Exercise 27 Page 172

The given system of equations consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the third equation, it will be easiest to isolate x. With a variable isolated in one of the equations, we can substitute its equivalent expression into the remaining equations. In the final step of the simplification of these substitutions, our goal is to have yet another variable isolated. This time, 3z was isolated in the first equation. We can now substitute its equivalent expression into the second equation.

3z=2+y 10+3z+y=8 x=5+z

Substitute

(II): 3z= 2+y

3z=2+y 10+( 2+y)+y=8 x=5+z

▼

(II): Solve for y

RemovePar

(II): Remove parentheses

3z=2+y 10+2+y+y=8 x=5+z

AddTerms

(II): Add terms

3z=2+y 12+2y=8 x=5+z

SubEqn

(II): LHS-12=RHS-12

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

DivEqn

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

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

The value of y is -2. Substituting -2 for y into the first equation, we can find the value of z.

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

Substitute

(I): y= -2

3z=2+( -2) y=-2 x=5+z

AddNeg

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

3z=0 y=-2 x=5+z

DivEqn

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

z=0 y=-2 x=5+z

Now that we know the value of z, we are able to find the value of x.

z=0 y=-2 x=5+z

Substitute

(III): z= 0

z=0 y=-2 x=5+ 0

AddTerms

(III):Add terms

z=0 y=-2 x=5

The solution to the system is the point ( 5, -2, 0). This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.

x-y+2z=7 & (I) 2x+y+z=8 & (II) x-z=5 & (III)

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

5-( -2)+2( 0)? =7 2( 5)+( -2)+ 0? =8 5- 0? =5

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

5-(-2)+0? =7 10+(-2)+0? =8 5-0? =5

(I), (II): Remove parentheses

5+2+0? =7 10-2+0? =8 5-0? =5

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

7=7 ✓ 8=8 ✓ 5=5 ✓

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