Question

# Solve the given equations 4x – 19y +13 = 0, 13x – 23y = –19.

Hint: Solving the equations means, we have to find the values of the variable in the given equations. We can use the substitution method to find the variables.

The given equations are
$4x - 19y + 13 = 0$ .... (1)

$13x - 23y = - 19$ .... (2)

From equation (2) we get x value as $x = \dfrac{{23y - 19}}{{13}}$

Substituting this x value in equation (1), we get

$4\left( {\dfrac{{23y - 19}}{{13}}} \right) - 19y + 13 = 0$

Multiplying the above equation with 13 on both sides,

$\Rightarrow 4 \times 13\left( {\dfrac{{23y - 19}}{{13}}} \right) - 19 \times 13y + 13 \times 13 = 0 \times 13$

On simplification,

\eqalign{ & \Rightarrow 4\left( {23y - 19} \right) - 247y + 169 = 0 \cr & \Rightarrow \left( {92y - 76} \right) - 247y + 169 = 0 \cr & \Rightarrow - 155y + 93 = 0 \cr & \Rightarrow 155y = 93 \cr & \Rightarrow y = \dfrac{{93}}{{155}} = \dfrac{3}{5} \cr}

Substituting this ‘y’ value in equation (2) to get ‘x’ value,

$\Rightarrow 13x - 23\left( {\dfrac{3}{5}} \right) = - 19$

Simplifying the above equation,

\eqalign{ & \Rightarrow 13x = - 19 + \dfrac{{69}}{5} \cr & \Rightarrow 13x = \dfrac{{ - 26}}{5} \cr & \Rightarrow x = \dfrac{{ - 26}}{{5 \times 13}} \cr & \Rightarrow x = \dfrac{{ - 2}}{5} \cr}

$\therefore$ The solution of given equations is $x = \dfrac{{ - 2}}{5},y = \dfrac{3}{5}$

Note: We have to change one equation completely into a one variable equation with the help of the second equation. Then we will get the value of one variable, we can substitute that value in any of the previous equations to get the value of the remaining variable.