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 a + 2 b + 2 c = - 6 a + 4 b = 1 4 a + b = - 11

▼

(II):

Solve by elimination

MultEqn

$(II):$ $LHS \cdot (- 4) = RHS \cdot (- 4)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 - 4 a - 16 b = - 4 4 a + b = - 11

SysEqnAdd

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 - 4 a - 16 b + (4 a + b) = - 4 + (- 11) 4 a + b = - 11

AddSubTerms

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 - 15 b = - 15 4 a + b = - 11

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 4 a + b = - 11

Now that we know that

b = 1,

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

a .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 4 a + b = - 11

Substitute

$(III):$ $b = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 4 a + 1 = - 11

▼

(III):

Solve for

a

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 4 a = - 12

DivEqn

$(III):$ $LHS / 4 = RHS / 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 a = - 3

The value of

a

is

- 3 .

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

c .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a + 2 b + 2 c = - 6 b = 1 a = - 3

SubstituteII

$(I):$ $a = - 3$ , $b = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 (- 3) + 2 (1) + 2 c = - 6 b = 1 a = - 3

▼

(I):

Solve for

c

Multiply

$(I):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 6 + 2 + 2 c = - 6 b = 1 a = - 3

AddTerms

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 4 + 2 c = - 6 b = 1 a = - 3

AddEqn

$(I):$ $LHS + 4 = RHS + 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 c = - 2 b = 1 a = - 3

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ c = - 1 b = 1 a = - 3

The solution to the system is

(- 3, 1, - 1) .

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = - 3 3 b - c = 4 2 a - b - 2 c = - 5 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ - 3 + 1 + (- 1) = ? - 3 3 (1) - (- 1) = ? 4 2 (- 3) - 1 - 2 (- 1) = ? - 5

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ - 3 + 1 + (- 1) = ? - 3 3 - (- 1) = ? 4 - 6 - 1 + 2 = ? - 5

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 3 = - 3 ✓ 4 = 4 ✓ - 5 = - 5 ✓

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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