Two questions on solving for x (precalc)??

Question

Solve for x:
1. 2xy^2 - y^2 - (1 - 3y) = x
2. x - y^2 + x(2zy^2 + 2y) = xz^2 + z

These are really impossble for me to figure out. I keep getting stuck on both of these problems. For #1, I set one side equal to zero but I got stuck on the fourth line. For #2, I again set one side equal to zero but could not factor the expression on one side.

Thanks for the help in advance! :)

Answer 1

1. 2xy^2 - y^2 -(1-3y) =x
-y^2+3y-1 = x-2xy^2 factor out the x
(-y^2+3y-1) = x (1-2y^2) and divide
(-y^2+3y-1) /(1-2y^2) = x

2. x - y^2 + x(2zy^2 + 2y) = xz^2 + z
put all the x variable to one side
x(2zy^2 + 2y) + x - xz^2 = y^2+z factor
x( 2zy^2 + 2y + z^2 +1) =(y^2+z)
x = (y^2+z)/( 2zy^2 + 2y + z^2 +1)

Answer 2

1. 2xy^2 - y^2 - (1 - 3y) = x

2xy^2 - y^2 - 1 + 3y = x

2xy^2 - x = y^2 - 3y + 1

x(2y^2 - 1) = y^2 - 3y + 1

x = (y^2 - 3y + 1)/(2y^2 - 1)


2. x - y^2 + x(2zy^2 + 2y) = xz^2 + z

x - y^2 + 2xy^2z + 2xy - xz^2 = z

x + 2xy^2z + 2xy - xz^2 = z + y^2

x(1 + 2y^2z + 2y - z^2) = z + y^2

x = (z + y^2)/(1 + 2y^2z + 2y - z^2)

Answer 3

Y1 = sin (4x) Y2 = ln (x + 2) Graph (window: xmin = 0, xmax = pi; ymin = -a million , ymax = a million) Mode: radians locate intersection. x-coordinate of intersection(s) may be the answer(s) you won't be in a position to unravel this algebraically it quite is going to require the two graphing or another analytic approach There are 2 ideas interior the needed era: x = .23 and x = .50 (to 2 decimal places) there's a third intersection ( answer), besides the undeniable fact that it quite is below x = 0

Answer 4

y^2 + (1-3y) = 2xy^2 - x
y^2 + (1-3y) = x(2y^2 -1 )
x = (y^2 -3y + 1)/(2y^2 -1)


x - xz^2 + x(2y^2 + 2y) = z + y^2
x(1 - z^2 + 2y^2 - 2y) = z + y^2
x = (z + y^2)/(1-z^2 + 2y^2 - 2y)

Answer 5

2xy² - y² + 3y - x - 1 = 0.

2xy² - x = y² - 3y + 1.

x(2y² - 1) = y² - 3y + 1.

x = (y² - 3y + 1)/(2y² - 1).