Question

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

Answer 1

Answer:

  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

__

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

Answer 2

Answer:

0 < k < 2

Step-by-step explanation:

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

• x > 0 and y > 0

• x = 2k, k > 0

And replacing x with 2k in the second equation:

• 3x + 2y = 12
• 3*2k + 2y = 12
• 2y= 12 - 6k
• y = 6 - 3k

Since y > 0:

• 6 - 3k> 0
• 2 - k > 0
• k < 2

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

• 0 < k < 2