I have some difficult algebra problems

1.) 4/(p-3) - 2/p = 0

2.) Find the equation of the line that goes through points (3, -3) and (-3, 3).

3.) b + 2/m + y/zm^2

1.) 4/(p-3) - 2/p = 0

So you know that 4/(p-3) is equal to 2/p

4/(p-3) = 2/p

Multiply everything by p(p - 3)

4p = 2(p - 3)
4p = 2p - 6
2p = -6

p = -3

To check the answer, substitute -3 for p in the original equation

4/(-3 - 3) - 2/(-3) = 0
4/-6 + 2/3 = 0
-2/3 + 2/3 = 0
0 = 0

2.) Equation through the points (3, -3) and (-3, 3)

slope = (y2 - y1)/(x2 - x1)
slope = (3 + 3)/(-3 - 3)
slope = 6/-6
slope = -1

Using either of the points will find the equation, using (y - y1) = m(x - x1), where m is the slope, y1 is the y coordinate of the points, and x1 is the x coordinate of the point.

(y - (-3)) = -1(x - 3)
y + 3 = -x + 3
y = -x

3.) b + 2/m + y/zm²

I'm not sure if tthat'ssupposed to be (b + 2)/m or b + (2/m), so bare with me. The first one goes:

The LCD is zm², so multiply everything by powers of 1.

b(zm²/zm²) + 2(zm)/zm² + y/zm²

(bzm² + 2zm + y)/zm²

The second one goes:

(b + 2)/m + y/zm²

The LCD is zm², so multiply the first fraction by zm/zm

zm(b + 2)/zm² + y/zm²

(zmb + 2zm + y)/zm²

hey buddy im not sure about question 2 but if its this like im thinking it would be like this :

2-3