How do solve system in natural numbers x,y,z? x+y+z=14 x+yz=19

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may consider the following approach, hence, you should try subtract the top equation x + y + z = 14, from the bottom equation, x + yz = 19, such that:

`x + yz - x - y - z = 19 - 14 => yz - y - z = 5 => yz - y - z +1 = 5 + 1 => yz - y - z +1 = 6`

You may replace the product `(y - 1)(z - 1)` for `yz - y - z + 1` such that:

`(y - 1)(z - 1) = 6`

Using the condition provided by the problem, x,y,z are natural numbers,you may consider the following 4 cases such that:

  1. `1. {(y - 1 = 1),(z - 1 = 6):}`

` 2. {(y - 1 = 2),(z - 1 = 3):} `` ``3. {(y - 1 = 3),(z - 1 = 2):} `` ``4. {(y - 1 = 6),(z - 1 = 1):} `

Solving the first system yields:

`{(y - 1 = 1),(z - 1 = 6):} => {(y = 2),(z = 7):}`

Replacing 2 for `y` and 7 for `z` in equation `x + y + z = 14` yields:

`x = 14 - 2 - 7 => x = 5`

Solving the second system yields:

`{(y - 1 = 2),(z - 1 = 3):} => {(y = 2 + 1),(z = 3 + 1):} => {(y = 3),(z = 4):}`

Replacing 3 for `y` and 4 for `z` in equation` x + y + z = 14` yields:

`x = 14 - 3 - 4 => x = 7`

Solving the third system yields:

`{(y - 1 = 3),(z - 1 = 2):} => {(y = 4),(z = 3):}`

Replacing 4 for `y` and 3 for `z` in equation `x + y + z = 14` yields:

`x = 14 - 4 - 3 => x = 7 `

Solving the fourth system yields:

`{(y - 1 = 6),(z - 1 = 1):} => {(y = 7),(z = 2):}`

Replacing 7 for `y` and 2 for `z` in equation `x + y + z = 14` yields:

`x = 14 - 7 - 2 => x = 5`

Hence, evaluating the solutions to the given system of equations, under the given conditions, yields `(5,2,7), (7,3,4),(7,4,3),(5,7,2).`

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