Exercise 15 Page 171

The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. 3x+3y+6z=9 & (I) 2x+y+3z=7 & (II) x+2y-z=-10 & (III) Currently, none of the terms in this system will cancel out. However, if we multiply (III) by 3 the coefficient of z in this equation will be the additive inverse of the coefficient of z in the second equation; they will add to be 0. 3x+3y+6z=9 2x+y+3z=7 3(x+2y-z)=3(-10) ⇓ 3x+3y+6z=9 2x+y + 3z=7 3x+6y - 3z=-30 We can start by adding the third equation to the second equation to eliminate the z-terms. Having eliminated the z-variable from the second equation, we can continue by creating additive inverse coefficients for z in the first and third equations. Then, we can add or subtract these equations to eliminate z from the first equation.

3x+3y+6z=9 5x+7y=-23 3x+6y-3z=-30

MultEqn

(III): LHS * 2=RHS* 2

3x+3y+6z=9 5x+7y=-23 6x+12y-6z=-60

SysEqnSub

(I): Subtract (III)

3x+3y+6z+( 6x+12y-6z)=9+( -60) 5x+7y=-23 6x+12y-6z=-60

▼

(I): Simplify

RemovePar

(I): Remove parentheses

3x+3y+6z+6x+12y-6z=9-60 5x+7y=-23 6x+12y-6z=-60

AddSubTerms

(I):Add and subtract terms

9x+15y=-51 5x+7y=-23 6x+12y-6z=-60

DivEqn

(I): .LHS /3.=.RHS /3.

3x+5y=-17 5x+7y=-23 6x+12y-6z=-60

Next, we 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.

3x+5y=-17 5x+7y=-23 6x+12y-6z=-60

▼

(II): Solve by elimination

MultEqn

(I): LHS * (-5)=RHS* (-5)

-15x-25y=85 5x+7y=-23 6x+12y-6z=-60

MultEqn

(II):LHS * 3=RHS* 3

-15x-25y=85 15x+21y=-69 6x+12y-6z=-60

SysEqnAdd

(II): Add (I)

-15x-25y=85 15x+21y+( -15x-25y)=-69+( 85) 6x+12y-6z=-60

RemovePar

(II): Remove parentheses

-15x-25y=85 15x+21y-15x-25y=-69+85 6x+12y-6z=-60

AddSubTerms

(II): Add and subtract terms

-15x-25y=85 -4y=16 6x+12y-6z=-60

DivEqn

(II): .LHS /(-4).=.RHS /(-4).

-15x-25y=85 y=-4 6x+12y-6z=-60

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

-15x-25y=85 y=-4 6x+12y-6z=-60

Substitute

(I): y= -4

-15x-25( -4)=85 y=-4 6x+12y-6z=-60

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(I): Solve for x

MultNegNegOnePar

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

-15x+100=85 y=-4 6x+12y-6z=-60

SubEqn

(I): LHS-100=RHS-100

-15x=-15 y=-4 6x+12y-6z=-60

DivEqn

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

x=1 y=-4 6x+12y-6z=-60

The value of x is 1. Let's substitute both values into the third equation to find z.

x=1 y=-4 6x+12y-6z=-60

SubstituteII

(III): x= 1, y= -4

x=1 y=-4 6( 1)+12( -4)-6z=-60

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(III): Solve for z

Multiply

(III): Multiply

x=1 y=-4 6-48-6z=-60

SubTerms

(III): Subtract terms

x=1 y=-4 -42-6z=-60

AddEqn

(III): LHS+42=RHS+42

x=1 y=-4 -6z=-18

DivEqn

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

x=1 y=-4 z=3

The solution to the system is ( 1, -4, 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.

3x+3y+6z=9 & (I) 2x+y+3z=7 & (II) x+2y-z=-10 & (III)

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

3( 1)+3( -4)+6( 3)? =9 2( 1)+( -4)+3( 3)? =7 1+2( -4)- 3? =-10

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

3-12+18? =9 2+(-4)+9? =7 1-8-3? =-10

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

9=9 ✓ 7=7 ✓ -10=-10 ✓

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