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Pearson Algebra 2 Common Core, 2011 View details

5. Quadratic Equations

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Exercise 69 Page 231

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

(7,1,- 1)

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The given system consists of equations of planes. Notice that the coefficient of z in the first equation is the additive inverse of the coefficient of z in the third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. x-9y+8z=- 10 & (I) x+y-z=9 & (II) - x-9z=2 & (III) ⇓ 1x-9y+8z=- 10 & (I) 1x+y-z=9 & (II) - 1x-9z=2 & (III) We can start by adding the third equation to the first equation to eliminate the x-terms.

x-9y+8z=- 10 & (I) x+y-z=9 & (II) - x-9z=2 & (III)

SysEqnAdd

(I): Add (III)

x-9y+8z+( - x-9z)=- 10+ 2 x+y-z=9 - x-9z=2

AddSubTerms

(I): Add and subtract terms

- 9y-z=- 8 x+y-z=9 - x-9z=2

We eliminated the x-variable from the first equation. Now, let's add the third equation to the second equation to eliminate the x-term once more.

- 9y-z=- 8 x+y-z=9 - x-9z=2

▼

Solve by elimination

SysEqnAdd

(II): Add (III)

- 9y-z=- 8 x+y-z+( - x-9z)=9+ 2 - x-9z=2

AddSubTerms

(II): Add and subtract terms

- 9y-z=- 8 y-10z=11 - x-9z=2

Next, we use our two equations that are only in terms of y and z to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time will be similar to when using it in a system with only two variables.

- 9y-z=- 8 y-10z=11 - x-9z=2

▼

Solve by elimination

MultEqn

(II): LHS * 9=RHS* 9

- 9y-z=- 8 9y-90z=99 - x-9z=2

SysEqnAdd

(II): Add (I)

- 9y-z=- 8 9y-90z+( - 9y-z)=99+( - 8) - x-9z=2

AddNeg

(II): a+(- b)=a-b

- 9y-z=- 8 9y-90z-9y-z=99-8 - x-9z=2

SubTerms

(II): Subtract terms

- 9y-z=- 8 - 91z=91 - x-9z=2

DivEqn

(II): .LHS /(- 91).=.RHS /(- 91).

- 9y-z=- 8 z=- 1 - x-9z=2

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

- 9y-z=- 8 z=- 1 - x-9z=2

Substitute

(I): z= - 1

- 9y-( - 1)=- 8 z=- 1 - x-9z=2

▼

(I): Solve for y

SubNeg

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

- 9y+1=- 8 z=- 1 - x-9z=2

SubEqn

(I): LHS-1=RHS-1

- 9y=- 9 z=- 1 - x-9z=2

DivEqn

(I): .LHS /(- 9).=.RHS /(- 9).

y=1 z=- 1 - x-9z=2

The value of y is 1. Let's substitute the value of z into the third equation to find the value of x.

y=1 z=- 1 - x-9z=2

Substitute

(III): z= -1

y=1 z=- 1 - x-9( - 1)=2

▼

(III): Solve for x

MultNegNegOnePar

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

y=1 z=- 1 - x+9=2

SubEqn

(III): LHS-9=RHS-9

y=1 z=- 1 - x=- 7

MultEqn

(III): LHS * (- 1)=RHS* (- 1)

y=1 z=- 1 x=7

The solution to the system is ( 7, 1, - 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.

x-9y+8z=- 10 & (I) x+y-z=9 & (II) - x-9z=2 & (III)

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

7-9( 1)+8( - 1)? =- 10 7+ 1-( - 1)? =9 - 7-9( - 1)? =2

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

7-9-8? =- 10 7+1+1? =9 - 7+9? =2

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

- 10=- 10 ✓ 9=9 ✓ 2=2 ✓

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.229

Practice and Problem-Solving Exercises p.229-231

Standardized Test Prep p.231

Mixed Review p.231

Exercise 69 Page 231

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