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Solve the system of equations: x+5y-2z=16 3x-3y+2z=12 2x+4y+z=20

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted February 24, 2013 at 1:27 AM via web

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Solve the system of equations:

x+5y-2z=16

3x-3y+2z=12

2x+4y+z=20

2 Answers | Add Yours

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 24, 2013 at 12:48 PM (Answer #1)

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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` .

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 24, 2013 at 5:23 PM (Answer #2)

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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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