I choose to eliminate z.
Since z is already eliminated from the
first equation we will use that first equation in its original form for
this step:
Since the z terms of the second and third equations are already opposites,
we can go right into adding them.
*z's dropped out
Let's first put those equations together:
Multiplying equation (4) by -1 and then adding the equations together
we get:
*y's have opposite coefficients
*y's dropped out
*Plug in 1/2 for x
*Inverse of add 1 is sub. 1
I choose equation (2) to plug in 1 for y that we found:
(1/2, 1, 2) is a solution to our system.
Let's first eliminate z using the first and second equations. This process is identical to how we approached it with the systems found in Tutorial 19: Solving a System of Linear Equations in Two variables . If I multiply 2 times the first equation, then the z terms will be opposites of each other and ultimately drop out.
Multiplying 2 times the first equation and then adding that to the
second equation we get:
*z's have opposite coefficients
*All three variables drop out AND
we have a FALSE statement
Final answer is no solution.
Last revised on July 10, 2011 by Kim Seward.
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