Solve the system of equations: x+5y-2z=16 3x-3y+2z=12 2x+4y+z=20
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calendarEducator since 2010
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The following system of equations has to be solved:
x+5y-2z=16 ...(1)
3x-3y+2z=12 ...(2)
2x+4y+z=20 ...(3)
(1) + (2)
=> x + 3x + 5y - 3y - 2z + 2z = 16 + 12
=> 4x + 2y = 28
=> 2x + y = 14
=> y = 14 - 2x ...(4)
2*(3) + (1)
=> 4x + x + 8y + 5y + 2z - 2z = 40 + 16
=> 5x + 13y = 56
substitute y = 14 - 2x
5x + 13*14 - 26x = 56
=> -21x = -126
=> x = 6
y = 14 - 12 = 2
x + 5y - 2z = 16
=> 6 + 10 - 2z = 16
=> z = 0
The solution of the given set of equations is x = 6, y = 2 and z = 0
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calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
You need to solve for `x,y` and `z` the given system of simultaneous equations, hence, you should add the first equation to the second equation, to eliminate z, such that:
`x + 5y - 2z + 3x - 3y + 2z = 16 + 12`
`4x + 2y = 28`
You need to multiply the third equation by 2 and then you need to add it to the first equation, such that:
`x + 5y - 2z + 2(2x + 4y + z) = 16 + 2*20`
`x + 5y - 2z + 4x + 8y + 2z = 56`
`5x + 13y = 56`
You need to solve for x and y the following system of equations, such that:
`{(4x + 2y = 28),(5x + 13y = 56):} => {(5(4x + 2y) = 5*28),(-4(5x + 13y) = -4*56):}`
`{(20x + 10y = 140),(-20x - 52y = -224):} => -42y = -84 => y = 2`
Substituting 2 for y in equation `4x + 2y = 28` yields:
`4x + 4 = 28 => 4x = 24 => x = 6`
Substituting 2 for y and 6 for x in equation `x + 5y - 2z = 16` yields:
`6 + 10 - 2z = 16 => -2z = 0 => z = 0`
Hence, evaluating the solution to the system of equations yields `x = 6, y = 2, z = 0` .