Question

For what values of ‘k’ will the following pair of linear equations will have infinetly
many solutions?
2x + 3y = k ; (k – 1)x + (2k – 1)y = 4k + 1

Answer 1

Answer:

For what values of ‘k’ will the following pair of linear equations

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Step-by-step explanation:

2x + 3y = k ; (k – 1)x + (2k – 1)y = 4k + 1

Answer 2

Step-by-step explanation:

Solution :

We have,

2x + 3y = 7

=> 2x + 3y - 7 = 0

(k + 1)x + (2k - 1)y = 4k + 1

=> (k + 1)x (2k - 1)y - (4k + 1) = 0

For infinitely,

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$

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

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

$= > 2(2k + 1) = 3(k + 1)$

$= > 4k - 2 = 3k + 3$

$= > 4k - 3k = 3 + 2$

$= > k = 5$

Hence :

The value of k is 5.