Solve y+1/2(y-1)leq1

$$\frac { y + 1 } { 2 ( y - 1 ) } \leq 1$$

Solve for y

$$y\in (-\infty,1)\cup [3,\infty)$$

Solution Steps

Use the distributive property to multiply 2 by y-1.

$$\frac{y+1}{2y-2}\leq 1$$

Denominator 2y-2 cannot be zero since division by zero is not defined. There are two cases.

$$2y-2>0 2y-2<0$$

Consider the case when 2y-2 is positive. Move -2 to the right hand side.

$$2y>2$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$y>1$$

The initial inequality does not change the direction when multiplied by 2y-2 for 2y-2>0.

$$y+1\leq 2y-2$$

Move the terms containing y to the left hand side and all other terms to the right hand side.

$$y-2y\leq -1-2$$

Combine like terms.

$$-y\leq -3$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$y\geq 3$$

Now consider the case when 2y-2 is negative. Move -2 to the right hand side.

$$2y<2$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$y<1$$

The initial inequality changes the direction when multiplied by 2y-2 for 2y-2<0.

$$y+1\geq 2y-2$$

Move the terms containing y to the left hand side and all other terms to the right hand side.

$$y-2y\geq -1-2$$

Combine like terms.

$$-y\geq -3$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$y\leq 3$$

Consider condition y<1 specified above.

$$y<1$$

The final solution is the union of the obtained solutions.

$$y\in (-\infty,1)\cup [3,\infty)$$