Exercise 9 Page 171

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

y

in the first and the third equations are 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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y + z = - 1 x + y + 3 z = - 3 2 x - y + 2 z = 0 (I) (II) (III)

We can start by adding the second equation to the first and the third equations to eliminate the

y - terms .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y + z = - 1 x + y + 3 z = - 3 2 x - y + 2 z = 0 (I) (II) (III)

$(I), (III):$ Add $(II)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y + z + (x + y + 3 z) = - 1 + (- 3) x + y + 3 z = - 3 2 x - y + 2 z + (x + y + 3 z) = 0 + (- 3)

$(I), (III):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y + z + x + y + 3 z = - 1 - 3 x + y + 3 z = - 3 2 x - y + 2 z + x + y + 3 z = 0 - 3

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

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

Next, we will use our two equations that are only in terms of

x

and

z

to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time it will be similar to when using it in a system with only two variables.

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

▼

(III):

Solve by elimination

MultEqn

$(I):$ $LHS \cdot (- 3) = RHS \cdot (- 3)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 3 x + 5 z = - 3

MultEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 6 x + 10 z = - 6

SysEqnAdd

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 6 x + 10 z + (- 6 x - 12 z) = - 6 + (12)

RemovePar

$(III):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 6 x + 10 z - 6 x - 12 z = - 6 + 12

AddSubTerms

$(III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 - 2 z = 6

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 z = - 3

Now that we know that

z = - 3,

we can substitute it into the first equation to find the value of

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 z = 12 x + y + 3 z = - 3 z = - 3

Substitute

$(I):$ $z = - 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x - 12 (- 3) = 12 x + y + 3 z = - 3 z = - 3

▼

(I):

Solve for

x

MultNegNegOnePar

$(I):$ $- a (- b) = a \cdot b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x + 36 = 12 x + y + 3 z = - 3 z = - 3

SubEqn

$(I):$ $LHS - 36 = RHS - 36$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 x = - 24 x + y + 3 z = - 3 z = - 3

DivEqn

$(I):$ $LHS / (- 6) = RHS / (- 6)$

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

The value of

x

is

4 .

Let's substitute both values into the second equation to find

y .

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

SubstituteII

$(II):$ $x = 4$ , $z = - 3$

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

▼

(II):

Solve for

y

MultPosNeg

$(II):$ $a (- b) = - a \cdot b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 4 4 + y - 9 = - 3 z = - 3

SubTerm

$(II):$ Subtract term

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

AddEqn

$(II):$ $LHS + 5 = RHS + 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 4 y = 2 z = - 3

The solution to the system is

(4, 2, - 3) .

This is the singular point at which all three planes intersect. Now we can check our solution by substituting the values into the system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y + z = - 1 x + y + 3 z = - 3 2 x - y + 2 z = 0

$(I), (II), (III):$ Substitute values

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 4 - 2 + (- 3) = ? - 1 4 + 2 + 3 (- 3) = ? - 3 2 (4) - 2 + 2 (- 3) = ? 0

$(I), (II), (III):$ Multiply

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 4 - 2 - 3 = ? - 1 4 + 2 - 9 = ? - 3 8 - 2 - 6 = ? 0

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 1 = - 1 ✓ - 3 = - 3 ✓ 0 = 0 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.