Exercise 10 Page 171

The given system consists of equations of planes. Notice that the coefficient of x in the first equation is the additive inverse of the coefficients of x in the second and the third equations; they will add to be 0. Let's use the Elimination Method to find a solution to this system. x-y-2z=4 & (I) - x+2y+z=1 & (II) - x+y-3z=11 & (III) We can start by adding the first equation to the second and the third equations to eliminate the x-terms. Having eliminated the y-variable from the second and the third equations, we can continue by solving for z in the third equation.

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

DivEqn

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

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

Next, we will substitute the value of z into the second equation and solve for y.

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

Substitute

(II):z= -3

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

▼

(II):Solve for y

SubNeg

(II):a-(- b)=a+b

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

SubEqn

(II):LHS-3=RHS-3

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

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

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

SubstituteII

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

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

▼

(I): Solve for x

MultNegNegOnePar

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

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

AddTerms

(I): Add terms

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

SubEqn

(I): LHS-4=RHS-4

x=0 y=2 z=-3

The solution to the system is ( 0, 2, -3). This is the singular point at which all three planes intersect. Now, we can check our solution by substituting the values into the system.

x-y-2z=4 & (I) - x+2y+z=1 & (II) - x+y-3z=11 & (III)

(I), (II), (III): Substitute values

0- 2-2( -3)? =4 - 0+2( 2)+( -3)? =1 - 0+ 2-3( -3)? =11

(I), (II), (III): Multiply

0-2+6? =4 0+4+(-3)? =1 0+2+9? =11

(I), (II), (III): Add and subtract terms

4=4 ✓ 1=1 ✓ 11=11 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.