The given system consists of equations of planes. Notice that the coefficient of

x

in the first equation is the additive inverse of the coefficients of

x

in the second and the third equations; they will add to be

0 .

Let's use the Elimination Method to find a solution to this system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y - 2 z = 4 - x + 2 y + z = 1 - x + y - 3 z = 11 (I) (II) (III)

We can start by adding the first equation to the second and the third equations to eliminate the

x - terms .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y - 2 z = 4 - x + 2 y + z = 1 - x + y - 3 z = 11 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y - 2 z = 4 - x + 2 y + z + (x - y - 2 z) = 1 + (4) - x + y - 3 z + (x - y - 2 z) = 11 + (4)

$(II), (III):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y - 2 z = 4 - x + 2 y + z + x - y - 2 z = 1 + 4 - x + y - 3 z + x - y - 2 z = 11 + 4

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

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

Having eliminated the

y -

variable from the second and the third equations, we can continue by solving for

z

in the third equation.

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

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

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

Next, we will substitute the value of

z

into the second equation and solve for

y .

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

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

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

▼

(II):

Solve for

y

$(II):$ $a - (- b) = a + b$

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

$(II):$ $LHS - 3 = RHS - 3$

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

The value of

y

is

2 .

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

x .

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

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

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

▼

(I):

Solve for

x

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

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

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x + 4 = 4 y = 2 z = - 3

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 0 y = 2 z = - 3

The solution to the system is

(0, 2, - 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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - y - 2 z = 4 - x + 2 y + z = 1 - x + y - 3 z = 11 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 0 - 2 - 2 (- 3) = ? 4 - 0 + 2 (2) + (- 3) = ? 1 - 0 + 2 - 3 (- 3) = ? 11

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 0 - 2 + 6 = ? 4 0 + 4 + (- 3) = ? 1 0 + 2 + 9 = ? 11

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 = 4 ✓ 1 = 1 ✓ 11 = 11 ✓

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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