Systems of Linear Equations a Case Study

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We subtract the second equation from the first and we obtain 2X-3Y=0

We multiply the second equation with 2 and add it to the third. We obtain 16X+4Y+2z=1206 and, when added to the third, 36X-6Y=1200 or 6X-Y=200.

We now have a new system of equations with 2 unknown variables, as such:

2x-3Y=0 and 6X-Y=200. We multiply the first equation by 3 and have 6X-9Y=0. We subtract the first equation from the second and the result is 8Y=200 and Y=25. X=37.5 and Z=203

c. Multiply the 1 stequation by 3 and the second by 5. The result is 66X+15Y+21Z=36 and 50X+15Y+10Z=25. We subtract the equations and the result is 16X+11Z=11.


We multiply the second equation with 2 and the third with 3. The result is

20X+6Y+4Z=10 and 27X+6Y+36Z=42. We subtract the equation and the result is

7X+32Z=32. We have a new system of equations 16X+11Z=11 and 7X+32Z=32. We multiply the 1st by 7 and the second by 16. The result is 112X+77Z=77 and 112X+512Z=512. Subtracting, the result is 435Z=435 and Z=1; X=0 and Y=1

Systems of Equations. On the Internet at http://www.sosmath.com/soe/SE/SE.html. Last retrieved on September 19, 2010.....

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