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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 5 y = - 17 5 x + 7 y = - 23 6 x + 12 y - 6 z = - 60

▼

(II):

Solve by elimination

MultEqn

$(I):$ $LHS \cdot (- 5) = RHS \cdot (- 5)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 5 x + 7 y = - 23 6 x + 12 y - 6 z = - 60

MultEqn

$(II):$ $LHS \cdot 3 = RHS \cdot 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 15 x + 21 y = - 69 6 x + 12 y - 6 z = - 60

SysEqnAdd

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 15 x + 21 y + (- 15 x - 25 y) = - 69 + (85) 6 x + 12 y - 6 z = - 60

RemovePar

$(II):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 15 x + 21 y - 15 x - 25 y = - 69 + 85 6 x + 12 y - 6 z = - 60

AddSubTerms

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 - 4 y = 16 6 x + 12 y - 6 z = - 60

DivEqn

$(II):$ $LHS / (- 4) = RHS / (- 4)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 y = - 4 6 x + 12 y - 6 z = - 60

Now that we know that

y = - 4,

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

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 y = 85 y = - 4 6 x + 12 y - 6 z = - 60

Substitute

$(I):$ $y = - 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x - 25 (- 4) = 85 y = - 4 6 x + 12 y - 6 z = - 60

▼

(I):

Solve for

x

MultNegNegOnePar

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x + 100 = 85 y = - 4 6 x + 12 y - 6 z = - 60

SubEqn

$(I):$ $LHS - 100 = RHS - 100$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 15 x = - 15 y = - 4 6 x + 12 y - 6 z = - 60

DivEqn

$(I):$ $LHS / (- 15) = RHS / (- 15)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 6 x + 12 y - 6 z = - 60

The value of

x

is

1 .

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

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 6 x + 12 y - 6 z = - 60

SubstituteII

$(III):$ $x = 1$ , $y = - 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 6 (1) + 12 (- 4) - 6 z = - 60

▼

(III):

Solve for

z

Multiply

$(III):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 6 - 48 - 6 z = - 60

SubTerms

$(III):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 - 42 - 6 z = - 60

AddEqn

$(III):$ $LHS + 42 = RHS + 42$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 - 6 z = - 18

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 1 y = - 4 z = 3

The solution to the system is

(1, - 4, 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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 x + 3 y + 6 z = 9 2 x + y + 3 z = 7 x + 2 y - z = - 10 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 3 (1) + 3 (- 4) + 6 (3) = ? 9 2 (1) + (- 4) + 3 (3) = ? 7 1 + 2 (- 4) - 3 = ? - 10

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 3 - 12 + 18 = ? 9 2 + (- 4) + 9 = ? 7 1 - 8 - 3 = ? - 10

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 = 9 ✓ 7 = 7 ✓ - 10 = - 10 ✓

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

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