For what value(s) of k will the following system of linear equati...

Question

For what value(s) of k will the following system of linear equations have no solution? infinitely many solutions? x - 2y = 3-2x + 4y = k

Accepted Answer

Find the values of few at which the system will have infinitely many solutions and no solution where we have four X plus seven, Y equals 28 negative 12 X minus 21 Y equals Q. Now to find infinitely many solutions, we need to have two equations in the system that have the same slope and same Y intercept. Let's first set our equations equals to Y equals, instead of in this form, we can first move our four X over which should get seven, Y equals negative four X plus 28. For our first equation then divide both sides by seven to get Y equals negative 47 X plus four. We can do the same on our second equation moving the 12 X over which is negative. We get negative 21 Y equals 12 X plus Q divide both sides by negative 21 to give us Y equals native 47 X minus Q divided by 21. Now, we have the same slope and we need to solve for Q by setting our Y intercepts equal to each other. Once you get four equals negative Q divided by 21. Now we can solve for Q multiply 21 on both sides to get equals negative Q and divided by negative one to get Q equals negative 84. This will be our answer for infinitely many solutions. Now for no solution, we need two equations that have the same slope but different Y intercepts. We know we have infinite mini when Q equals negative 84. So for no solution, Q should be not equals to negative 84. We now have both of our answers. Q equals negative 84 Q does not equal negative 84. This corresponds to answer B OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

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For what value(s) of k will the following system of linear equations have no solution? infinitely many solutions? x - 2y = 3-2x + 4y = k

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