Find the values for k so that the intersection of x = 2k and 3x + 2y = 12 lies in the first quadrant.

0 < k < 2

Step-by-step explanation:

In the first quadrant both of x and y get positive values, so

And replacing x with 2k in the second equation:

Since y > 0:

Combining both k > 0 and k < 2, we get:

  0 < k < 2

Step-by-step explanation:

The intersection will lie in the first quadrant when the solution has the characteristics: x > 0, y > 0.

For x > 0, we require ...

  x = 2k

  x > 0

  2k > 0 . . . . substitute for x

  k > 0 . . . . . divide by 2

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For y > 0, we require ...

  y = (12 -3x)/2

  y = (12 -3(2k))/2 = 6 -3k . . . . . substituted for x and simplify

  y > 0

  6 -3k > 0 . . . . . . substitute for y

  2 -k > 0 . . . . . . . divide by 3

  k < 2 . . . . . . . . . add k

So, the values of k that result in the intersection being in the first quadrant are ...

  0 < k < 2