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2 years ago

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

Mathematics

2 answers:

RUDIKE [14]2 years ago

5 0

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

RUDIKE [14]2 years ago

3 0

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

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