For what value of k, the following pair of linear equations has infinite number of solutions:
a) (k-1)x-y=5; (k+1)x+(1-k)y=(3k+1)

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Solution

Dear Student
The given system of equations:

(k-1)x-y =5
= (k-1)x-y-5 =0 ------(1)

(k+1)x+(1-k)y = 3k+1
(k+1)x+(1-k)y -(3k+1) =0 -------(2)

The equations are of the following form:
a₁x+b₁y+c₁ =0, a₂x+b₂y+c₂ =0

For an infinite number of solutions we must have:

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \frac{k - 1}{k + 1} = \frac{- 1}{- (k - 1)} = \frac{- 5}{- (3 k + 1)} \frac{k - 1}{k + 1} = \frac{1}{(k - 1)} = \frac{5}{(3 k + 1)} N o w C a s e 1 \frac{k - 1}{k + 1} = \frac{1}{(k - 1)} {(k - 1)}^{2} = k + 1 k^{2} - 2 k + 1 = k + 1 k^{2} - 3 k = 0 k = 0 o r 3 C a s e 2 \frac{1}{(k - 1)} = \frac{5}{(3 k + 1)} 3 k + 1 = 5 (k - 1) 3 k + 1 = 5 k - 5 2 k = 6 k = 3 C a s e 3 \frac{k - 1}{k + 1} = \frac{5}{(3 k + 1)} (3 k + 1) (k - 1) = 5 (k + 1) 3 k^{2} + k - 3 k - 1 = 5 k + 5 3 k^{2} - 2 k - 5 k - 1 - 5 = 0 3 k^{2} - 7 k - 6 = 0 3 k^{2} - 9 k + 2 k - 6 = 0 3 k (k - 3) + 2 (k - 3) = 0 (k - 3) (3 k + 2) = 0 k = 3 o r k = \frac{- 2}{3} H e n c e, t h e g i v e n s y s t e m o f e q u a t i o n s h a s a n i n f i n i t e n u m b e r o f s o l u t i o n s w h e n k i s e q u a l t o 3$

Regards

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Similar questions

(k - 1) x - y = 5, (k + 1) x + (1 - k) y = (3 k + 1) .

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

$5 x + 2 y = 2 k, 2 (k + 1) x + k y = (3 k + 4) .$

x + (k + 1) y = 5

(k + 1) x + 9 y = 8 k - 1

(3 k + 1) x + 3 y - 5 = 0 a n d 2 x - 3 y + 5 = 0

K .

5 x + 2 y = 2 k and 2 (k + 1) x + ky = (3 k + 4)

k = \frac{2}{3}

k = \frac{- 2}{3}

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