Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
1. Defining and Using Sequences and Series
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Exercise 62 Page 416

When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.

(2,1,- 1)

Practice makes perfect
The given system 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 second equation it will be easiest to isolate x.
2x-y-3z=6 & (I) x+y+4z=- 1 & (II) 3x-2z=8 & (III)
2x-y-3z=6 & (I) x=- 1-y-4z & (II) 3x-2z=8 & (III)
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.
2x-y-3z=6 x=- 1-y-4z 3x-2z=8

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

2( - 1-y-4z)-y-3z=6 x=- 1-y-4z 3( - 1-y-4z)-2z=8
â–Ľ
(I), (III): Simplify
- 2-2y-8z-y-3z=6 x=- 1-y-4z 3(- 1-y-4z)-2z=8
- 2-2y-8z-y-3z=6 x=- 1-y-4z - 3-3y-12z-2z=8

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

- 2-3y-11z=6 x=- 1-y-4z - 3-3y-14z=8
- 3y-11z=8 x=- 1-y-4z - 3-3y-14z=8
- 3y=8+11z x=- 1-y-4z - 3-3y-14z=8
- 3y=8+11z x=- 1-y-4z - 3y-14z=11
- 3y=8+11z x=- 1-y-4z - 3y=11+14z
This time, - 3y was isolated in the third equation. We can now substitute its equivalent expression into the first equation.
- 3y=8+11z x=- 1-y-4z - 3y=11+14z
11 +14z=8+11z x=- 1-y-4z - 3y=11+14z
â–Ľ
(I): Solve by substitution
14z=- 3+11z x=- 1-y-4z - 3y=11+14z
3z=- 3 x=- 1-y-4z - 3y=11+14z
z=- 1 x=- 1-y-4z - 3y=11+14z
The value of z is - 1. Substituting - 1 for z into the third equation, we can find the value of y.
z=- 1 x=- 1-y-4z - 3y=11+14z
z=- 1 x=- 1-y-4z - 3y=11+14( - 1)
â–Ľ
(III): Solve by substitution
z=- 1 x=- 1-y-4z - 3y=11-14
z=- 1 x=- 1-y-4z - 3y=- 3
z=- 1 x=- 1-y-4z y=1
Now that we know the values of y and z, we are able to find the value of x.
z=- 1 x=- 1-y-4z y=1
z=- 1 x=- 1- 1-4( - 1) y=1
â–Ľ
(II): Simplify
z=- 1 x=- 1-1+4 y=1
z=- 1 x=2 y=1
The solution to the system is the point ( 2, 1, - 1). This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.
2x-y-3z=6 & (I) x+y+4z=- 1 & (II) 3x-2z=8 & (III)

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

2( 2)- 1-3( - 1)? =6 & (I) 2+ 1+4( - 1)? =- 1 & (II) 3( 2)-2( - 1)? =8 & (III)
â–Ľ
(II): Simplify

(I), (II), (III): Multiplication Property of -1

2(2)-1+3? =6 & (I) 2+1-4? =- 1 & (II) 3(2)+2? =8 & (III)

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

4-1+3? =6 & (I) 2+1-4? =- 1 & (II) 6+2? =8 & (III)

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

6=6 - 1=- 1 8=8
Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.