Question

solve by cross multiplication method 2x-5y-12=0 and 4x+6y-14=0​

Answer 1

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

Given pair of linear equations are .

$\rm :\longmapsto\:2x - 5y - 12 = 0 - - - (1)$

and

$\rm :\longmapsto\:4x + 6y - 14 = 0 - - - (2)$

Now,

Equations (1) and (2) can be rewritten as

$\rm :\longmapsto\:2x - 5y = 12 - - - (3)$

and

$\rm :\longmapsto\:2x + 3y = 7 - - - (4)$

Now, using Cross multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - 5 & \sf 12 & \sf 2 & \sf - 5\\ \\ \sf 3 & \sf 7 & \sf 2 & \sf 3\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{x}{ - 35 - 36} = \dfrac{y}{24 - 14} = \dfrac{ - 1}{6 - ( - 10)}$

$\rm :\longmapsto\:\dfrac{x}{ - 71} = \dfrac{y}{10} = \dfrac{ - 1}{16}$

$\rm :\longmapsto\:\dfrac{x}{ - 71} = \dfrac{y}{10} = \dfrac{1}{ - 16}$

Taking first and third member, we get

$\rm :\longmapsto\:\dfrac{x}{ - 71} = \dfrac{1}{ - 16}$

$\bf\implies \:x = \dfrac{71}{16}$

Now, Taking second and third member, we get

$\rm :\longmapsto\: \dfrac{y}{10} = \dfrac{1}{ - 16}$

$\rm :\longmapsto\: y = \dfrac{10}{ - 16}$

$\rm :\longmapsto\: y = - \: \dfrac{10}{16}$

$\bf\implies \:y = - \: \dfrac{5}{8}$

So, Solution of equations

$\rm :\longmapsto\:2x - 5y - 12 = 0$

and

$\rm :\longmapsto\:4x + 6y - 14 = 0$

is

$\bf\implies \: \boxed{ \bf{x = \dfrac{71}{16} }}$

and

$\bf\implies \: \boxed{ \bf{y = - \: \dfrac{5}{8} }}$

Verification :-

The given first equation is

$\rm :\longmapsto\:2x - 5y - 12 = 0$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2 \times \dfrac{71}{16} - 5 \times \dfrac{ - 5}{8} - 12 = 0$

$\rm :\longmapsto\: \dfrac{142}{16} + \dfrac{25}{8} - 12 = 0$

$\rm :\longmapsto\: \dfrac{142 + 50 - 192}{16} = 0$

$\rm :\longmapsto\: \dfrac{192 - 192}{16} = 0$

$\rm :\longmapsto\: \dfrac{0}{16} = 0$

$\bf\implies \:0 = 0$

Hence, Verified

Consider, second equation,

$\rm :\longmapsto\:4x + 6y - 14 = 0$

On substituting the values of x and y, we have

$\rm :\longmapsto\:4 \times \dfrac{71}{16} + 6 \times \dfrac{ - 5}{8} - 14 = 0$

$\rm :\longmapsto\: \dfrac{71}{4} - \dfrac{15}{4} - 14 = 0$

$\rm :\longmapsto\: \dfrac{71 - 15 - 56}{4} = 0$

$\rm :\longmapsto\: \dfrac{56 - 56}{4} = 0$

$\rm :\longmapsto\: \dfrac{0}{4} = 0$

$\bf\implies \:0 = 0$

Hence, Verified