Exercise 17 Page 171

The given system consists of equations of planes. Notice that the coefficient of y in the first equation is the additive inverse of the coefficient of y in the third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. x - 2y+3z=12 & (I) 2x-y-2z=5 & (II) 2x + 2y-z=4 & (III) We can start by adding the third equation to the first equation to eliminate the z-terms. Having eliminated the y-variable from the first equation, we can continue by creating additive inverse coefficients for y in the second and third equations. Then, we can add or subtract these equations to eliminate y from the second equation. Next, we use our two equations that are only in terms of x and z to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time will be similar to when using it in a system with only two variables.

3x+2z=16 6x-5z=14 2x+2y-z=4

▼

(II): Solve by elimination

MultEqn

(I): LHS * 5=RHS* 5

15x+10z=80 6x-5z=14 2x+2y-z=4

MultEqn

(II):LHS * 2=RHS* 2

15x+10z=80 12x-10z=28 2x+2y-z=4

SysEqnAdd

(II): Add (I)

15x+10z=80 12x-10z+( 15x+10z)=28+( 80) 2x+2y-z=4

RemovePar

(II): Remove parentheses

15x+10z=80 12x-10z+15x+10z=28+80 2x+2y-z=4

AddTerms

(II): Add terms

15x+10z=80 27x=108 2x+2y-z=4

DivEqn

(II): .LHS /27.=.RHS /27.

15x+10z=80 x=4 2x+2y-z=4

Now that we know that x=4, we can substitute it into the first equation to find the value of z.

15x+10z=80 x=4 2x+2y-z=4

Substitute

(I): x= 4

15( 4)+10z=80 x=4 2x+2y-z=4

▼

(I): Solve for z

Multiply

(I): Multiply

60+10z=80 x=4 2x+2y-z=4

SubEqn

(I): LHS-60=RHS-60

10z=20 x=4 2x+2y-z=4

DivEqn

(I): .LHS /10.=.RHS /10.

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

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

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

SubstituteII

(III): x= 4, z= 2

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

▼

(III): Solve for y

Multiply

(III): Multiply

z=2 x=4 8+2y- 2=4

SubTerm

(III): Subtract term

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

SubEqn

(III): LHS-6=RHS-6

z=2 x=4 2y=-2

DivEqn

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

z=2 x=4 y=-1

The solution to the system is ( 4, -1, 2). 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-2y+3z=12 & (I) 2x-y-2z=5 & (II) 2x+2y-z=4 & (III)

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

4-2( -1)+3( 2)? =12 2( 4)-( -1)-2( 2)? =5 2( 4)+2( -1)- 2? =4

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

4+2+6? =12 8-(-1)-4? =5 8-2-2? =4

SubNeg

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

4+2+6? =12 8+1-4? =5 8-2-2? =4

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

12=12 ✓ 5=5 ✓ 4=4 ✓

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