Can someone correct me if I trypedthese incorrectly andslolve them PLEASE
5/(y-2)=y+2
(x/3)-(2/3)=(1/y-5)
(y+2)/7=1(y-5)
(x^2)/(x-4)-7/(x-4)=0
x^2/(x+3)-5/(x+3)=0
To solve these equations, we need to isolate the variable on one side of the equation. Let's go through each equation step by step to find their solutions:
1. 5/(y-2) = y+2
To get rid of the denominator, we can multiply both sides of the equation by (y-2):
5 = (y+2)(y-2)
Expanding the right side:
5 = y^2 - 4
Rearranging the equation:
y^2 - 4 - 5 = 0
y^2 - 9 = 0
Now we have a quadratic equation. Factoring it, we get:
(y - 3)(y + 3) = 0
Therefore, y can be either 3 or -3.
2. (x/3) - (2/3) = 1/(y-5)
Multiplying both sides of the equation by 3(y-5) to eliminate the denominators:
x(y-5) - 2(y-5) = 3
Expanding and rearranging:
xy - 5x - 2y + 10 = 3
xy - 2y = 5x - 7
Factoring out y:
y(x - 2) = 5x - 7
Dividing both sides by (x - 2):
y = (5x - 7)/(x - 2)
This expression represents y in terms of x.
3. (y+2)/7 = 1(y-5)
Distributing 1 to (y - 5):
(y + 2)/7 = y - 5
Now, we can multiply both sides of the equation by 7 to eliminate the fraction:
y + 2 = 7y - 35
Rearranging:
6y = 37
Dividing both sides by 6:
y = 37/6
4. (x^2)/(x-4) - 7/(x-4) = 0
Multiplying both sides by (x-4) to eliminate the denominator:
x^2 - 7 = 0
Rearranging:
x^2 = 7
Taking the square root of both sides (considering both positive and negative roots):
x = ±√7
5. x^2/(x+3) - 5/(x+3) = 0
Multiplying both sides by (x+3) to eliminate the denominator:
x^2 - 5 = 0
Rearranging:
x^2 = 5
Taking the square root of both sides (considering both positive and negative roots):
x = ±√5
These are the step-by-step explanations for solving each of your equations.