Question

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Answer 1

Answer:

See the above answer

Step-by-step explanation:

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Answer 2

Given Question :-

Solve the simultaneous linear equations :-

2x + 3y = xy

and

3x + 4y = 2xy

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

Given pair of equations are

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

and

$\rm :\longmapsto\:3x + 4y = 2xy$

Now, here if we put x = 0 in either of the two equations, we get y = 0

So, one solution is x = 0 and y = 0.

To find the other solution,

Let assume that x and y are non - zero.

So,

Divide both equations by xy, we get

$\rm :\longmapsto\:\dfrac{2}{y} + \dfrac{3}{x} = 1 - - - (1)$

and

$\rm :\longmapsto\:\dfrac{3}{y} + \dfrac{4}{x} = 2 - - - (2)$

Now, multiply equation (1) by 3 and equation (2) by 2, we get

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

and

$\rm :\longmapsto\:\dfrac{6}{y} + \dfrac{8}{x} = 4 - - - (4)$

On Subtracting equation (4) from equation (3) we get

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

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

On substituting x = - 1, in equation (1) we get

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

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

$\rm :\longmapsto\:\dfrac{2}{y} = 4$

$\rm :\longmapsto\:\dfrac{2}{4} = y$

$\bf\implies \:y = \dfrac{1}{2}$

$\red{\:\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: So-\begin{cases} &\sf{x \: = - \: 1} \\ \\ &\sf{y \: = \: \dfrac{1}{2} } \end{cases}\end{gathered}\end{gathered}}$

Verification :-

Consider equation

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

On substituting the values of x and y, we get

$\rm :\longmapsto\:2( - 1) + 3\bigg[\dfrac{1}{2} \bigg] = - (1)\bigg[\dfrac{1}{2} \bigg]$

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

$\rm :\longmapsto\: \dfrac{ - 4 + 3}{2} = - \dfrac{1}{2}$

$\bf :\longmapsto\: - \dfrac{1}{2} = - \dfrac{1}{2}$

Hence, Verified

Hence, Solution of equations are

$\red{\:\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: So-\begin{cases} &\sf{x \: = - \: 1} \\ \\ &\sf{y \: = \: \dfrac{1}{2} } \end{cases}\end{gathered}\end{gathered}}$

and

$\red{\:\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: So-\begin{cases} &\sf{x \: = \: 0} \\ \\ &\sf{y \: = \: 0 } \end{cases}\end{gathered}\end{gathered}}$