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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Systems With Three Variables

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Exercise 36 Page 172

Can you manipulate the coefficients of any variable terms such that they could be eliminated?

(-2,-1,12)

Practice makes perfect

The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. Notice that in the third equation there is no c-term. 4a+2b+c=2 & (I) 5a-3b+2c=17 & (II) a-5b=3 & (III) Currently, none of the terms in this system will cancel out. However, if we multiply (I) by -2 the coefficient of c in this equation will be the additive inverse of the coefficient of c in the second equation; they will add to be 0. -2(4a+2b+c)=-2(2) 5a-3b+2c=17 a-5b=3 ⇓ -8a-4b - 2c=-4 5a-3b + 2c=17 a-5b=3 We can start by adding the second equation to the first equation to eliminate the c-terms.

-8a-4b-2c=-4 & (I) 5a-3b+2c=17 & (II) a-5b=3 & (III)

(I): Add (II)

-8a-4b-2c+( 5a-3b+2c)=-4+( 17) 5a-3b+2c=17 a-5b=3

(I): Remove parentheses

-8a-4b-2c+5a-3b+2c=-4+17 5a-3b+2c=17 a-5b=3

(I): Add and subtract terms

-3a-7b=13 5a-3b+2c=17 a-5b=3

Next, we will use our two equations that are only in terms of a and b 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.

-3a-7b=13 5a-3b+2c=17 a-5b=3

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(III): Solve by elimination

(III): LHS * 3=RHS* 3

-3a-7b=13 5a-3b+2c=17 3a-15b=9

(III): Add (I)

-3a-7b=13 5a-3b+2c=17 3a-15b+( -3a-7b)=9+( 13)

(III): Remove parentheses

-3a-7b=13 5a-3b+2c=17 3a-15b-3a-7b=9+13

(III): Add and subtract terms

-3a-7b=13 5a-3b+2c=17 -22b=22

(III): .LHS /(-22).=.RHS /(-22).

-3a-7b=13 5a-3b+2c=17 b=-1

Now that we know that b=-1, we can substitute it into the first equation to find the value of a.

-3a-7b=13 5a-3b+2c=17 b=-1

(I): b= -1

-3a-7( -1)=13 5a-3b+2c=17 b=-1

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(I): Solve for a

MultNegNegOnePar

(I): - a(- b)=a* b

-3a+7=13 5a-3b+2c=17 b=-1

(I): LHS-7=RHS-7

-3a=6 5a-3b+2c=17 b=-1

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

a=-2 5a-3b+2c=17 b=-1

The value of a is -2. Let's substitute the value of a and b into the second equation to find c.

a=-2 5a-3b+2c=17 b=-1

(II): a= -2, b= -1

a=-2 5( -2)-3( -1)+2c=17 b=-1

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(II): Solve for c

(II): Multiply

a=-2 -10+3+2c=17 b=-1

(II): Add terms

a=-2 -7+2c=17 b=-1

(II): LHS+7=RHS+7

a=-2 2c=24 b=-1

(II):.LHS /2.=.RHS /2.

a=-2 c=12 b=-1

The solution to the system is (-2,-1,12). This is the singular point at which all three planes intersect.

Subchapter links

Lesson Check p.171

Practice and Problem-Solving Exercises p.171-173

Standardized Test Prep p.173

Mixed Review p.173