Find the value of $k$ , infinitely many solutions
$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k$

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Solution

Consider the given equations.

$2 x + 3 y = 7$

$(k - 1) x + (k + 2) y = 3 k$

The general equations

$a_{1} x + b_{1} y = c_{1}$

$a_{2} x + b_{2} y = c_{2}$

So,

$a_{1} = 2, b_{1} = 3, c_{1} = 7$

$a_{2} = k - 1, b_{2} = k + 2, c_{2} = 3 k$

We know that the condition of infinite solution

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Therefore,

$\frac{2}{k - 1} = \frac{3}{k + 2} = \frac{7}{3 k}$

$\Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2}$

$\Rightarrow 2 k + 4 = 3 k - 3$

$\Rightarrow k = 7$

Hence, the value of $k$ is $7$ .

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Similar questions

Find the value of k for which the following pair of linear equations have infinitely many solutions:

$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k .$

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(k - 1) x + (k + 2) y = 3 k

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