Exercise 33 Page 172

The given system consists of equations of planes. Notice that the coefficient of x in the first equation is the additive inverse of the coefficient of x in the second equation; they will add to be 0. Therefore, let's use the Elimination Method to find a solution to this system. x-3y+2z=11 & (I) - x+4y+3z=5 & (II) 2x-2y-4z=2 & (III) We can start by adding the second equation to the first equation to eliminate the x-terms. Having eliminated the x-variable from the first equation, we can continue by creating additive inverse coefficients for x in the second and third equations. Then, we can add or subtract these equations to eliminate x from the second equation.

y+5z=16 - x+4y+3z=5 2x-2y-4z=2

MultEqn

(II): LHS * 2=RHS* 2

y+5z=16 -2x+8y+6z=10 2x-2y-4z=2

SysEqnAdd

(II): Add (III)

y+5z=16 -2x+8y+6z+( 2x-2y-4z)=10+( 2) 2x-2y-4z=2

▼

(II): Simplify

RemovePar

(II): Remove parentheses

y+5z=16 -2x+8y+6z+2x-2y-4z=10+2 2x-2y-4z=2

AddSubTerms

(II): Add and subtract terms

y+5z=16 6y+2z=12 2x-2y-4z=2

DivEqn

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

y+5z=16 3y+z=6 2x-2y-4z=2

Next, we will 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 it will be similar to when using it in a system with only two variables.

y+5z=16 3y+z=6 2x-2y-4z=2

▼

(II): Solve by elimination

MultEqn

(II): LHS * (-5)=RHS* (-5)

y+5z=16 -15y-5z=-30 2x-2y-4z=2

SysEqnAdd

(II): Add (I)

y+5z=16 -15y-5z+( y+5z)=-30+( 16) 2x-2y-4z=2

RemovePar

(II): Remove parentheses

y+5z=16 -15y-5z+y+5z=-30+16 2x-2y-4z=2

AddTerms

(II): Add terms

y+5z=16 -14y=-14 2x-2y-4z=2

DivEqn

(II): .LHS /(-14).=.RHS /(-14).

y+5z=16 y=1 2x-2y-4z=2

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

y+5z=16 y=1 2x-2y-4z=2

Substitute

(I): y= 1

1+5z=16 y=1 2x-2y-4z=2

▼

(I): Solve for z

SubEqn

(I): LHS-1=RHS-1

5z=15 y=1 2x-2y-4z=2

DivEqn

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

z=3 y=1 2x-2y-4z=2

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

z=3 y=1 2x-2y-4z=2

SubstituteII

(III): y= 1, z= 3

z=3 y=1 2x-2( 1)-4( 3)=2

▼

(III): Solve for x

Multiply

(III): Multiply

z=3 y=1 2x-2-12=2

SubTerms

(III): Subtract terms

z=3 y=1 2x-14=2

AddEqn

(III): LHS+14=RHS+14

z=3 y=1 2x=16

DivEqn

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

z=3 y=1 x=8

The solution to the system is (8,1,3). This is the singular point at which all three planes intersect.