Exercise 13 Page 171

The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. 6q-r+2s=8 & (I) 2q+3r-s=-9 & (II) 4q+2r+5s=1 & (III) Currently, none of the terms in this system will cancel out. However, if we multiply (I) by 2 the coefficient of r in this equation will be the additive inverse of the coefficient of r in the third equation; they will add to be 0. 2(6q-r+2s)=2(8) 2q+3r-s=-9 4q+2r+5s=1 ⇓ 12q - 2r+4s=16 2q+3r-s=-9 4q + 2r+5s=1 We can start by adding the first equation to the third equation to eliminate the r-terms. Having eliminated the r-variable from the third equation, we can continue by creating additive inverse coefficients for r in the first and second equations. Then, we can add or subtract these equations to eliminate r from the second equation.

12q-2r+4s=16 2q+3r-s=-9 16q+9s=17

MultEqn

(I): LHS * 3/2=RHS* 3/2

18q-3r+6s=24 2q+3r-s=-9 16q+9s=17

SysEqnAdd

(II): Add (I)

18q-3r+6s=24 2q+3r-s+( 18q-3r+6s)=-9+( 24) 16q+9s=17

RemovePar

(II): Remove parentheses

18q-3r+6s=24 2q+3r-s+18q-3r+6s=-9+24 16q+9s=17

AddSubTerms

(II): Add and subtract terms

18q-3r+6s=24 20q+5s=15 16q+9s=17

DivEqn

(II): .LHS /5.=.RHS /5.

18q-3r+6s=24 4q+s=3 16q+9s=17

Next, we will use our two equations that are only in terms of q and s 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.

18q-3r+6s=24 4q+s=3 16q+9s=17

▼

(II): Solve by elimination

MultEqn

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

18q-3r+6s=24 -16q-4s=-12 16q+9s=17

SysEqnAdd

(II): Add (III)

18q-3r+6s=24 -16q-4s+( 16q+9s)=-12+( 17) 16q+9s=17

RemovePar

(II): Remove parentheses

18q-3r+6s=24 -16q-4s+16q+9s=-12+17 16q+9s=17

AddTerms

(II): Add terms

18q-3r+6s=24 5s=5 16q+9s=17

DivEqn

(II): .LHS /5.=.RHS /5.

18q-3r+6s=24 s=1 16q+9s=17

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

18q-3r+6s=24 s=1 16q+9s=17

Substitute

(III): s= 1

18q-3r+6s=24 s=1 16q+9( 1)=17

▼

(III): Solve for q

IdPropMult

(III): Identity Property of Multiplication

18q-3r+6s=24 s=1 16q+9=17

SubEqn

(III): LHS-9=RHS-9

18q-3r+6s=24 s=1 16q=8

DivEqn

(III): .LHS /16.=.RHS /16.

18q-3r+6s=24 s=1 q= 816

ReduceFrac

(III):a/b=.a /8./.b /8.

18q-3r+6s=24 s=1 q= 12

The value of q is 12. Let's substitute both values into the first equation to find r.

18q-3r+6s=24 s=1 q= 12

SubstituteII

(I): q= 1/2, s= 1

18( 12)-3r+6( 1)=24 s=1 q= 12

▼

(I): Solve for r

Multiply

(I): Multiply

9-3r+6=24 s=1 q= 12

AddTerms

(I): Add terms

15-3r=24 s=1 q= 12

SubEqn

(I): LHS-15=RHS-15

-3r=9 s=1 q= 12

DivEqn

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

r=-3 s=1 q= 12

The solution to the system is ( 12, -3, 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.

12q-2r+4s=16 & (I) 2q+3r-s=-9 & (II) 4q+2r+5s=1 & (III)

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

12( 12)-2( -3)+4( 1)? =16 2( 12)+3( -3)- 1? =-9 4( 12)+2( -3)+5( 1)? =1

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

6+6+4? =16 1-9-1? =-9 2-6+5? =1

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

16=16 ✓ -9=-9 ✓ 1=1 ✓

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