The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. Notice that in the third equation there is no

x -

term.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - y + 2 z = - 6 - 2 x + 3 y - z = 8 2 y + 3 z = - 5 (I) (II) (III)

By manipulating the first equation and adding it to the second equation, we can solve for

y .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - y + 2 z = - 6 - 2 x + 3 y - z = 8 2 y + 3 z = - 5 (I) (II) (III)

▼

(II):

Solve by elimination

MultEqn

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

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

SysEqnAdd

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

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

RemovePar

$(II):$ Remove parentheses

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

AddTerms

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - y + 2 z = - 6 5 y = 10 2 y + 3 z = - 5

DivEqn

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

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

Now that we know that

y = 2,

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

z .

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

Substitute

$(III):$ $y = 2$

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

▼

(III):

Solve for

z

Multiply

$(III):$ Multiply

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

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - y + 2 z = - 6 y = 2 3 z = - 9

DivEqn

$(III):$ $LHS / 3 = RHS / 3$

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

The value of

z

is

- 3 .

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

x .

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

SubstituteII

$(I):$ $y = 2$ , $z = - 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - 2 + 2 (- 3) = - 6 y = 2 z = - 3

▼

(I):

Solve for

x

MultPosNeg

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

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

SubTerm

$(I):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x - 8 = - 6 y = 2 z = - 3

AddEqn

$(I):$ $LHS + 8 = RHS + 8$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x = 2 y = 2 z = - 3

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{2}{4} y = 2 z = - 3

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = \frac{1}{2} y = 2 z = - 3

The solution to the system is

(\frac{1}{2}, 2, - 3) .

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

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