Question

Find the value of k for which the following pair of linear equation have infinitely many solution 2 X + 3 Y =7 ;(k-1)x +(k+2)y=3k

Answer 1

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Answer 2

given equation

2x+3y=7

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

for infinite many solution

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

comparing this we get

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

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

$2k + 4 = 3k - 3...........9k = 7k + 14$

$4 + 3 = 3k - 2k........9k - 7k = 14$

$7 = k........2k = 14$

$7 = k........k = 7$

the value of k for which the given equation have infinite many solutions is

7