Exercise 35 Page 172

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 third equation there is no x-term. 4x-y+2z=-6 & (I) -2x+3y-z=8 & (II) 2y+3z=-5 & (III) By manipulating the first equation and adding it to the second equation, we can solve for y. Now that we know that y=2, we can substitute it into the third equation to find the value of z.

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

Substitute

(III): y= 2

4x-y+2z=-6 y=2 2( 2)+3z=-5

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(III): Solve for z

Multiply

(III): Multiply

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

SubEqn

(III): LHS-4=RHS-4

4x-y+2z=-6 y=2 3z=-9

DivEqn

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

4x-y+2z=-6 y=2 z=-3

The value of z is -3. Let's substitute both values into the first equation to find x.

4x-y+2z=-6 y=2 z=-3

SubstituteII

(I): y= 2, z= -3

4x- 2+2( -3)=-6 y=2 z=-3

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(I): Solve for x

MultPosNeg

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

4x-2-6=-6 y=2 z=-3

SubTerm

(I): Subtract term

4x-8=-6 y=2 z=-3

AddEqn

(I): LHS+8=RHS+8

4x=2 y=2 z=-3

DivEqn

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

x= 24 y=2 z=-3

ReduceFrac

a/b=.a /2./.b /2.

x= 12 y=2 z=-3

The solution to the system is ( 12,2,-3). This is the singular point at which all three planes intersect.