Question

Solve the following equations:
$
  {x^2} + 4{y^2} - 15x = 10\left( {3y - 8} \right),{\text{ }}xy = 6. \\
  {\text{a}}{\text{. }}\left( {4,5} \right) \\
  {\text{b}}{\text{. }}\left( {2,3} \right) \\
  {\text{c}}{\text{. }}\left( {3,2} \right) \\
  {\text{d}}{\text{. }}\left( {4,3} \right) \\
$

Answer 1

Hint: - Substitute the value of $x{\text{ or }}y$ from the second equation into the first equation and convert the equation in terms of y then solve the equation using a hit and trial method.

Complete step-by-step solution -
Given equations is
${x^2} + 4{y^2} - 15x = 10\left( {3y - 8} \right)..........\left( 1 \right)$
$xy = 6........................\left( 2 \right)$
From equation 2
$x = \dfrac{6}{y}............\left( 3 \right)$
Put this value of $x$ in equation 1
${\left( {\dfrac{6}{y}} \right)^2} + 4{y^2} - 15\left( {\dfrac{6}{y}} \right) = 10\left( {3y - 8} \right)$
$ \Rightarrow \dfrac{{36}}{{{y^2}}} + 4{y^2} - \dfrac{{90}}{y} = \left( {30y - 80} \right)$
Now multiply by ${y^2}$ throughout we have,
$ \Rightarrow 36 + 4{y^4} - 90y = 30{y^3} - 80{y^2}$
$ \Rightarrow 4{y^4} - 30{y^3} + 80{y^2} - 90y + 36 = 0$
Now factorize the above equation
Solving by hit and trial method.
Put $y = 1$, we have
$4 - 30 + 80 - 90 + 36 = 0$
Hence, $\left( {y - 1} \right)$ be the root of the equation.
Put $y = 2$, we have
$64 - 240 + 320 - 180 + 36 = 420 - 420 = 0$
Hence, $\left( {y - 2} \right)$ be the root of the equation.
Put $y = 3$, we have
$324 - 810 + 720 - 270 + 36 = 1080 - 1080 = 0$
Hence, $\left( {y - 3} \right)$ be the root of the equation.
Now put $y = \dfrac{3}{2}$, we have
$ \Rightarrow 4{\left( {\dfrac{3}{2}} \right)^4} - 30{\left( {\dfrac{3}{2}} \right)^3} + 80{\left( {\dfrac{3}{2}} \right)^2} - 90\left( {\dfrac{3}{2}} \right) + 36$
$ = \dfrac{{81}}{4} - \dfrac{{405}}{4} + \dfrac{{720}}{4} - \dfrac{{270}}{2} + 36$
$ = \dfrac{{81 - 405 + 720 - 540 + 144}}{4} = \dfrac{{945 - 945}}{4} = 0$
Hence, $\left( {y - \dfrac{3}{2}} \right)$ be the root of the equation.
Therefore the equation becomes
$\left( {y - 1} \right)\left( {y - 2} \right)\left( {y - 3} \right)\left( {y - \dfrac{3}{2}} \right) = 0$
Therefore $y = 1,2,3,\dfrac{3}{2}$
So, from equation 3
$x = \dfrac{6}{y}$
Therefore $x = 6,3,2,4$
So, the solutions of the given equation is $\left( {x,y} \right) = \left( {6,1} \right),\left( {3,2} \right),\left( {2,3} \right),\left( {4,\dfrac{3}{2}} \right)$
Hence, the correct option is b and c.

Note: - In such types of question always put the value of $x{\text{ or }}y$ from the simple equation into complex equation, then simplify the equation using hit and trial method and calculate the values of $x{\text{ and }}y$ which is the required answer, also check whether there is multiple options are correct or not.