Exercise 4 Page 171

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 second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. 2x-y + z=-2 & (I) x+3y - z=10 & (II) x+2z=-8 & (III) We can start by adding the second equation to the first equation to eliminate the z-terms. Having eliminated the z-variable from the first equation, we can continue 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.

3x+2y=8 x+3y-z=10 x+2z=-8

MultEqn

(II): LHS * 2=RHS* 2

3x+2y=8 2x+6y-2z=20 x+2z=-8

SysEqnAdd

(II): Add (III)

3x+2y=8 2x+6y-2z+( x+2z)=20+( -8) x+2z=-8

RemovePar

(II): Remove parentheses

3x+2y=8 2x+6y-2z+x+2z=20-8 x+2z=-8

AddTerms

(II):Add terms

3x+2y=8 3x+6y=12 x+2z=-8

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.

3x+2y=8 3x+6y=12 x+2z=-8

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(II): Solve by elimination

SysEqnSub

(II): Subtract (I)

3x+2y=8 3x+6y-( 3x+2y)=12-( 8) x+2z=-8

Distr

(II): Distribute -1

3x+2y=8 3x+6y-3x-2y=12-8 x+2z=-8

SubTerms

(II): Subtract terms

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

DivEqn

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

3x+2y=8 y=1 x+2z=-8

Now that we know that y=1, we can substitute it into the first equation to find the value of x.

3x+2y=8 y=1 x+2z=-8

Substitute

(I): y= 1

3x+2( 1)=8 y=1 x+2z=-8

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(I): Solve for x

IdPropMult

(I): Identity Property of Multiplication

3x+2=8 y=1 x+2z=-8

SubEqn

(I): LHS-2=RHS-2

3x=6 y=1 x+2z=-8

DivEqn

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

x=2 y=1 x+2z=-8

The value of x is 2. Let's substitute the value of x into the third equation to find z.

x=2 y=1 x+2z=-8

Substitute

(III): x= 2

x=2 y=1 2+2z=-8

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(III): Solve for z

SubEqn

(III): LHS-2=RHS-2

x=2 y=1 2z=-10

DivEqn

(III):.LHS /2.=.RHS /2.

x=2 y=1 z=-5

The solution to the system is (2,1,-5). This is the singular point at which all three planes intersect.