 Hello and welcome to the session. Let's discuss the following question. It says, find the value of k for which the following system of linear equations has infinite number of solutions. So let's now move on to the solution. The given system of equations is x plus k plus 1 into y is equal to 5. And k plus 1 into x plus 9y is equal to 8k minus 1. Now here, a1 that is the coefficient of, in the first linear equation is 1 and a2 is k plus 1. So a1 upon a2 is 1 upon k plus 1. Similarly, b1 upon b2 is k plus 1 upon 9 and c1 upon c2 is 5 upon 8k minus 1. We have to find the value of k for which this system of equation has infinite number of solutions. So for infinite number of solutions, we have a1 upon a2 is equal to b1 upon b2 is equal to c1 upon c2. 1 upon k plus 1 is equal to k plus 1 upon 9 is equal to 5 upon 8k minus 1. Now we'll solve these for k. So first we'll take 1 and 2. So taking 1 and 2, 1 upon k plus 1 is equal to k plus 1 upon 9. So this implies 9 is equal to k plus 1 whole square. So this implies k plus 1 is equal to plus minus 3 and this implies k is equal to plus minus 3 minus 1. So this implies k is equal to plus 3 minus 1 or k is equal to minus 3 minus 1. So this implies k is equal to 2 or minus 4. Now taking second and third, we have k plus 1 upon 9 is equal to 5 upon 8k minus 1. So this implies 8k minus 1 into k plus 1 is equal to 45. Now 8k into k is 8k square. 8k into 1 is 8k minus 1 into k is minus 1k. Minus 1 into plus 1 is minus 1 is equal to 45. So this implies 8k square plus 7k minus 1 minus 45 is equal to 0. And this implies 8k square plus 7k minus 46 is equal to 0. Now we have to factorize this quadratic equation and we have to solve this for k. So let us factorize this 8k square plus 23k minus 16k minus 46 is equal to 0. So this implies taking k common from the first two terms we have k into 8k plus 23 taking 16 common taking 2 common minus 2 common from the last two terms we have minus 2 into 8k plus 23 is equal to 0. Now this implies k minus 2 into 8k plus 23 is equal to 0 taking 8k plus 23 common we have this. So this implies k minus 2 is equal to 0 or 8k plus 23 is equal to 0. So this implies k is equal to 2 or k is equal to minus 23 upon 8 taking first and third we have 1 upon k plus 1 is equal to 5 upon 8k minus 1. So we have 8k minus 1 is equal to 5 into k plus 1. So this implies 8k minus 1 is equal to 5k plus 5. So this implies 8k minus 5k is equal to 5 plus 1. Now 8k minus 5k is 3k is equal to 6. So this implies k is equal to 6 by 3 that is 2. Now we see that k is equal to 2 is the common value of k in all the three cases thus equal to 2 is solution. This completes the question and the session while for now take care have a good day.