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प्रश्न
Find the value of k for which the following pair of linear equations has infinitely many solutions.
2x + 3y = 7, (k +1) x+ (2k -1) y = 4k + 1
उत्तर
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 many solutions
`a_1/a_2 = b_1/b_2 = c_1/c_2`
⇒ `(2)/(k+1) = (3)/((2k -1)) = (-7)/-(4k +1)`
⇒ `(2)/(k+1) = (3)/(2k -1)`
⇒ `2(2k + 1) = 3 (k+1)`
⇒`4k - 2 = 3k + 3`
⇒`4k - 3k = 3 +2`
`k = 5`
or
⇒ `(2)/(k+1) = (-7)/-(4k +1)`
⇒ `2(4k + 1) = 7 (k+1)`
⇒ `8k + 2 = 7k + 2`
⇒`8k - 7k = 7- 2`
`k = 5`
Hence, the value of k is 5 for which given equations have infinitely many solutions.
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