Question

1/x-1=2/x+1 x/x-5=x-2/x-6 3/z-2=2/z-3 y+1/y-1=y-5/y-3

Answer 1

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Answer 2

To solve these equations, we will need to isolate the variable on one side of the equation.

1. 1/x - 1 = 2/x + 1

   We can start by combining the two terms on the right side of the equation:

   1/x - 2/x = 1 + 1

   Simplifying, we get:

   -1/x = 2

   Multiplying both sides by -1, we get:

   1/x = -2

   Finally, we can isolate x by taking the reciprocal of both sides:

   x = -1/2

   Therefore, the solution to the equation is x = -1/2.

2. x/(x-5) = (x-2)/(x-6)

   We can start by cross-multiplying:

   x(x-6) = (x-2)(x-5)

   Expanding both sides, we get:

   x^2 - 6x = x^2 - 7x + 10

   Simplifying, we get:

   x = 10

   However, we must check if this solution is valid. We can do this by plugging x = 10 back into the original equation:

   10/(10-5) = (10-2)/(10-6)

   2 = 2

   Since the equation is true for x = 10, the solution is x = 10.

3. 3/(z-2) = 2/(z-3)

   We can start by cross-multiplying:

   3(z-3) = 2(z-2)

   Expanding both sides, we get:

   3z - 9 = 2z - 4

   Simplifying, we get:

   z = 5

   However, we must check if this solution is valid. We can do this by plugging z = 5 back into the original equation:

   3/(5-2) = 2/(5-3)

   1 = 1

   Since the equation is true for z = 5, the solution is z = 5.

4. y+1/(y-1) = y-5/(y-3)

   We can start by cross-multiplying:

   (y+1)(y-3) = (y-5)(y-1)

   Expanding both sides, we get:

   y^2 - 2y - 3 = y^2 - 6y + 5

   Simplifying, we get:

   4y = 8

   Dividing both sides by 4, we get:

   y = 2

   However, we must check if this solution is valid. We can do this by plugging y = 2 back into the original equation:

   2+1/(2-1) = 2-5/(2-3)

   3 = 3

   Since the equation is true for y = 2, the solution is y = 2.

Therefore, the solutions to the equations are:

x = -1/2, x = 10, z = 5, y = 2.