Exercise 67 Page 231

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. 7x-2y-5z=24 & (I) - x+3y+4z=- 10 & (II) x-y-z=4 & (III) ⇓ 7x-2y-5z=24 & (I) - 1x+3y+4z=- 10 & (II) 1x-y-z=4 & (III) We can start by adding the third equation to the second equation to eliminate the x-terms. Having eliminated the x-variable from the second equation, we can continue by creating additive inverse coefficients for x in the first and third equations. Then we can add or subtract these equations to eliminate x from the second equation.

7x-2y-5z=24 2y+3z=- 6 x-y-z=4

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Solve by elimination

MultEqn

(III): LHS * (- 7)=RHS* (- 7)

7x-2y-5z=24 2y+3z=- 6 - 7x+7y+7z=- 28

SysEqnAdd

(III): Add (I)

7x-2y-5z=24 2y+3z=- 6 - 7x+7y+7z+ 7x-2y-5z=- 28+ 24

AddSubTerms

(III): Add and subtract terms

7x-2y-5z=24 2y+3z=- 6 5y+2z=- 4

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.

7x-2y-5z=24 2y+3z=- 6 5y+2z=- 4

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Solve by elimination

MultEqn

(II): LHS * (- 2)=RHS* (- 2)

7x-2y-5z=24 - 4y-6z=12 5y+2z=- 4

MultEqn

(III): LHS * 3=RHS* 3

7x-2y-5z=24 - 4y-6z=12 15y+6z=- 12

SysEqnAdd

(II): Add (III)

7x-2y-5z=24 - 4y-6z+ 15y+6z=12+( - 12) 15y+6z=- 12

AddNeg

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

7x-2y-5z=24 - 4y-6z+15y+6z=12-12 15y+6z=- 12

AddSubTerms

(II): Add and subtract terms

7x-2y-5z=24 11y=0 15y+6z=- 12

DivEqn

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

7x-2y-5z=24 y=0 15y+6z=- 12

Now that we know that y=0, we can substitute it into the third equation to find the value of z.

7x-2y-5z=24 y=0 15y+6z=- 12

Substitute

(III): y= 0

7x-2y-5z=24 y=0 15( 0)+6z=- 12

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(III): Solve for z

ZeroPropMult

(III): Zero Property of Multiplication

7x-2y-5z=24 y=0 0+6z=- 12

IdPropAdd

(III): Identity Property of Addition

7x-2y-5z=24 y=0 6z=- 12

DivEqn

(III): .LHS /6.=.RHS /6.

7x-2y-5z=24 y=0 z=- 2

The value of z is - 2. Let's substitute both values into the first equation to find x.

7x-2y-5z=24 y=0 z=- 2

SubstituteII

(III): y= 0, z= - 2

7x-2( 0)-5( - 2)=24 y=0 z=- 2

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(I): Solve for x

ZeroPropMult

(I): Zero Property of Multiplication

7x-0-5(- 2)=24 y=0 z=- 2

MultNegNegOnePar

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

7x-0+10=24 y=0 z=- 2

AddSubTerms

(I): Add and subtract terms

7x+10=24 y=0 z=- 2

SubEqn

(I): LHS-10=RHS-10

7x=14 y=0 z=- 2

DivEqn

(I): .LHS /7.=.RHS /7.

x=2 y=0 z=- 2

The solution to the system is ( 2, 0, - 2). This is the singular point at which all three planes intersect. Now we can check our solution by substituting the values into the system.

7x-2y-5z=24 - x+3y+4z=- 10 x-y-z=4

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

7( 2)-2( 0)-5( - 2)? =24 - 2+3( 0)+4( - 2)? =- 10 2- 0-( - 2)? =4

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

14-0+10? =24 - 2+0-8? =- 10 2-0+2? =4

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

24=24 ✓ - 10=- 10 ✓ 4=4 ✓

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