Question

Solve the following equations:
$
  4x - 3y = 1, \\
  12xy + 13{y^2} = 25. \\
$

Answer 1

Hint: - Substitute the value from ${1^{st}}$equation into${2^{nd}}$equation.

Given equations is
$
  4x - 3y = 1.............................\left( 1 \right) \\
  12xy + 13{y^2} = 25...........................\left( 2 \right) \\
$
From equation 1
$y = \dfrac{{4x - 1}}{3}...................\left( 3 \right)$
Put this value of$y$in equation 2
$
  12x\left( {\dfrac{{4x - 1}}{3}} \right) + 13{\left( {\dfrac{{4x - 1}}{3}} \right)^2} = 25 \\
  4x\left( {4x - 1} \right) + \dfrac{{13}}{9}{\left( {4x - 1} \right)^2} = 25 \\
$
Multiply by 9 in equation
$
  36x\left( {4x - 1} \right) + 13{\left( {4x - 1} \right)^2} = 225 \\
  144{x^2} - 36x + 13\left( {16{x^2} + 1 - 8x} \right) = 225 \\
  352{x^2} - 140x - 212 = 0 \\
$
Divide by 4 in the equation
$88{x^2} - 35x - 53 = 0$
Divide the equation by 88.
$
  {x^2} - \dfrac{{35}}{{88}}x - \dfrac{{53}}{{88}} = 0 \\
  {x^2} - x + \dfrac{{53}}{{88}}x - \dfrac{{53}}{{88}} = 0 \\
$
So, factorize this equation


$
  \left( {x - 1} \right)\left( {x + \dfrac{{53}}{{88}}} \right) = 0 \\
   \Rightarrow x - 1 = 0 \Rightarrow x = 1 \\
   \Rightarrow x + \dfrac{{53}}{{88}} \Rightarrow x = - \dfrac{{53}}{{88}} \\
$
Now, from equation 3
$y = \dfrac{{4x - 1}}{3}$
When
$
  x = 1 \\
   \Rightarrow y = \dfrac{{4 - 1}}{3} = \dfrac{3}{3} = 1 \\
$
When
$
  x = - \dfrac{{53}}{{88}} \\
  y = \dfrac{{4\left( { - \dfrac{{53}}{{88}}} \right) - 1}}{3} = \dfrac{{ - \dfrac{{53}}{{22}} - 1}}{3} = \dfrac{{ - 75}}{{22 \times 3}} = - \dfrac{{25}}{{22}} \\
$
So, the required solution for the given equation is $\left( {1,1} \right),{\text{ }}\left( { - \dfrac{{53}}{{88}}, - \dfrac{{25}}{{22}}} \right)$

Note: - whenever we face such types of question always put the value of$x$or$y$ from simple equation into complex equation, then simplify the equation and find out the value of $x$or$y$, then put these values in the first equation we will get the required solution of the equations.