Exercise 5 Page 41

The given system consists of equations of planes. Notice that the coefficient of y in the first equation is the same as the coefficient of y in the third equation; they will cancel out each other. Let's use the Elimination Method to find a solution to this system. x- y+5z=3 & (I) 2x+3y-z=2 & (II) -4x- y-9z=-8 & (III) We can start by subtracting the first equation from the third equation to eliminate the y-terms. Having eliminated the y-variable from the third equation, we can continue by creating additive inverse coefficients for y in the first and second equations. Then we can add or subtract these equations to eliminate y from the second equation.

x-y+5z=3 2x+3y-z=2 -5x-14z=-11

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Solve by elimination

MultEqn

(I): LHS * 3=RHS* 3

3x-3y+15z=9 2x+3y-z=2 -5x-14z=-11

SysEqnAdd

(II): Add (I)

3x-3y+15z=9 2x+3y-z+( 3x-3y+15z)=2+( 9) -5x-14z=-11

AddSubTerms

(II): Add and subtract terms

3x-3y+15z=9 5x+14z=11 -5x-14z=-11

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

3x-3y+15z=9 5x+14z=11 -5x-14z=-11

SysEqnAdd

(III):Add (II)

3x-3y+15z=9 5x+14z=11 -5x-14z+( 5x+14z)=-11+( 11)

AddTerms

(III):Add terms

3x-3y+15z=9 5x+14z=11 0=0

We ended with an identity: 0 = 0. This means that the third and second equations are the same plane. Therefore, there are infinite number of solutions to the system. However, we can describe the solutions using z-terms by solving for x and y.

3x-3y+15z=9 5x+14z=11 0=0

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(II):Solve for x

SubEqn

(II):LHS-14z=RHS-14z

3x-3y+15z=9 5x=11-14z 0=0

DivEqn

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

3x-3y+15z=9 x= 11-14z5 0=0

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(I):Solve for y

DivEqn

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

x-y+5z=3 x= 11-14z5 0=0

AddEqn

(I):LHS+y=RHS+y

x+5z=3+y x= 11-14z5 0=0

SubEqn

LHS-3=RHS-3

x+5z-3=y x= 11-14z5 0=0

RearrangeEqn

Rearrange equation

y=x+5z-3 x= 11-14z5 0=0

Substitute

(I):x= 11-14z5

y= 11-14z5+5z-3 x= 11-14z5 0=0

NumberToFrac

(I): a = 5* a/5

y= 11-14z5+ 25z5- 155 x= 11-14z5 0=0

AddSubFrac

(I): Add and subtract fractions

y= 11-14z+25z-155 x= 11-14z5 0=0

AddSubTerms

(I):Add and subtract terms

y= 11z-45 x= 11-14z5 0=0

We described all the variables using only z-terms. Therefore, any ordered triple of the form of ( 11-14z5, 11z-45, z) will satisfy this system of equations. We can check the solution by substituting any number for z. Let's try for z=0.

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

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

11-14z5- 11z-45+5 z? =3 2( 11-14z5)+3( 11z-45)- z? =2 -4( 11-14z5)- 11z-45-9 z? =-8

(I), (II), (III): y=

11-14( )5- 11( )-45+5( )? =3 2( 11-14( )5)+3( 11( )-45)- ? =2 -4( 11-14( )5)- 11( )-45-9( )? =-8

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(I), (II), (III): Simplify

(I), (II), (III): Zero Property of Multiplication

115- -45+0? =3 2( 115)+3( -45)-0? =2 -4( 115)-( -45)-0? =-8

(I), (II), (III): Put minus sign in front of fraction

115+ 45+0? =3 2( 115)-3( 45)-0? =2 -4( 115)+ 45-0? =-8

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

115+ 45+0? =3 225- 125-0? =2 - 445+ 45-0? =-8

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

155+0? =3 105-0? =2 - 405-0? =-8

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

3+0? =3 2-0? =2 -8-0? =-8

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

3=3 2=2 -8=-8

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