Question

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

Answer 1

k = any value between 0 and 2

Explanation:

We can start by considering the graph of 3x + 2y = 12. If you convert this line into point - slope form it will have a negative slope, and hence will extend infinitely in the negative direction. Therefore if we calculate the x - intercept, k will be bound by 0 to the x - intercept if their intersection lies in the first quadrant.

3x + 2y = 12,

3x + 2(0) = 12,

3x = 12,

x = 4 - this is the x - intercept

As x = 4 is the x - intercept, k can be any value between 0 and 4, and will lie in the first quadrant. But remember that we have the equation " x = 2k. " k will only be between 0 and 4 is the equation is " x = k. "

- Therefore k will instead be any positive value between 0 and 2.

Answer 2

0< k < 2

Explanation:

x = 2k

3x + 2y = 12

Substitute the first equation into the second equation

3(2k) +2y = 12

6k+2y = 12

Subtract 6k from each side

2y = 12 -6k

Divide by 2

y = 6 - 3k

For this to be in the first quadrant y > 0

6 - 3k > 0

6 > 3k

Divide by 3

2 > k

The lower limit is when it reaches the axis or when k=0

0 <k <2

Checking for x

x = 2k

If 0<k <2 then it will be in the first quadrant