step, and solve for the remaining variable.

Let's do it!

Step $1$

For simplicity, we will start by isolating the

x -

variable in Equation (I).

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

SubEqn

$(I):$ $LHS - y - z = RHS - y - z$

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

Step $2$

Now in Equation (II) and Equation (III) we will substitute the obtained expression,

1 - y - z,

for

x .

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

$(II), (III):$ $x = 1 - y - z$

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

▼

(II), (III):

Simplify

SubEqn

$(II):$ $LHS - 1 = RHS - 1$

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

Distr

$(III):$ Distribute $- 1$

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

AddEqn

$(III):$ $LHS + 1 = RHS + 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 - y - z - y - z - y - z = 2 y + z - y + z = 0

$(II), (III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 - y - z - 2 y - 2 z = 2 2 z = 0

Note that now Equation (II) and Equation (III) form a system of two equations in two variables.

{- 2 y - 2 z = 2 (II) 2 z = 0 (III)

Step $3$

Let's now solve the obtained

2 \times 2

system. We will start by solving Equation (III), since it only contains one variable.

{- 2 y - 2 z = 2 (II) 2 z = 0 (III)

DivEqn

$(III):$ $LHS / 2 = RHS / 2$

{- 2 y - 2 z = 2 z = 0

We found that

z = 0 .

To find the value of

y,

we will substitute

0

for

z

in Equation (II).

{- 2 y - 2 z = 2 (II) z = 0 (III)

Substitute

$(II):$ $z = 0$

{- 2 y - 2 (0) = 2 z = 0

▼

(II):

Solve for

y

ZeroProdProp

$(II):$ Use the Zero Product Property

{- 2 y - 0 = 2 z = 0

SubTerm

$(II):$ Subtract term

{- 2 y = 2 z = 0

DivEqn

$(II):$ $LHS / (- 2) = RHS / (- 2)$

{y = - 1 z = 0

We found that

y = - 1 .

Step $4$

Finally, we will substitute

- 1

and

0

for

y

and

z,

respectively, in Equation (I). Then, we will solve for the remaining variable,

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 - y - z y = - 1 z = 0 (I) (II) (III)

SubstituteII

$(I):$ $y = - 1$ , $z = 0$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 - (- 1) - 0 y = - 1 z = 0

▼

(I):

Simplify right-hand side

SubNeg

$(I):$ $a - (- b) = a + b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 + 1 - 0 y = - 1 z = 0

AddSubTerms

$(I):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 2 y = - 1 z = 0