Exercise 34 Page 172

The given system consists of equations of planes. Notice that the coefficient of y in the second equation is the additive inverse of the coefficient of y in the third equation; they will add to be 0. Therefore, let's use the Elimination Method to find a solution to this system. x+2y+z=4 & (I) 2x - y+4z=-8 & (II) -3x + y-2z=-1 & (III) We can start by adding the second equation to the third equation to eliminate the y-terms. Having eliminated the y-variable from the third equation, we can continue by creating additive inverse coefficients for y in the first and second equations. Then, we can add or subtract these equations to eliminate y from the second equation.

x+2y+z=4 2x-y+4z=-8 - x+2z=-9

MultEqn

(II): LHS * 2=RHS* 2

x+2y+z=4 4x-2y+8z=-16 - x+2z=-9

SysEqnAdd

(II): Add (I)

x+2y+z=4 4x-2y+8z+( x+2y+z)=-16+( 4) - x+2z=-9

RemovePar

(II): Remove parentheses

x+2y+z=4 4x-2y+8z+x+2y+z=-16+4 - x+2z=-9

AddTerms

(II): Add terms

x+2y+z=4 5x+9z=-12 - x+2z=-9

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.

x+2y+z=4 5x+9z=-12 - x+2z=-9

▼

(II): Solve by elimination

MultEqn

(III): LHS * 5=RHS* 5

x+2y+z=4 5x+9z=-12 -5x+10z=-45

SysEqnAdd

(II): Add (III)

x+2y+z=4 5x+9z+( -5x+10z)=-12+( -45) -5x+10z=-45

RemovePar

(II): Remove parentheses

x+2y+z=4 5x+9z-5x+10z=-12-45 -5x+10z=-45

AddTerms

(II): Add terms

x+2y+z=4 19z=-57 -5x+10z=-45

DivEqn

(II): .LHS /19.=.RHS /19.

x+2y+z=4 z=-3 -5x+10z=-45

Now that we know that z=-3, we can substitute it into the third equation to find the value of x.

x+2y+z=4 z=-3 -5x+10z=-45

Substitute

(III): z= -3

x+2y+z=4 z=-3 -5x+10( -3)=-45

▼

(III): Solve for x

MultPosNeg

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

x+2y+z=4 z=-3 -5x-30=-45

AddEqn

(III): LHS+30=RHS+30

x+2y+z=4 z=-3 -5x=-15

DivEqn

(III): .LHS /(-5).=.RHS /(-5).

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

The value of x is 3. Let's substitute both values into the first equation to find y.

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

SubstituteII

(I): x= 3, z= -3

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

▼

(I): Solve for y

RemovePar

(I): Remove parentheses

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

SubTerm

(I): Subtract term

2y=4 z=-3 x=3

DivEqn

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

y=2 z=-3 x=3

The solution to the system is (3,2,-3). This is the singular point at which all three planes intersect.