Exercise 3 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+3y - 2z=1 & (I) - x-y + 2z=5 & (II) 3x+2y-3z=-6 & (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.

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

MultEqn

(II): LHS * 3=RHS* 3

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

MultEqn

(III): LHS * 2=RHS* 2

x+2y=6 -3x-3y+6z=15 6x+4y-6z=-12

SysEqnAdd

(II):Add (III)

x+2y=6 -3x-3y+6z+( 6x+4y-6z)=15+( -12) 6x+4y-6z=-12

RemovePar

(II): Remove parentheses

x+2y=6 -3x-3y+6z+6x+4y-6z=15-12 6x+4y-6z=-12

AddSubTerms

(II):Add and subtract terms

x+2y=6 3x+y=3 6x+4y-6z=-12

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.

x+2y=6 3x+y=3 6x+4y-6z=-12

▼

(I): Solve by elimination

MultEqn

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

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

SysEqnAdd

(I):Add (II)

x+2y+( -6x-2y)=6+( -6) -6x-2y=-6 6x+4y-6z=-12

RemovePar

(I): Remove parentheses

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

SubTerms

(I):Subtract terms

-5x=0 -6x-2y=-6 6x+4y-6z=-12

DivEqn

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

x=0 -6x-2y=-6 6x+4y-6z=-12

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

x=0 -6x-2y=-6 6x+4y-6z=-12

Substitute

(II): x= 0

x=0 -6( 0)-2y=-6 6x+4y-6z=-12

ZeroPropMult

(II): Zero Property of Multiplication

x=0 -2y=-6 6x+4y-6z=-12

DivEqn

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

x=0 y=3 6x+4y-6z=-12

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

x=0 y=3 6x+4y-6z=-12

SubstituteII

(III): x= 0, y= 3

x=0 y=3 6( 0)+4( 3)-6z=-12

▼

(III): Solve for z

Multiply

(III): Multiply

x=0 y=3 12-6z=-12

SubEqn

(III): LHS-12=RHS-12

x=0 y=3 -6z=-24

DivEqn

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

x=0 y=3 z=4

The solution to the system is (0,3,4). This is the singular point at which all three planes intersect.