Question-143130

Question Number 143130 by liberty last updated on 10/Jun/21

Commented by liberty last updated on 10/Jun/21

Answered by EDWIN88 last updated on 10/Jun/21

let y = px

{\begin{cases} x + px + \frac{x^{2}}{p^{2} x^{2}} = 7 \\ (x - px) \frac{x^{2}}{p^{2} x^{2}} = 12 \end{cases} \Rightarrow {\begin{cases} x (1 + p) + \frac{1}{p^{2}} = 7 \\ x (\frac{1 - p}{p^{2}}) = 12 \end{cases}

{\begin{cases} x = \frac{{7 p}^{2} - 1}{p^{2} (1 + p)} \\ x = \frac{{12 p}^{2}}{1 - p} \end{cases} \Rightarrow \frac{{12 p}^{2}}{1 - p^{2}} = \frac{{7 p}^{2} - 1}{p^{2} (1 + p)}

\Rightarrow {12 p}^{4} (1 + p) = ({7 p}^{2} - 1) (1 - p^{2})

\Rightarrow {12 p}^{4} + {12 p}^{5} = - {7 p}^{4} + {8 p}^{2} - 1

\Rightarrow {12 p}^{5} + {19 p}^{4} - {8 p}^{2} + 1 = 0

\Rightarrow (p + 1) ({12 p}^{4} + {7 p}^{3} - {7 p}^{2} - p + 1) = 0

\Rightarrow (p + 1) (p + 1) ({12 p}^{3} - {5 p}^{2} - 2 p + 1) = 0

\Rightarrow {(p + 1)}^{2} ({12 p}^{3} - {5 p}^{2} - 2 p + 1) = 0

we get p = - 1 (rejected)

{12 p}^{3} - {5 p}^{2} - 2 p + 1 = 0

\Rightarrow p \approx - 0.4279

then y = - 0 . 4279 x

\Rightarrow x = \frac{12 {(- 0.4279)}^{2}}{1 - (- 0.4279)} = 1.386

\Rightarrow y = - 0.43 \times 1.55 = - 0.558

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