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Pearson Algebra 2 Common Core, 2011 View details

5. Systems With Three Variables

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Exercise 18 Page 171

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

(-122/11,72/11,71/11)

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 first equation there is no z-term. x+2y=2 & (I) 2x+3y-z=-9 & (II) 4x+2y+5z=1 & (III) We can start by creating additive inverse coefficients for z in the second and third equations. Then, we can add or subtract these equations to eliminate z from the second equation.

x+2y=2 2x+3y-z=-9 4x+2y+5z=1

MultEqn

(II): LHS * 5=RHS* 5

x+2y=2 10x+15y-5z=-45 4x+2y+5z=1

SysEqnAdd

(II): Add (III)

x+2y=2 10x+15y-5z+( 4x+2y+5z)=-45+( 1) 4x+2y+5z=1

RemovePar

(II): Remove parentheses

x+2y=2 10x+15y-5z+4x+2y+5z=-45+1 4x+2y+5z=1

AddTerms

(II): Add terms

x+2y=2 14x+17y=-44 4x+2y+5z=1

Next, we will 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 it will be similar to when using it in a system with only two variables.

x+2y=2 14x+17y=-44 4x+2y+5z=1

▼

(II): Solve by elimination

MultEqn

(I): LHS * (-14)=RHS* (-14)

-14x-28y=-28 14x+17y=-44 4x+2y+5z=1

SysEqnAdd

(II): Add (I)

-14x-28y=-28 14x+17y+( -14x-28y)=-44+( -28) 4x+2y+5z=1

RemovePar

(II): Remove parentheses

-14x-28y=-28 14x+17y-14x-28y=-44-28 4x+2y+5z=1

AddTerms

(II): Add terms

-14x-28y=-28 -11y=-72 4x+2y+5z=1

DivEqn

(II): .LHS /(-11).=.RHS /(-11).

-14x-28y=-28 y= 7211 4x+2y+5z=1

Now that we know that y= 7211, we can substitute it into the first equation to find the value of x.

-14x-28y=-28 y= 7211 4x+2y+5z=1

DivEqn

(I):.LHS /(-14).=.RHS /(-14).

x+2y=2 y= 7211 4x+2y+5z=1

Substitute

(I): y= 72/11

x+2( 7211)=2 y= 7211 4x+2y+5z=1

▼

(I): Solve for x

MoveLeftFacToNum

(I): a*b/c= a* b/c

x+ 14411=2 y= 7211 4x+2y+5z=1

NumberToFrac

(I):a = 11* a/11

x+ 14411= 11(2)11 y= 7211 4x+2y+5z=1

Multiply

(I): Multiply

x+ 14411= 2211 y= 7211 4x+2y+5z=1

SubEqn

(I): LHS-144/11=RHS-144/11

x= -12211 y= 7211 4x+2y+5z=1

The value of x is -12211. Let's substitute both values into the third equation to find z.

x= -12211 y= 7211 4x+2y+5z=1

SubstituteII

(III): x= -122/11, y= 72/11

x= -12211 y= 7211 4( -12211)+2( 7211)+5z=1

▼

(III): Solve for z

MoveLeftFacToNum

(III): a*b/c= a* b/c

x= -12211 y= 7211 -48811+ 14411+5z=1

AddFrac

(III): Add fractions

x= -12211 y= 7211 -34411+5z=1

OneToFrac

(III):Rewrite 1 as 11/11

x= -12211 y= 7211 -34411+5z= 1111

AddEqn

(III): LHS+344/11=RHS+344/11

x= -12211 y= 7211 5z= 35511

DivEqn

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

x= -12211 y= 7211 z= 35511 / 5

DivFracSL

(III):.a/b /c.= a/b* c

x= -12211 y= 7211 z= 35511(5)

Multiply

(III):Multiply

x= -12211 y= 7211 z= 35555

ReduceFrac

(III):a/b=.a /5./.b /5.

x= -12211 y= 7211 z= 7111

MoveNegNumToFrac

(I):Put minus sign in front of fraction

x=- 12211 y= 7211 z= 7111

The solution to the system is ( - 12211, 7211, 7111). 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+2y=2 2x+3y-z=-9 4x+2y+5z=1

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

- 12211+2( 7211)? =2 2( - 12211)+3( 7211)- 7111? =-9 4( - 12211)+2( 7211)+5( 7111)? =1

(I), (II), (III): a*b/c= a* b/c

- 12211+ 14411? =2 - 24411+ 21611- 7111? =-9 - 48811+ 14411+ 35511? =1

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

2211? =2 - 9911? =-9 1111? =1

(I), (II), (III): Calculate quotient

2=2 ✓ -9=-9 ✓ 1=1 ✓

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

Exercise 18 Page 171

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