Question

6. 2х + 5y = 1, 2x + Зу = 3. Solve by the cross multiplication method​.


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Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:2x + 5y = 1$

and

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

By using cross multiplication method,

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

Now,

$\rm :\longmapsto\:\dfrac{x}{15 - 3} = \dfrac{y}{2 - 6} = \dfrac{ - 1}{6 - 10}$

$\rm :\longmapsto\:\dfrac{x}{12} = \dfrac{y}{ - 4} = \dfrac{ - 1}{ - 4}$

$\rm :\longmapsto\:\dfrac{x}{12} = \dfrac{y}{ - 4} = \dfrac{ 1}{4}$

On multiply by 4 each term, we get

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

$\bf\implies \:x = 3$

and

$\bf\implies \:y = - \: 1$

Check the solution :-

Assume the first equation,

$\rm :\longmapsto\:2x + 5y = 1$

On substitute the value of x = 3 and y = - 1,

$\rm :\longmapsto\:2 \times 3 + 5 \times ( - 1) = 1$

$\rm :\longmapsto\:6 - 5 = 1$

$\bf\implies \:1 = 1$

Hence, verified

Answer 2

Answer:

-1

Step-by-step explanation:

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