Exercise 12 Page 171

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

2a+2b+2c=-6 a+4b=1 4a+b=-11

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(II): Solve by elimination

MultEqn

(II): LHS * (-4)=RHS* (-4)

2a+2b+2c=-6 -4a-16b=-4 4a+b=-11

SysEqnAdd

(II): Add (III)

2a+2b+2c=-6 -4a-16b+( 4a+b)=-4+( -11) 4a+b=-11

AddSubTerms

(II): Add and subtract terms

2a+2b+2c=-6 -15b=-15 4a+b=-11

DivEqn

(II): .LHS /(-15).=.RHS /(-15).

2a+2b+2c=-6 b=1 4a+b=-11

Now that we know that b=1, we can substitute it into the third equation to find the value of a.

2a+2b+2c=-6 b=1 4a+b=-11

Substitute

(III): b= 1

2a+2b+2c=-6 b=1 4a+ 1=-11

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(III): Solve for a

SubEqn

(III): LHS-1=RHS-1

2a+2b+2c=-6 b=1 4a=-12

DivEqn

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

2a+2b+2c=-6 b=1 a=-3

The value of a is -3. Let's substitute both values into the first equation to find c.

2a+2b+2c=-6 b=1 a=-3

SubstituteII

(I): a= -3, b= 1

2( -3)+2( 1)+2c=-6 b=1 a=-3

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(I): Solve for c

Multiply

(I): Multiply

-6+2+2c=-6 b=1 a=-3

AddTerms

(I): Add terms

-4+2c=-6 b=1 a=-3

AddEqn

(I): LHS+4=RHS+4

2c=-2 b=1 a=-3

DivEqn

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

c=-1 b=1 a=-3

The solution to the system is ( -3, 1, -1). This is the singular point at which all three planes intersect. Now, we can check our solution by substituting the values into the system.

a+b+c=-3 & (I) 3b-c=4 & (II) 2a-b-2c=-5 & (III)

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

-3+ 1+( -1)? =-3 3( 1)-( -1)? =4 2( -3)- 1-2( -1)? =-5

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

-3+1+(-1)? =-3 3-(-1)? =4 -6-1+2? =-5

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

-3=-3 ✓ 4=4 ✓ -5=-5 ✓

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