Question

These are two simultaneous equations.
Solve using cross multiplication. ​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\: - x + 2y = 1$

and

$\rm :\longmapsto\:5x - 9y = - 3$

So, using Cross multiplication method, we have

$\red{\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 2 & \sf 1 & \sf - 1 & \sf 2\\ \\ \sf - 9 & \sf - 3 & \sf 5 & \sf - 9\\ \end{array}} \\ \end{gathered}}$

So, we get

$\rm :\longmapsto\:\dfrac{x}{ - 6 - ( - 9)} = \dfrac{y}{5 - 3} = \dfrac{ - 1}{9 - 10}$

$\rm :\longmapsto\:\dfrac{x}{ - 6 + 9} = \dfrac{y}{2} = \dfrac{ - 1}{ - 1}$

$\rm :\longmapsto\:\dfrac{x}{3} = \dfrac{y}{2} = \dfrac{1}{ 1}$

$\rm \implies\:\boxed{ \tt{ \: x = 3 \: \: and \: \: y = 2 \: }}$

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Verification :-

Consider Equation (1)

$\rm :\longmapsto\: - x + 2y = 1$

On substituting the values of x and y, we get

$\rm :\longmapsto\: - 3 + 2 \times 2 = 1$

$\rm :\longmapsto\: - 3 + 4 = 1$

$\bf\implies \:1 = 1$

Hence, Verified

Consider Equation (2)

$\rm :\longmapsto\:5x - 9y = - 3$

On substituting the values of x and y, we get

$\rm :\longmapsto\:5 \times 3 - 9 \times 2= - 3$

$\rm :\longmapsto\:15 - 18= - 3$

$\bf\implies \: - 3 = - 3$

Hence, Verified