The given system consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the first equation, it will be easiest to isolate

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 4 y + z = 6 2 x + 5 y - z = 7 2 x - y - z = 1 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 4 y + z = - x + 6 2 x + 5 y - z = 7 2 x - y - z = 1

AddEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 2 x + 5 y - z = 7 2 x - y - z = 1

With a variable isolated in one of the equations, we can substitute its equivalent expression into the remaining equations. In the final step of the simplification of these substitutions, our goal is to have yet another variable isolated.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 2 x + 5 y - z = 7 2 x - y - z = 1

$(II), (III):$ $z = 6 - x + 4 y$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 2 x + 5 y - (6 - x + 4 y) = 7 2 x - y - (6 - x + 4 y) = 1

▼

(II), (III):

Simplify

$(II), (III):$ Distribute $(- 1)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 2 x + 5 y - 6 + x - 4 y = 7 2 x - y - 6 + x - 4 y = 1

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 3 x + y - 6 = 7 3 x - 5 y - 6 = 1

AddEqn

$(II):$ $LHS + 6 = RHS + 6$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y 3 x + y = 13 3 x - 5 y - 6 = 1

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 3 x - 5 y - 6 = 1

This time, the

y -

variable was isolated in the second equation. We can now substitute its equivalent expression into the third equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 3 x - 5 y - 6 = 1

Substitute

$(III):$ $y = 13 - 3 x$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 3 x - 5 (13 - 3 x) - 6 = 1

▼

(III):

Solve by substitution

Distr

$(III):$ Distribute $(- 5)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 3 x - 65 + 15 x - 6 = 1

AddSubTerms

$(III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 18 x - 71 = 1

AddEqn

$(III):$ $LHS + 71 = RHS + 71$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x 18 x = 72

DivEqn

$(III):$ $LHS / 18 = RHS / 18$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x x = 4

The value of

x

is

4 .

Substituting

4

for

x

into the second equation, we can find the value of

y .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 x x = 4

Substitute

$(II):$ $x = 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 3 (4) x = 4

Multiply

$(II):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 - x + 4 y y = 13 - 12 x = 4

SubTerm

$(II):$ Subtract term

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

Now that we know the values of

x

and

y,

we are able to find the value of

z .

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

SubstituteII

$(I):$ $x = 4$ , $y = 1$

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

▼

(I):

Simplify right-hand side

IdPropMult

$(I):$ Identity Property of Multiplication

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

AddSubTerms

$(I):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ z = 6 y = 1 x = 4

The solution to the system is the point

(4, 1, 6) .

This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x - 4 y + z = 6 2 x + 5 y - z = 7 2 x - y - z = 1 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 4 - 4 (1) + 6 = ? 6 2 (4) + 5 (1) - 6 = ? 7 2 (4) - 1 - 6 = ? 1

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 4 - 4 + 6 = ? 6 8 + 5 - 6 = ? 7 8 - 1 - 6 = ? 1

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 6 = 6 ✓ 7 = 7 ✓ 1 = 1 ✓

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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