Solving systems of linear equations using elimination method

Systems of Linear Equations

A system of equations is a set of several equations that share the same unknowns

With more equations present, the identified value for a variable in one equation must correspond and be valid in the other equations of the system as well and thus be verifiable.

We subtract the first equation from the first one and obtain an equation with only one unknown, as such: 2X = 2, which means that x = 2/2 = 1. With that, replacing X with its value, 1, in the first equation, Y = 9

We multiply the second equation by 2 and the first by 3. This means that the new system is 6X+15Y=57 and 6X+6Y=30. We subtract the second equation from the first to obtain 9Y=27, with Y=3. As such, X=

We multiply the second equation by 2 and the result is 4X+6Y=32. We subtract the first equation from the second 5Y=10, Y=2. X=

d. We multiply the first equation with 2 and the result is 24X+2Y=348. We add the second equation to the first and the result is 32X = 384. X = 12 and Y=30

2. a. Bob: 2,000X+10,000Y=$372,000

Frank: 8,000X+6,000Y=$400,000

b. We multiply Bob's equation with 4 and the result is 8,000X+40,000Y=$1,4888,000. We subtract the second equation from the first. The result is 34,000Y=1,088,000 and Y= $32. X= $26

3. a. We express X as X=15-Y (from the second equation). This will lead to a system of only two equations, as such:

15-Y+2Y+Z=22, with Y+Z=7 and

45-3Y+Y+Z=37, with -2Y+Z=-8

We subtract the first equation from the second and the result is -3Y=-15, with Y=5; X=10 and Z=2

b. 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.