Exercise 41 Page 172

The given system consists of equations of planes. Notice that the coefficient of z in the first equation is the additive inverse of the coefficients of z in the second and third equations; they will add to be 0. Therefore, let's use the Elimination Method to find a solution to this system. 4x-y + z=-5 & (I) - x+y - z=5 & (II) 2x - z-1=y & (III) We can start by adding the first equation to the second and third equations to eliminate the z-terms. Now that we know that x=0, we can substitute it into the third equation to find the value of y.

4x-y+z=-5 x=0 6x-1-y=y-5

Substitute

(III): x= 0

4x-y+z=-5 x=0 6( 0)-1-y=y-5

▼

(III): Solve for y

ZeroPropMult

(III): Zero Property of Multiplication

4x-y+z=-5 x=0 -1-y=y-5

AddEqn

(III): LHS+y=RHS+y

4x-y+z=-5 x=0 -1=2y-5

AddEqn

(III): LHS+5=RHS+5

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

DivEqn

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

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

RearrangeEqn

(III):Rearrange equation

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

The value of y is 2. Let's substitute the values of x and y into the first equation to find z.

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

SubstituteII

(I): x= 0, y= 2

4( 0)- 2+z=-5 x=0 y=2

▼

(I): Solve for z

ZeroPropMult

(I): Zero Property of Multiplication

-2+z=-5 x=0 y=2

AddEqn

(I): LHS+2=RHS+2

z=-3 x=0 y=2

The solution to the system is (0,2,-3). This is the singular point at which all three planes intersect.