Question

4x-3y-8=0,6x-y-29/3=0 by cross multiplication method​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:4x - 3y = 8 - - (1)$

and

$\rm :\longmapsto\:6x - y - \dfrac{29}{3} = 0$

can be rewritten as

$\rm :\longmapsto\:6x - y = \dfrac{29}{3}$

$\rm :\longmapsto\:18x - 3y = 29 - - (2)$

Using Cross Multiplication method

$\: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered} \begin{array}{|c|c|c|c|} \bf{2} & \bf{3}& \bf{1}& \bf{2} \\ \\ - 3&8&4& - 3\\ \\ - 3&29&18& - 3\end{array}\end{gathered}$

$\rm :\longmapsto\:\dfrac{x}{ - 87 + 24} = \dfrac{y}{144 - 116} = \dfrac{ - 1}{ - 12 + 54}$

$\rm :\longmapsto\:\dfrac{x}{ -63} = \dfrac{y}{28} = \dfrac{ - 1}{42}$

$\rm :\longmapsto\:\dfrac{x}{ -63} = \dfrac{y}{28} = \dfrac{ 1}{ - 42}$

On taking first and third member, we have

$\rm :\longmapsto\:\dfrac{x}{ -63} = \dfrac{ 1}{ - 42}$

$\rm :\longmapsto\:x = \dfrac{ - 63}{ - 42}$

$\bf\implies \:x = \dfrac{3}{2}$

On taking second and third member, we have

$\rm :\longmapsto\: \dfrac{y}{28} = \dfrac{ 1}{ - 42}$

$\rm :\longmapsto\:y = \dfrac{28}{ - 42}$

$\bf\implies \:y = - \: \dfrac{2}{3}$

Additional Information :-

There are 4 methods to solve this type of pair of linear equations.

• 1. Method of Substitution

• 2. Method of Eliminations

• 3. Method of Cross Multiplication

• 4. Graphical Method