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Pearson Algebra 2 Common Core, 2011

PA

Pearson Algebra 2 Common Core, 2011 View details

5. Systems With Three Variables

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Exercise 29 Page 172

When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.

$(1, 3, 2)$

Practice makes perfect

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 .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 5 x - y + z = 4 x + 2 y - z = 5 2 x + 3 y - 3 z = 5 (I) (II) (III)

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

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

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.

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

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

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

▼

(I), (III):

Simplify

$(I):$ Distribute $5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 25 - 10 y + 5 z - y + z = 4 x = 5 - 2 y + z 2 (5 - 2 y + z) + 3 y - 3 z = 5

$(III):$ Distribute $2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 25 - 10 y + 5 z - y + z = 4 x = 5 - 2 y + z 10 - 4 y + 2 z + 3 y - 3 z = 5

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z + 25 = 4 x = 5 - 2 y + z - y - z + 10 = 5

$(I):$ $LHS - 25 = RHS - 25$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z = - 21 x = 5 - 2 y + z - y - z + 10 = 5

$(III):$ $LHS - 10 = RHS - 10$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z = - 21 x = 5 - 2 y + z - y - z = - 5

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z = - 21 x = 5 - 2 y + z - z = y - 5

$(III):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z = - 21 x = 5 - 2 y + z z = - y + 5

This time, the

z -

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 z = - 21 x = 5 - 2 y + z z = - y + 5

$(I):$ $z = - y + 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y + 6 (- y + 5) = - 21 x = 5 - 2 y + z z = - y + 5

▼

(I):

Solve by substitution

$(I):$ Distribute $6$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 11 y - 6 y + 30 = - 21 x = 5 - 2 y + z z = - y + 5

$(I):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 17 y + 30 = - 21 x = 5 - 2 y + z z = - y + 5

$(I):$ $LHS - 30 = RHS - 30$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 17 y = - 51 x = 5 - 2 y + z z = - y + 5

$(I):$ $LHS / (- 17) = RHS / (- 17)$

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

The value of

y

is

3 .

Substituting

3

for

y

into the third equation, we can find the value of

z .

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

$(III):$ $y = 3$

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

$(III):$ Add terms

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

Now that we know the values of

y

and

z,

we are able to find the value of

x .

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

$(II):$ $y = 3$ , $z = 2$

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

▼

(II):

Simplify

$(II):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 3 x = 5 - 6 + 2 z = 2

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 3 x = 1 z = 2

The solution to the system is the point

(1, 3, 2) .

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 5 x - y + z = 4 x + 2 y - z = 5 2 x + 3 y - 3 z = 5 (I) (II) (III)

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 5 (1) - 3 + 2 = ? 4 1 + 2 (3) - 2 = ? 5 2 (1) + 3 (3) - 3 (2) = ? 5

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

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 5 - 3 + 2 = ? 4 1 + 6 - 2 = ? 5 2 + 9 - 6 = ? 5

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 = 4 5 = 5 5 = 5

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

Subchapter links

Lesson Check p.171

Practice and Problem-Solving Exercises p.171-173

Standardized Test Prep p.173

Mixed Review p.173