McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
4. Systems of Equations in Three Variables
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Exercise 1 Page 165

Can you manipulate the coefficients of any variable terms such that they could be eliminated?

(-2,-3,5)

Practice makes perfect
The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. Notice that in the third equation, there is no b-term. -3a-4b+2c=28 & (I) a+3b-4c=-31 & (II) 2a+3c=11 & (III) We can start with creating additive inverse coefficients for b in the first and second equations. Then we can add or subtract these equations to eliminate b from the second equation.
-3a-4b+2c=28 & (I) a+3b-4c=-31 & (II) 2a+3c=11 & (III)
-9a-12b+6c=84 a+3b-4c=-31 2a+3c=11
-9a-12b+6c=84 4a+12b-16c=-124 2a+3c=11
-9a-12b+6c=84 4a+12b-16c+( -9a-12b+6c)=-124+( 84) 2a+3c=11
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(II): Simplify
-9a-12b+6c=84 4a+12b-16c-9a-12b+6c=-124+84 2a+3c=11
-9a-12b+6c=84 -5a-10c=-40 2a+3c=11
-9a-12b+6c=84 a+2c=8 2a+3c=11
Next, we use our two equations that only contains a and c 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.
-9a-12b+6c=84 a+2c=8 2a+3c=11
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(III): Solve by elimination
-9a-12b+6c=84 -2a-4c=-16 2a+3c=11
-9a-12b+6c=84 -2a-4c=-16 2a+3c+( -2a-4c)=11+( -16)
-9a-12b+6c=84 -2a-4c=-16 - c=-5
-9a-12b+6c=84 -2a-4c=-16 c=5
Now that we know that c=5, we can substitute it into the second equation to find the value of a.
-9a-12b+6c=84 -2a-4c=-16 c=5
-9a-12b+6c=84 -2a-4( 5)=-16 c=5
â–Ľ
(II): Solve for a
-9a-12b+6c=84 -2a-20=-16 c=5
-9a-12b+6c=84 -2a=4 c=5
-9a-12b+6c=84 a=-2 c=5
The value of a is -2. Let's substitute both values into the first equation to find b.
-9a-12b+6c=84 a=-2 c=5
-9( -2)-12b+6( 5)=84 a=-2 c=5
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(I): Solve for b
18-12b+30=84 a=-2 c=5
48-12b=84 a=-2 c=5
-12b=36 a=-2 c=5
b=-3 a=-2 c=5
The solution to the system is (-2,-3,5). This is the singular point at which all three planes intersect.