Question

Solve the following equations: $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$ and $x\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)+8=0$.
A) (2,4)
B) (2,6)
C) (3,4)
D) (4,7)

Answer 1

Hint: We will write $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$ as $\left( {{y}^{2}}-3xy-{{x}^{2}} \right)=-\dfrac{24}{y}$ and $x\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)+8=0$ as $\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)=-\dfrac{8}{x}$ and then subtract the obtained equation to get a new equation then we will perform some basic mathematical operations to find the value of x and y respectively.
Complete step-by-step answer:
It is given in the question to solve $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$ and $x\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)+8=0$. Now we can write $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$ as $\left( {{y}^{2}}-3xy-{{x}^{2}} \right)=-\dfrac{24}{y}......(i)$ and $x\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)+8=0$ as $\left( {{y}^{2}}-4xy+2{{x}^{2}} \right)=-\dfrac{8}{x}......(ii)$. On subtracting equation (i) and equation (ii), we get $\left( \left( {{y}^{2}}-3xy-{{x}^{2}} \right)=-\dfrac{24}{y} \right)-\left( \left( {{y}^{2}}-4xy+2{{x}^{2}} \right)=-\dfrac{8}{x} \right)=xy-3{{x}^{2}}=\left( \dfrac{8}{x}-\dfrac{24}{y} \right)$ which can be written as $x\left( y-3x \right)=8\dfrac{\left( y-3x \right)}{xy}$.
Now, on cross multiplying both sides, we get ${{x}^{2}}y\left( y-3x \right)=8\left( y-3x \right)$, transposing $8\left( y-3x \right)$ from RHS to LHS, we get ${{x}^{2}}y\left( y-3x \right)-8\left( y-3x \right)=0$, that is $\left( {{x}^{2}}y-8 \right)\left( y-3x \right)=0$. Therefore, we get values of y as $y=3x,\dfrac{8}{{{x}^{2}}}$.
Now putting the value of y in equation $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$as y=3x, we get - $3x\left( {{\left( 3x \right)}^{2}}-3x\left( 3x \right)-{{x}^{2}} \right)+24=0$ therefore we get $-3{{x}^{3}}=-24$ or ${{x}^{3}}=8,x=2$. Therefore we get the value of y as $y=3x=3\times 2=6$.
Now, putting the value of $y=\dfrac{8}{{{x}^{2}}}$ in equation $y\left( {{y}^{2}}-3xy-{{x}^{2}} \right)+24=0$, we get $\dfrac{8}{{{x}^{2}}}\left( {{\left( \dfrac{8}{{{x}^{2}}} \right)}^{2}}-3x\dfrac{8}{{{x}^{2}}}-{{x}^{2}} \right)+24=0$, taking 8 outside as common term and multiplying with ${{x}^{2}}$ we get $\dfrac{64}{{{x}^{4}}}-\dfrac{24}{x}-{{x}^{2}}+3{{x}^{2}}=0$ solving further $\dfrac{64}{{{x}^{4}}}-\dfrac{24}{x}+2{{x}^{2}}=0$ or $\dfrac{32}{{{x}^{4}}}-\dfrac{12}{x}+{{x}^{2}}=0$. Now taking LCM of the terms, we get ${{x}^{6}}-12{{x}^{3}}+32=0$.
Now splitting the middle term we get ${{x}^{6}}-4{{x}^{3}}-8{{x}^{3}}+32=0$, taking ${{x}^{3}}$ as common we get ${{x}^{3}}\left( {{x}^{3}}-4 \right)-8\left( {{x}^{3}}-4 \right)=0$ or we get \[\left( {{x}^{3}}-4 \right)\left( {{x}^{3}}-8 \right)=0\]. Therefore values of x are $x=2,x=\sqrt[3]{4}$ Also $y=\dfrac{8}{{{x}^{2}}}$, therefore putting value of x in this equation, we get $y=\dfrac{8}{{{2}^{2}}}=\dfrac{8}{4}=2$ and \[y=\dfrac{8}{{{\left( \sqrt[3]{4} \right)}^{2}}}=2\sqrt[3]{4}\] .
Now on the basis of obtained values of x and y, we get (x,y): (2,6) as the correct answer and the given option b) is the correct answer.
Note: Usually students do mistakes in splitting term $12{{x}^{3}}$ taking common because of higher degree also many times students take wrong sign with terms which results in formation of wrong answer thus it is recommended to do calculation carefully in order to avoid silly mistakes.