# Values for N

• May 8th 2007, 06:46 AM
Dragon
Values for N
What are the values of n for which the system 3x+ny=3,2x-4y=1 has a solution x<0,y<0
• May 8th 2007, 06:56 AM
topsquark
Quote:

Originally Posted by Dragon
What are the values of n for which the system 3x+ny=3,2x-4y=1 has a solution x<0,y<0

3x + ny = 3
2x - 4y = 1

Solve the bottom equation for y:
y = (1/2)x - (1/4)

and insert this into the top equation:
3x + n[(1/2)x - (1/4)] = 3

Now solve for x:
(3 + n/2)x - n/4 = 3

(6 + n)/2 * x = n/4 + 3

x = 2*(n/4 + 3)/(n + 6)

x = (n/2 + 6)/(n + 6)

To simplify this a bit, multiply the numerator and denominator by 2:
x = (n + 12)/(2n + 12)

So
y = (1/2)[(n + 12)/(2n + 12)] - (1/4)

y = (n + 12)/(4n + 24) - (1/4)

y = (4n + 12 - (4n + 24))/[4(4n + 24)]

y = 9/(4n + 24)

So
x = (n + 12)/(2n + 12)
y = 9/(4n + 24)

For what values of n are both x and y negative? Well, both the numerators and denominators have to have the opposite sign of each other. But since the "9" in the y equation is always positive, we require that 4n + 24 < 0, or n < -6.

Now, for -12 < n < -6 the numerator of x is positive and the denominator negative, so x is also negative here. But for n < -12 the numerator of x is negative, so x is positive. Thus

-12 < n < -6

-Dan

-Dan