Question

Find the value of k, if the system of linear equation (3k+4)x+2y= (-k+3) ; (2k-5)x+4y=7-k has no solution

Answer 1

(3k+4)x+2y= (-k+3). (2k-5)x+4y=7-k

comparing with comparing with

a1x^2+b1c+c1=0 a2x^2+b2x+c2=0

a1 = 3k+4. a2 =2k-5

b1=2 b2 =4

c1 = -(-k+3 ) c2 = -(7-k)

for no solution

a1/a2=b1/b2 ≠c1/c2

3k+4/2k-5 = 2/4≠ -(-k+3) / -(7-k)

4(3k+4) = 2( 2k-5)

12k+16=4k-10

12k-4k =-16-10

8k = -26

k= -26/8

k = -13/4

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