equation

[Valentas]

\(\displaystyle \begin{cases} x^{2}-5y^{2}+4xy=0\\x^{2}-8y^{2}-7xy=52\end{cases}\)

\(\displaystyle \begin{cases}|y-8|=1\\y+x=\frac{3x-y}{4}\end{cases}\)

 

[tkhunny]

Excellent sets of equations. Is there an assignment? What is your plan?

 

[soroban]

Hello, Valentas!

\(\displaystyle \begin{Bmatrix} x^2-5y^2+4xy &=&0 & [1] \\ x^2-8y^2-7xy &=& 52 & [2] \end{array}\)

\(\displaystyle \text{Subtract [1] - [2]: }\;3y^2 + 11xy \:=\:-52 \quad\Rightarrow\quad x \:=\:-\frac{3y^2+52}{11y}\quad {\bf[3]}\)

\(\displaystyle \text{Substitute into [1]: }\;\left(-\frac{3y^2+5}{11y}\right)^2 - 5y^2 + 4y\left(-\frac{3y^2+52}{11y}\right) \;=\;0\\)

. . \(\displaystyle \text{which simplifies to: }\;149y^4 - 45y^2 - 104 \;=\;0\)

. . \(\displaystyle \text{which factors: }\;(y^2-1)(149y^2+104) \;-\;0\)

. . \(\displaystyle \text{and has real roots: }\;y\:=\:\pm1\)

\(\displaystyle \text{Substitute into }{\bf[3]}:\;\;x \:=\:-\frac{3(\pm1)^2 + 52}{11(\pm1)} \;=\;\mp5\)


\(\displaystyle \text{Solutions: }\;(\pm5,\,\mp1)\)

\(\displaystyle \begin{Bmatrix}|y-8| &=& 1 & [1] \\ y+x &=& \dfrac{3x-y}{4} & [2] \end{array}\)

Is there a typo? . . . It seem too easy (almost).

\(\displaystyle \text{ From }{\bf[1]}\text{ we have: }\;\begin{Bmatrix}y-8 &=& 1 \\ y-8 &=& \text{-}1 \end{Bmatrix}\quad\Rightarrow\quad y \:=\:7,\,9\)

\(\displaystyle {\bf[2]}\text{ simplifies to: }\:x \:=\:-5y\)

. . \(\displaystyle \text{Hence: }\:x \;=\;-35,\:-45\)


\(\displaystyle \text{Solutions: }\-35,\,7),\-45,\,9)\)

 

[Valentas]

Thank you very much !

 

[tkhunny]

You are very welcome. Too bad we don't have any idea if you did anything or learned anything.

 

[mmm4444bot]

we don't have any idea if you did anything

I'm confident that I know exactly what was done with the completed assignment posted by Soroban.