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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 5 x + 9 z = - 12 - x + 2 z = - 9

▼

(II):

Solve by elimination

MultEqn

$(III):$ $LHS \cdot 5 = RHS \cdot 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 5 x + 9 z = - 12 - 5 x + 10 z = - 45

SysEqnAdd

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 5 x + 9 z + (- 5 x + 10 z) = - 12 + (- 45) - 5 x + 10 z = - 45

RemovePar

$(II):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 5 x + 9 z - 5 x + 10 z = - 12 - 45 - 5 x + 10 z = - 45

AddTerms

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 19 z = - 57 - 5 x + 10 z = - 45

DivEqn

$(II):$ $LHS / 19 = RHS / 19$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 z = - 3 - 5 x + 10 z = - 45

Now that we know that

z = - 3,

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

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 z = - 3 - 5 x + 10 z = - 45

Substitute

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 z = - 3 - 5 x + 10 (- 3) = - 45

▼

(III):

Solve for

x

MultPosNeg

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 z = - 3 - 5 x - 30 = - 45

AddEqn

$(III):$ $LHS + 30 = RHS + 30$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 2 y + z = 4 z = - 3 - 5 x = - 15

DivEqn

$(III):$ $LHS / (- 5) = RHS / (- 5)$

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

The value of

x

is

3 .

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

y .

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

SubstituteII

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

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

▼

(I):

Solve for

y

RemovePar

$(I):$ Remove parentheses

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

SubTerm

$(I):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 y = 4 z = - 3 x = 3

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 z = - 3 x = 3

The solution to the system is

(3, 2, - 3) .

This is the singular point at which all three planes intersect.

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