Big Ideas Math: Modeling Real Life, Grade 8
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3. Solving Systems of Linear Equations by Elimination
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Exercise 43 Page 218

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

x=-1, y=2, and z=1

Practice makes perfect
The given system consists of equations of planes. Notice that the coefficient of y in the first equation is the additive inverse of the coefficient of y in the third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. 2x - y +3z=-1 & (I) x+2y-4z=-1 & (II) y-2z=0 & (III) We can start by adding the third equation to the first equation to eliminate the y-terms.
2x-y+3z=-1 & (I) x+2y-4z=-1 & (II) y-2z=0 & (III)
2x-y+3z+( y-2z)=-1+( 0) x+2y-4z=-1 y-2z=0
2x-y+3z+y-2z=-1+0 x+2y-4z=-1 y-2z=0
2x+z=-1 x+2y-4z=-1 y-2z=0
Having eliminated the y-variable from the first equation, we can continue by creating additive inverse coefficients for y in the second and third equations. Then, we can add or subtract these equations to eliminate y from the second equation.
2x+z=-1 x+2y-4z=-1 y-2z=0
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Solve by elimination
2x+z=-1 x+2y-4z=-1 2(y-2z)=2(0)
2x+z=-1 x+2y-4z=-1 2y-2(2z)=2(0)
2x+z=-1 x+2y-4z=-1 2y-4z=2(0)
2x+z=-1 x+2y-4z=-1 2y-4z=0
2x+z=-1 x+2y-4z-( 2y-4z)=-1-( 0) 2y-4z=0
2x+z=-1 x+2y-4z-2y+4z=-1-0 2y-4z=0
2x+z=-1 x=-1 2y-4z=0
Now that we know that x=-1, we can substitute it into the first equation to find the value of z.
2x+z=-1 x=-1 2y-4z=0
2( -1)+z=-1 x=-1 2y-4z=0
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(I): Solve for z
-2+z=-1 x=-1 2y-4z=0
-2+z+2=-1+2 x=-1 2y-4z=0
z=1 x=-1 2y-4z=0
The value of z is 1. Let's substitute this value into the third equation to find y.
z=1 x=-1 2y-4z=0
z=1 x=-1 2y-4(1)=0
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(III): Solve for y
z=1 x=-1 2y-4=0
z=1 x=-1 2y-4+4=0+4
z=1 x=-1 2y=4
z=1 x=-1 2y/2=4/2
z=1 x=-1 y=2
The solution to the system is x=-1, y=2, and z=1. This is the singular point at which all three planes intersect.