Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. Systems With Three Variables
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Exercise 40 Page 172

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

(-1,2,0)

Practice makes perfect
The given system consists of equations of planes. We will use the Substitution Method to solve a system of equations. When using this method, it is necessary to isolate a variable. In the third equation, it will be easier to isolate x.
4y+2x=6-3z & (I) x+z-2y=-5 & (II) x-2z=3y-7 & (III)
4y+2x=6-3z x+z-2y=-5 x=3y-7+2z
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.
4y+2x=6-3z x+z-2y=-5 x=3y-7+2z

(I), (II): x= 3y-7+2z

4y+2( 3y-7+2z)=6-3z 3y-7+2z+z-2y=-5 x=3y-7+2z
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(I), (II): Simplify
4y+6y-14+4z=6-3z 3y-7+2z+z-2y=-5 x=3y-7+2z

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

10y-14+4z=6-3z y+3z-7=-5 x=3y-7+2z
10y-14+4z=6-3z y+3z=2 x=3y-7+2z
10y-14+4z=6-3z y=2-3z x=3y-7+2z
This time, the y-variable was isolated in the second equation. We can now substitute its equivalent expression into the first equation.
10y-14+4z=6-3z y=2-3z x=3y-7+2z
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(I): Solve by substitution
10( 2-3z)-14+4z=6-3z y=2-3z x=3y-7+2z
20-30z-14+4z=6-3z y=2-3z x=3y-7+2z
6-26z=6-3z y=2-3z x=3y-7+2z
-26z=-3z y=2-3z x=3y-7+2z
-23z=0 y=2-3z x=3y-7+2z
z=0 y=2-3z x=3y-7+2z
The value of z is 0. Substituting 0 for z into the second equation, we can find the value of y.
z=0 y=2-3z x=3y-7+2z
z=0 y=2-3( 0) x=3y-7+2z
z=0 y=2 x=3y-7+2z
Now that we know the values of y and z, we are able to find the value of x.
z=0 y=2 x=3y-7+2z
z=0 y=2 x=3( 2)-7+2( 0)
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(III): Simplify
z=0 y=2 x=6-7+0
z=0 y=2 x=-1
The solution to the system is the point (-1,2,0). This is the singular point at which all three planes intersect.