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Pearson Algebra 2 Common Core, 2011 View details

5. Quadratic Equations

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Exercise 68 Page 231

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

(- 2,1,5)

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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 second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. - 2x+9y - z=8 & (I) 3x-4y + z=- 5 & (II) 5x+5y-z=- 10 & (III) We can start by adding the second equation to the first equation to eliminate the z-terms.

- 2x+9y-z=8 & (I) 3x-4y+z=- 5 & (II) 5x+5y-z=- 10 & (III)

SysEqnAdd

(I): Add (II)

- 2x+9y-z+ 3x-4y+z=8+( - 5) 3x-4y+z=- 5 5x+5y-z=- 10

▼

(I): Add and subtract terms

AddNeg

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

- 2x+9y-z+3x-4y+z=8-5 3x-4y+z=- 5 5x+5y-z=- 10

AddSubTerms

(I): Add and subtract terms

x+5y=3 3x-4y+z=- 5 5x+5y-z=- 10

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

x+5y=3 3x-4y+z=- 5 5x+5y-z=- 10

▼

Solve by elimination

SysEqnAdd

(III): Add (II)

x+5y=3 3x-4y+z=- 5 5x+5y-z+ 3x-4y+z=- 10+( - 5)

AddNeg

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

x+5y=3 3x-4y+z=- 5 5x+5y-z+3x-4y+z=- 10-5

AddSubTerms

(III): Add and subtract terms

x+5y=3 3x-4y+z=- 5 8x+y=- 15

Next, we use our two equations that are only in terms of x and y 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.

x+5y=3 3x-4y+z=- 5 8x+y=- 15

▼

Solve by elimination

MultEqn

(III): LHS * (- 5)=RHS* (- 5)

x+5y=3 3x-4y+z=- 5 - 40x-5y=75

SysEqnAdd

(III): Add (I)

x+5y=3 3x-4y+z=- 5 - 40x-5y+ x+5y=75+ 3

AddTerms

(III): Add terms

x+5y=3 3x-4y+z=- 5 - 39x=78

DivEqn

(III): .LHS /(- 39).=.RHS /(- 39).

x+5y=3 3x-4y+z=- 5 x=- 2

Now that we know that x=- 2, we can substitute it into the first equation to find the value of y.

x+5y=3 3x-4y+z=- 5 x=- 2

Substitute

(I): x= - 2

- 2+5y=3 3x-4y+z=- 5 x=- 2

▼

(I): Solve for y

AddEqn

(I): LHS+2=RHS+2

5y=5 3x-4y+z=- 5 x=- 2

DivEqn

(I): .LHS /5.=.RHS /5.

y=1 3x-4y+z=- 5 x=- 2

The value of y is 1. Let's substitute both values into the second equation to find z.

y=1 3x-4y+z=- 5 x=- 2

SubstituteII

(II): x= - 2, y= 1

y=1 3( - 2)-4( 1)+z=- 5 x=- 2

▼

(II): Solve for z

MultPosNeg

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

y=1 - 6-4(1)+z=- 5 x=- 2

IdPropMult

(II): Identity Property of Multiplication

y=1 - 6-4+z=- 5 x=- 2

SubTerm

(II): Subtract term

y=1 - 10+z=- 5 x=- 2

AddEqn

(II): LHS+10=RHS+10

y=1 z=5 x=- 2

The solution to the system is ( - 2, 1, 5). This is the singular point at which all three planes intersect. Now we can check our solution by substituting the values into the system.

- 2x+9y-z=8 & (I) 3x-4y+z=- 5 & (II) 5x+5y-z=- 10 & (III)

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

- 2( - 2)+9( 1)- 5? =8 3( - 2)-4( 1)+ 5? =- 5 5( - 2)+5( 1)- 5? =- 10

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

4+9-5? =8 - 6-4+5? =- 5 - 10+5-5? =- 10

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

8=8 ✓ - 5=- 5 ✓ - 10=- 10 ✓

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 68 Page 231

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