Question

Find u and v using cross-multiplication method-
u + 3v = 2
3u + (2/3)*v = 1

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:u + 3v = 2 - - - (1)$

and

$\rm :\longmapsto\:3u + \dfrac{2}{3}v = 1$

can be rewritten as

$\rm :\longmapsto\:\dfrac{9u + 2v}{3} = 1$

$\rm :\longmapsto\:9u + 2v = 3 - - (2)$

Using Cross Multiplication method

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered} \begin{array}{|c|c|c|c|} \bf{2} & \bf{3}& \bf{1}& \bf{2} \\ \\ 3&2&1& 3\\ \\2&3&9&2\end{array}\end{gathered}$

$\rm :\longmapsto\:\dfrac{u}{9 - 4} = \dfrac{v}{18 - 3} = \dfrac{ - 1}{2 - 27}$

$\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{v}{15} = \dfrac{ - 1}{ - 25}$

$\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{v}{15} = \dfrac{1}{25}$

On taking first and third member, we have

$\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{1}{25}$

$\bf\implies \:u = \dfrac{1}{5}$

On taking second and third member, we have

$\rm :\longmapsto\: \dfrac{v}{15} = \dfrac{1}{25}$

$\bf\implies \:v = \dfrac{3}{5}$

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