5/21 is a repeating decimal but the first digits are 0.238095
First of all, we can observe that
So the expression becomes
This means that the expression is defined for every 
Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask
Since we can't accept 3 as an answer, the actual solution set is
![(-\infty,-2] \cup [2,3) \cup (3,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C-2%5D%20%5Ccup%20%5B2%2C3%29%20%5Ccup%20%283%2C%5Cinfty%29)
Answer:
M (-3, -5/2)
N (3, -1)
Step-by-step explanation:
Solve the first equation for x.
4y = x − 7
x = 4y + 7
Substitute into the second equation.
x² + xy = 4 + 2y²
(4y + 7)² + (4y + 7)y = 4 + 2y²
Simplify.
16y² + 56y + 49 + 4y² + 7y = 4 + 2y²
18y² + 63y + 45 = 0
2y² + 7y + 5 = 0
Factor.
(y + 1) (2y + 5) = 0
y = -1 or -5/2
Plug back into the first equation to find x.
x = 4(-1) + 7 = 3
x = 4(-5/2) + 7 = -3
M (-3, -5/2)
N (3, -1)
In Calculus - limits, there are several limits law, but the most common ones should be the following:
limx-->a [f(x) + g(x)] = L+M
limx-->a [f(x) - g(x)] = L-M
limx-->a f(x)/g(x) = L/M, if M does not equal to 0
Hope this helps!
