For what value of k, the following system of linear equations has infinitely many solutions 5x+2y=2k 2(k+1)x+ky=3k+4 - Maths - Real Numbers

Question

For what value of k, the following system of linear equations has
infinitely many solutions? 5x+2y=2k 2(k+1)x+ky=3k+4

Decoder_6 · Accepted Answer

Given equations are:
5x+2y=2k
2(k+1)x+ky=3k+4

⇒5x+2y-2k=0⇒2(k+1)+ky-(3k+4)=0For the equations to have infinitely many solutions,a1a2=b1b2=c1c2Here,a1=5a2=2(k+1)b1=2b2=kc1=-2kc2=-(3k+4)∴52(k+1)=2k=-2-(3k+4)⇒52(k+1)=2k⇒5k=4(k+1)⇒5k=4k+4⇒k=4

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For what value of k, the following system of linear equations has infinitely many solutions? 5x+2y=2k 2(k+1)x+ky=3k+4