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How solve system ? 4x^2-3y=xy^3 x^2+x^3y^2=2y

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home1h | Student, Undergraduate | (Level 2) eNoter

Posted June 18, 2013 at 2:22 PM via web

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How solve system ?

4x^2-3y=xy^3

x^2+x^3y^2=2y

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 18, 2013 at 3:13 PM (Answer #1)

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You need to solve for x and y the given system of equations, such that:

`{(4x^2-3y=xy^3),(x^2+x^3y^2=2y):}`

You need to re-write the bottom equation of the system, such that:

`{(4x^2-3y=xy^3),(x^2-2y=-x^3y^2):}`

You need to divide by the bottom equation, the top equation, such that:

`(4x^2-3y)/(x^2-2y)=xy^3/(-x^3y^2)`

Reducing duplicate factors to the left side, yields:

`(4x^2-3y)/(x^2-2y)=y/(-x^2)`

You should force factor `x^2` to the left side, such that:

`(x^2(4 - 3y/x^2))/(x^2(1 - 2y/x^2))=y/(-x^2)`

Reducing duplicate factors yields:

`(4 - 3y/x^2)/(1 - 2y/x^2) = y/(-x^2)`

You should come up with the following substitution, such that:

`y/(x^2) = t`

Changing the variables, yields:

`(4 - 3t)/(1 - 2t) = -t => 4 - 3t = -t*(1 - 2t)`

`4 - 3t = -t + 2t^2`

You need to move the members to one side such that:

`2t^2 - t + 3t - 4 = 0 => 2t^2 + 2t - 4 = 0 => 2(t^2 + t - 2) = 0`

`t^2 + t - 2 = 0`

Using quadratic formula yields:

`t_(1,2) = (-1+-sqrt(1 + 8))/2`

`t_(1,2) = (-1+-sqrt9)/2 => t_(1,2) = (-1+-3)/2`

`t_1 = 1 ; t_2 = -2`

Replacing back `y/(x^2) ` for `t_1` yields:

`y/(x^2) = 1 => y = x^2`

You need to replace `y` for `x^2` in the bottom equation, such that:

`y + y^3*x - 2y = 0 => y^3*x - y = 0`

Factoring out y yields:

`y(y^2*x - 1) = 0 => y = 0 => x = 0`

`y^2*x - 1 = 0 => y^2*x = 1 => y^2 = 1/x`

Since `y = x^2` yields:

`x^4 = 1/x => x^4 - 1/x = 0 => (x^5 - 1)/x = 0 => x^5 - 1 = 0`

Using the special product `a^n - b^n = (a - b)(a^n + ... + b^n)` yields:

`x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)`

`x^5 - 1 = 0 => (x - 1)(x^4 + x^3 + x^2 + x + 1) = 0`

Using zero product rule yields:

`(x - 1)(x^4 + x^3 + x^2 + x + 1)= 0 => {(x - 1 = 0),(x^4 + x^3 + x^2 + x + 1!= 0):}`

`x = 1 => y = 1^2 = 1`

Hence, evaluating the solutions of the system if `y = x^2` yields `(x = 0, y = 0)` or `(x = 1, y = 1)` .

Replacing back `y/(x^2)` for `t_2` yields:

`y/(x^2) = -2 => y = -2x^2 => x^2 = -y/2` invalid `(x,y !in R)`

Hence, evaluating the solutions of the system, using substitution, yields `x = 0, y = 0` or `x = 1, y = 1.`

x^2-2y

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