Exercise 3 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. -2x+ y+4z=5 & (I) x+3y-z=2 & (II) 4x+ y-6z=11 & (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.

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

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Solve by elimination

MultEqn

(I): LHS * (-3)=RHS* (-3)

6x-3y-12z=-15 x+3y-z=2 3x-5z=3

SysEqnAdd

(II): Add (I)

6x-3y-12z=-15 x+3y-z+( 6x-3y-12z)=2+( -15) 3x-5z=3

AddSubTerms

(II): Add and subtract terms

6x-3y-12z=-15 7x-13z=-13 3x-5z=3

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.

6x-3y-12z=-15 7x-13z=-13 3x-5z=3

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Solve by elimination

MultEqn

(II): LHS * (-3)=RHS* (-3)

6x-3y-12z=-15 -21x+39z=39 3x-5z=3

MultEqn

(III):LHS * 7=RHS* 7

6x-3y-12z=-15 -21x+39z=39 21x-35z=21

SysEqnAdd

(II): Add (III)

6x-3y-12z=-15 -21x+39z+( 21x-35z)=39+( 21) 21x-35z=21

AddSubTerms

(II): Add and subtract terms

6x-3y-12z=-15 4z=60 21x-35z=21

DivEqn

(II): .LHS /4.=.RHS /4.

6x-3y-12z=-15 z=15 21x-35z=21

Now that we know that z=15, we can substitute it into the third equation to find the value of x.

6x-3y-12z=-15 z=15 21x-35z=21

DivEqn

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

6x-3y-12z=-15 z=15 3x-5z=3

Substitute

(III): z= 15

6x-3y-12z=-15 z=15 3x-5( 15)=3

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(III): Solve for x

Multiply

(III): Multiply

6x-3y-12z=-15 z=15 3x-75=3

AddEqn

(III): LHS+75=RHS+75

6x-3y-12z=-15 z=15 3x=78

DivEqn

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

6x-3y-12z=-15 z=15 x=26

The value of x is 26. Let's substitute both values into the first equation to find y.

6x-3y-12z=-15 z=15 x=26

DivEqn

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

-2x+y+4z=5 z=15 x=26

SubstituteII

(I): x= 26, z= 15

-2( 26)+y+4( 15)=5 z=15 x=26

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(I): Solve for y

Multiply

(I): Multiply

-52+y+60=5 z=15 x=26

AddTerms

(I): Add terms

y+8=5 z=15 x=26

SubEqn

(I): LHS-8=RHS-8

y=-3 z=15 x=26

The solution to the system is ( 26, -3, 15). 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+y+4z=5 & (I) x+3y-z=2 & (II) 4x+y-6z=11 & (III)

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

-2( 26)+( -3)+4( 15)? =5 26+3( -3)- 15? =2 4( 26)+( -3)-6( 15)? =11

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

-52+(-3)+60? =5 26-9-15? =2 104+(-3)-90? =11

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

5=5 2=2 11=11

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