Exercise 19 Page 171

The given system consists of equations of planes. Notice that the coefficient of z in the second equation is the additive inverse of the coefficient of z in the third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. 3x+2y+2z=-2 & (I) 2x+y - z=-2 & (II) x-3y + z=0 & (III) We can start by adding the second equation to the third equation to eliminate the z-terms. Having eliminated the z-variable from the third equation, we can continue by creating additive inverse coefficients for z in the first and second equations. Then, we can add or subtract these equations to eliminate z from the second equation. Next, we will 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+2y+2z=-2 7x+4y=-6 3x-2y=-2

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(II): Solve by elimination

MultEqn

(III): LHS * 2=RHS* 2

3x+2y+2z=-2 7x+4y=-6 6x-4y=-4

SysEqnAdd

(II): Add (III)

3x+2y+2z=-2 7x+4y+( 6x-4y)=-6+( -4) 6x-4y=-4

RemovePar

(II): Remove parentheses

3x+2y+2z=-2 7x+4y+6x-4y=-6-4 6x-4y=-4

AddSubTerms

(II): Add and subtract terms

3x+2y+2z=-2 13x=-10 6x-4y=-4

DivEqn

(II): .LHS /13.=.RHS /13.

3x+2y+2z=-2 x=- 1013 6x-4y=-4

Now that we know that x=- 1013, we can substitute it into the third equation to find the value of y.

3x+2y+2z=-2 x=- 1013 6x-4y=-4

Substitute

(III): x= -10/13

3x+2y+2z=-2 x=- 1013 6( - 1013)-4y=-4

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(III): Solve for y

MoveLeftFacToNum

(III): a*b/c= a* b/c

3x+2y+2z=-2 x=- 1013 - 6013-4y=-4

NumberToFrac

(III):a = 13* a/13

3x+2y+2z=-2 x=- 1013 - 6013-4y=- 13(4)13

Multiply

(III): Multiply

3x+2y+2z=-2 x=- 1013 - 6013-4y=- 5213

AddEqn

(III): LHS+60/13=RHS+60/13

3x+2y+2z=-2 x=- 1013 -4y= 813

DivEqn

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

3x+2y+2z=-2 x=- 1013 y= 813/(-4)

DivFracSL

(III): .a/b /c.= a/b* c

3x+2y+2z=-2 x=- 1013 y= 813*(-4)

ReduceFrac

(III):a/b=.a /4./.b /4.

3x+2y+2z=-2 x=- 1013 y= 2-13

MoveNegDenomToFrac

(III):Put minus sign in front of fraction

3x+2y+2z=-2 x=- 1013 y=- 213

The value of y is - 213. Let's substitute both values into the first equation to find z.

3x+2y+2z=-2 x=- 1013 y=- 213

SubstituteII

(I): x= -10/13, y= -2/13

3( - 1013)+2( - 213)+2z=-2 x=- 1013 y=- 213

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(I): Solve for z

MoveLeftFacToNum

(I): a*b/c= a* b/c

- 3013- 413+2z=-2 x=- 1013 y=- 213

SubFrac

(I): Subtract fractions

- 3413+2z=-2 x=- 1013 y=- 213

NumberToFrac

(I):a = 13* a/13

- 3413+2z=- 13(2)13 x=- 1013 y=- 213

Multiply

(I):Multiply

- 3413+2z=- 2613 x=- 1013 y=- 213

AddEqn

(I): LHS+34/13=RHS+34/13

2z= 813 x=- 1013 y=- 213

DivEqn

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

z= 813/2 x=- 1013 y=- 213

DivFracSL

(I):.a/b /c.= a/b* c

z= 813*2 x=- 1013 y=- 213

ReduceFrac

(I):a/b=.a /2./.b /2.

z= 413 x=- 1013 y=- 213

The solution to the system is ( - 1013, - 213, 413). 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+2y+2z=-2 & (I) 2x+y-z=-2 & (II) x-3y+z=0 & (III)

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

3( - 1013)+2( - 213)+2( 413)? =-2 2( - 1013)+( - 213)- 413? =-2 - 1013-3( - 213)+ 413? =0

(I), (II), (III): a*b/c= a* b/c

- 3013- 413+ 813? =-2 - 2013+(- 213)- 413? =-2 - 1013+ 613+ 413? =0

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

- 2613? =-2 - 2613? =-2 - 013? =0

(I), (II), (III): Calculate quotient

-2=-2 ✓ -2=-2 ✓ 0=0 ✓

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