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

1 thought on “Find u and v using cross-multiplication method-<br /> u + 3v = 2<br /> 3u + (2/3)*v = 1”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given pair of linear equations are

   [tex]\rm :\longmapsto\:u + 3v = 2 – – – (1)[/tex]

   and

   [tex]\rm :\longmapsto\:3u + \dfrac{2}{3}v = 1 [/tex]

   can be rewritten as

   [tex]\rm :\longmapsto\:\dfrac{9u + 2v}{3} = 1 [/tex]

   [tex]\rm :\longmapsto\:9u + 2v = 3 – – (2)[/tex]

   Using Cross Multiplication method

   [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \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}[/tex]

   [tex]\rm :\longmapsto\:\dfrac{u}{9 – 4} = \dfrac{v}{18 – 3} = \dfrac{ – 1}{2 – 27} [/tex]

   [tex]\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{v}{15} = \dfrac{ – 1}{ – 25} [/tex]

   [tex]\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{v}{15} = \dfrac{1}{25} [/tex]

   On taking first and third member, we have

   [tex]\rm :\longmapsto\:\dfrac{u}{5} = \dfrac{1}{25} [/tex]

   [tex]\bf\implies \:u = \dfrac{1}{5} [/tex]

   On taking second and third member, we have

   [tex]\rm :\longmapsto\: \dfrac{v}{15} = \dfrac{1}{25} [/tex]

   [tex]\bf\implies \:v = \dfrac{3}{5} [/tex]

   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

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