12. Find the number of solutions to each system.
a. 4xy+1=0
4x-y+30
b. 2xy+4=0 4x-2y+8=0
a. The given system of equations is:
4xy + 1 = 0
4x - y + 30 = 0
To find the number of solutions, we need to find the intersection point(s) of the two equations.
First, let's isolate y in the first equation:
4xy + 1 = 0
4xy = -1
y = -1/(4x)
Now substitute this value of y in the second equation:
4x - (-1/(4x)) + 30 = 0
4x + 1/(4x) + 30 = 0
Multiplying through by 4x to clear the denominator:
16x^2 + 1 + 120x = 0
This is a quadratic equation in terms of x. However, upon solving it, we find that it does not have any real solutions. Therefore, the given system of equations has 0 solutions.
b. The given system of equations is:
2xy + 4 = 0
4x - 2y + 8 = 0
We can solve this system using substitution or elimination method. Let's solve it using the substitution method.
First, isolate y in the first equation:
2xy + 4 = 0
2xy = -4
y = -2/x
Now replace y in the second equation with -2/x:
4x - 2(-2/x) + 8 = 0
4x + 4/x + 8 = 0
Multiplying through by x to clear the denominator:
4x^2 + 4 + 8x = 0
Simplifying:
4x^2 + 8x + 4 = 0
x^2 + 2x + 1 = 0
(x + 1)^2 = 0
The quadratic equation has a single solution x = -1.
Now substitute this value of x back into either of the original equations (let's take the first equation):
2xy + 4 = 0
2(-1)y + 4 = 0
-2y + 4 = 0
-2y = -4
y = 2
Therefore, the system of equations has a unique solution x = -1, y = 2.
shorten them both
a. The given system of equations, 4xy + 1 = 0 and 4x - y + 30 = 0, has 0 solutions.
b. The given system of equations, 2xy + 4 = 0 and 4x - 2y + 8 = 0, has a unique solution x = -1, y = 2.