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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 3 y + z = 6 2 x - 2 y - 4 z = 2

▼

(II):

Solve by elimination

MultEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 - 15 y - 5 z = - 30 2 x - 2 y - 4 z = 2

SysEqnAdd

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 - 15 y - 5 z + (y + 5 z) = - 30 + (16) 2 x - 2 y - 4 z = 2

RemovePar

$(II):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 - 15 y - 5 z + y + 5 z = - 30 + 16 2 x - 2 y - 4 z = 2

AddTerms

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 - 14 y = - 14 2 x - 2 y - 4 z = 2

DivEqn

$(II):$ $LHS / (- 14) = RHS / (- 14)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 y = 1 2 x - 2 y - 4 z = 2

Now that we know that

y = 1,

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

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y + 5 z = 16 y = 1 2 x - 2 y - 4 z = 2

Substitute

$(I):$ $y = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 + 5 z = 16 y = 1 2 x - 2 y - 4 z = 2

▼

(I):

Solve for

z

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 5 z = 15 y = 1 2 x - 2 y - 4 z = 2

DivEqn

$(I):$ $LHS / 5 = RHS / 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x - 2 y - 4 z = 2

The value of

z

is

3 .

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

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x - 2 y - 4 z = 2

SubstituteII

$(III):$ $y = 1$ , $z = 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x - 2 (1) - 4 (3) = 2

▼

(III):

Solve for

x

Multiply

$(III):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x - 2 - 12 = 2

SubTerms

$(III):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x - 14 = 2

AddEqn

$(III):$ $LHS + 14 = RHS + 14$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 2 x = 16

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 3 y = 1 x = 8

The solution to the system is

(8, 1, 3) .

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

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