# Solve the following equations:

$

4x - 3y = 1, \\

12xy + 13{y^2} = 25. \\

$

Last updated date: 19th Mar 2023

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Total views: 309.6k

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Answer

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309.6k+ views

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.

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.

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