Question

Find the value of K for which the given Equations have infinitely many solutions

2x - 3y = 7

(K + 1)x + (1 - 2k)y = ( 5k - 4)

Answer 1

$<b>==$

[Hello]

Given Equation:-

2x-3y=7

(k+1)x+(1-2y)=5k-4

Where as:-

a1=)2

a2=)k+1
========

b1=)-3

b2=(1-2k)
=========

c1=) 7

c2=) 5k-4

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For Infinitely many Solution we have must be:-

$\bold{ \huge{ \fbox{ \frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}}}}$
$\bold{ \huge{ \fbox{ \frac{2}{k + 1} = \frac{ - 3}{1 - 2k} = \frac{7}{5k - 4}}}}$
$\bold{ \huge{ \fbox{ \frac{2}{k + 1} = \frac{7}{5k - 4}}}}$

2(5k-4)= 7(k+1)

10k-8=) 7k+7

10k-7k=) 7+8

3k=) 15

$\bold{ \huge{k = \frac{15}{3} =) 5}}$

$\bold{ \huge{ thanks}}$

Answer 2

(a)

Given Equation is 2x - 3y = 7.

= > 2x - 3y - 7 = 0.

Here, a1 = 2, b1 = -3, c1 = -7.

(b)

Given Equation is (k + 1)x + (1 - 2k)y = (5k - 4).

= > (k + 1)x + (1 - 2k)y - (5k - 4) = 0

Here, a2 = (k + 1), b2 = (1 - 2k), c2 = -(5k - 4).

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Given that the equation have infinitely many solutions.

= > (a1/a2) = (b1/b2) = (c1/c2)

= > (2/k + 1) = (-3/1 - 2k) = (7/5k - 4)

(i)

= > (2/k + 1) = (-3/1 - 2k)

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

= > 2 - 4k = -3k - 3

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

= > k = 5.

(ii)

= > (-3/1 - 2k) = (7/5k - 4)

= > -3(5k - 4) = 7(1 - 2k)

= > -15k + 12 = 7 - 14k

= > 12 - 7 = -14k + 15k

= > k = 5.

Therefore, the value of k = 5.

Hope this helps!