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 second equation, it will be easiest to isolate

x .

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

$(II):$ $LHS - 2 z = RHS - 2 z$

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

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

$(I), (III):$ $x = 5 - 2 z$

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

▼

(I), (III):

Simplify

RemovePar

$(I):$ Remove parentheses

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

Distr

$(III):$ Distribute $2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 5 - 2 z + y + z = 2 x = 5 - 2 z 10 - 4 z + y - z = - 1

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 5 - z + y = 2 x = 5 - 2 z 10 - 5 z + y = - 1

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - z + y = - 3 x = 5 - 2 z 10 - 5 z + y = - 1

AddEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z 10 - 5 z + y = - 1

This time, the

y -

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z 10 - 5 z + y = - 1

Substitute

$(III):$ $y = z - 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z 10 - 5 z + (z - 3) = - 1

▼

(III):

Solve for

z

RemovePar

$(III):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z 10 - 5 z + z - 3 = - 1

AddSubTerms

$(III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z 7 - 4 z = - 1

SubEqn

$(III):$ $LHS - 7 = RHS - 7$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = z - 3 x = 5 - 2 z - 4 z = - 8

DivEqn

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

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

The value of

z

is

2 .

Substituting

2

for

z

into the first and the second equations, we can find the values of

x

and

y .

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

$(I), (II):$ $z = 2$

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

▼

(I), (II):

Simplify

Multiply

$(II):$ Multiply

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

$(I), (II):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = - 1 x = 1 z = 2

The solution to the system is the point

(1, - 1, 2) .

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

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

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

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

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

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

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

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

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 = 2 ✓ 5 = 5 ✓ - 1 = - 1 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.

Exercise 28 Page 172

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