Question

sᴏʟᴠᴇ ʙʏ ᴄʀᴏss ᴍᴜʟᴛɪᴘʟɪᴄᴀᴛɪᴏɴ:

2x+3y=6
6x-5y=4

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Answer 1

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Answer 2

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

Given pair of linear equations are

$\rm :\longmapsto\:2x + 3y = 6$

and

$\rm :\longmapsto\:6x - 5y = 4$

Using Cross Multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 3 & \sf 6 & \sf 2 & \sf 3\\ \\ \sf - 5 & \sf 4 & \sf 6 & \sf - 5\\ \end{array}} \\ \end{gathered}$

Now,

$\rm :\longmapsto\:\dfrac{x}{12 - ( - 30)} = \dfrac{y}{36 - 8} = \dfrac{ - 1}{ - 10 - 18}$

$\rm :\longmapsto\:\dfrac{x}{12 + 30} = \dfrac{y}{28} = \dfrac{ - 1}{ - 28}$

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

Taking first and third member, we get

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

$\bf\implies \:x = \dfrac{42}{28} = \dfrac{3}{2}$

Taking second and third member, we get

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

$\bf\implies \:y = \dfrac{28}{28} = 1$

Hence,

$\purple{\bf :\longmapsto\:x = \dfrac{3}{2} }$

and

$\purple{\bf :\longmapsto\:y =1}$

Verification :-

Consider first equation,

$\rm :\longmapsto\:2x + 3y = 6$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2 \times \dfrac{3}{2} + 3 \times 1 = 6$

$\rm :\longmapsto\:3 + 3 = 6$

$\rm :\longmapsto\:6 = 6$

Hence, Verified

Consider second equation,

$\rm :\longmapsto\:6x - 5y = 4$

On substituting the values of x and y, we get

$\rm :\longmapsto\:6 \times \dfrac{3}{2} - 5 \times 1 = 4$

$\rm :\longmapsto\:9 - 5 = 4$

$\rm :\longmapsto\:4 = 4$

Hence, Verified