Answer:
0< k < 2
Step-by-step 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
Answer:
k = any value between 0 and 2
Step-by-step 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.
