Can someone please show me how to solve these for me?
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
I assume that your second equation has y as a typo. I will use that as an example.
(x/3)-(2/3)=(1/x-5)
Multiply both sides by 3. I will also assume that (1/x-5) is the same as (1/x)-5 rather than 1/(x-5).
x-2=3/x-15
Get the x terms on one side and the numerical terms on the other by adding 2 and subtracting 3/x from both sides.
x-3/x=17
Multiply both sides by x.
x^2-3=17x
x^2-17x-3=0
Unfortunately, I cannot carry the process further. However, it should make the process clear.
I hope this helps. Thanks for asking.
To solve the first equation, 5/(y-2) = y+2, you can follow these steps:
1. Multiply both sides of the equation by (y-2) to get rid of the denominator:
5 = (y-2)(y+2)
2. Expand the right side of the equation:
5 = y^2 - 4
3. Rearrange the equation to get all terms on one side:
y^2 - 4 - 5 = 0
4. Simplify:
y^2 - 9 = 0
This equation is a quadratic equation, so you can use factoring, completing the square, or the quadratic formula to find the solutions.
Similarly, you can follow these steps to solve the rest of the equations you provided.