The given system consists of equations of planes. We will use the Substitution Method to solve this system because in the second and the third equations there are isolated variables. The first step will be to substitute the values of

ℓ

and

h

into the first equation and solve for

w .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 ℓ + 2 w + h = 72 ℓ = 3 w h = 2 w

SubstituteII

$(I):$ $ℓ = 3 w$ , $h = 2 w$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 (3 w) + 2 w + 2 w = 72 ℓ = 3 w h = 2 w

▼

(I):

Solve for

w

Multiply

$(I):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 6 w + 2 w + 2 w = 72 ℓ = 3 w h = 2 w

AddTerms

$(I):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 10 w = 72 ℓ = 3 w h = 2 w

DivEqn

$(I):$ $LHS / 10 = RHS / 10$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w = \frac{7 2}{1 0} ℓ = 3 w h = 2 w

The value of

w

is

\frac{7 2}{1 0} .

Substituting

\frac{7 2}{1 0}

for

w

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

ℓ

and

h .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w = \frac{7 2}{1 0} ℓ = 3 w h = 2 w

$(II), (III):$ $w = \frac{7 2}{1 0}$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w = \frac{7 2}{1 0} ℓ = 3 (\frac{7 2}{1 0}) h = 2 (\frac{7 2}{1 0})

$(II), (III):$ $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w = \frac{7 2}{1 0} ℓ = \frac{3 ( 7 2 )}{1 0} h = \frac{2 ( 7 2 )}{1 0}

$(II), (III):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w = \frac{7 2}{1 0} ℓ = \frac{2 1 6}{1 0} h = \frac{1 4 4}{1 0}

The solution to the system is the point

(\frac{7 2}{1 0}, \frac{2 1 6}{1 0}, \frac{1 4 4}{1 0}) .

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

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