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{x+y+z=100} {x+y=z-10} {3x+4y+5z=422}

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted February 26, 2013 at 1:30 AM via web

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{x+y+z=100}

{x+y=z-10}

{3x+4y+5z=422}

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 4, 2013 at 1:33 PM (Answer #3)

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You should express `x + y ` in terms of z, from the first equation, such that:

`x + y = 100 - z`

You should substitute 100 - z for x + y in the second equation, such that:

`100 - z = z - 10`

Moving the terms that contain z to one side yields:

`-2z = -100 - 10 => -z = 110/2 => z = 55`

Substituting 55 for z in the second and the third equation yields:

`{(x + y = 45),(3x + 4y = 422 - 275):}`

`{(x + y = 45),(3x + 4y = 147):}`

`{(-3x - 3y = -135),(3x + 4y = 147):} => y = 12`

`x = 45 - 12 => x = 33`

Hence, evaluating the solutions to the given system of equations yields` x = 33, y = 12, z = 55.`

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