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.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 4 q + s = 3 16 q + 9 s = 17

▼

(II):

Solve by elimination

MultEqn

$(II):$ $LHS \cdot (- 4) = RHS \cdot (- 4)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 - 16 q - 4 s = - 12 16 q + 9 s = 17

SysEqnAdd

$(II):$ Add $(III)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 - 16 q - 4 s + (16 q + 9 s) = - 12 + (17) 16 q + 9 s = 17

RemovePar

$(II):$ Remove parentheses

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 - 16 q - 4 s + 16 q + 9 s = - 12 + 17 16 q + 9 s = 17

AddTerms

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 5 s = 5 16 q + 9 s = 17

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 16 q + 9 s = 17

Now that we know that

s = 1,

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

q .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 16 q + 9 s = 17

Substitute

$(III):$ $s = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 16 q + 9 (1) = 17

▼

(III):

Solve for

q

IdPropMult

$(III):$ Identity Property of Multiplication

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 16 q + 9 = 17

SubEqn

$(III):$ $LHS - 9 = RHS - 9$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 16 q = 8

DivEqn

$(III):$ $LHS / 16 = RHS / 16$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 q = \frac{8}{1 6}

ReduceFrac

$(III):$ $\frac{a}{b} = \frac{a / 8}{b / 8}$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 q = \frac{1}{2}

The value of

q

is

\frac{1}{2} .

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

r .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 q - 3 r + 6 s = 24 s = 1 q = \frac{1}{2}

SubstituteII

$(I):$ $q = \frac{1}{2}$ , $s = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 18 (\frac{1}{2}) - 3 r + 6 (1) = 24 s = 1 q = \frac{1}{2}

▼

(I):

Solve for

r

Multiply

$(I):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 - 3 r + 6 = 24 s = 1 q = \frac{1}{2}

AddTerms

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 15 - 3 r = 24 s = 1 q = \frac{1}{2}

SubEqn

$(I):$ $LHS - 15 = RHS - 15$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 3 r = 9 s = 1 q = \frac{1}{2}

DivEqn

$(I):$ $LHS / (- 3) = RHS / (- 3)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ r = - 3 s = 1 q = \frac{1}{2}

The solution to the system is

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

This is the singular point at which all three planes intersect. Now, we can check our solution by substituting the values into the system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 12 q - 2 r + 4 s = 16 2 q + 3 r - s = - 9 4 q + 2 r + 5 s = 1 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 12 (\frac{1}{2}) - 2 (- 3) + 4 (1) = ? 16 2 (\frac{1}{2}) + 3 (- 3) - 1 = ? - 9 4 (\frac{1}{2}) + 2 (- 3) + 5 (1) = ? 1

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 6 + 6 + 4 = ? 16 1 - 9 - 1 = ? - 9 2 - 6 + 5 = ? 1

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 16 = 16 ✓ - 9 = - 9 ✓ 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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