|y-1|=2y

Remove the absolute value term. This creates a ± on the right side of the equation because |x|=±x.

y-1=±2y

Set up the positive portion of the ± solution.

y-1=2y

Add 1 to both sides of the equation.

y=2y+1

Subtract 2y from both sides of the equation.

y-2y=1

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

1,y,1

Since 1,y,1 contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part 1,1,1 then find LCM for the variable part y1.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number 1 is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of 1,1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for y1 is y itself.

y1=y

y occurs 1 time.

The LCM of y1 is the result of multiplying all prime factors the greatest number of times they occur in either term.

y

y

Multiply each term by y and simplify.

Multiply each term in y-2y=1 by y in order to remove all the denominators from the equation.

y⋅y-2y⋅y=1⋅y

Simplify each term.

Multiply y by y.

y2-2y⋅y=1⋅y

Cancel the common factor of y.

Move the leading negative in -2y into the numerator.

y2+-2y⋅y=1⋅y

Cancel the common factor.

y2+-2y⋅y=1⋅y

Rewrite the expression.

y2-2=1⋅y

y2-2=1⋅y

y2-2=1⋅y

Multiply y by 1.

y2-2=y

y2-2=y

Solve the equation.

Subtract y from both sides of the equation.

y2-2-y=0

Factor y2-2-y using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -2 and whose sum is -1.

-2,1

Write the factored form using these integers.

(y-2)(y+1)=0

(y-2)(y+1)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

y-2=0

y+1=0

Set the first factor equal to 0 and solve.

Set the first factor equal to 0.

y-2=0

Add 2 to both sides of the equation.

y=2

y=2

Set the next factor equal to 0 and solve.

Set the next factor equal to 0.

y+1=0

Subtract 1 from both sides of the equation.

y=-1

y=-1

The final solution is all the values that make (y-2)(y+1)=0 true.

y=2,-1

y=2,-1

y=2,-1

Set up the negative portion of the ± solution.

y-1=-2y

Add 2y to both sides of the equation.

y-1+2y=0

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

1,1,y,1

Since 1,1,y,1 contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part 1,1,1,1 then find LCM for the variable part y1.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number 1 is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of 1,1,1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for y1 is y itself.

y1=y

y occurs 1 time.

The LCM of y1 is the result of multiplying all prime factors the greatest number of times they occur in either term.

y

y

Multiply each term by y and simplify.

Multiply each term in y-1+2y=0 by y in order to remove all the denominators from the equation.

y⋅y-1⋅y+2y⋅y=0⋅y

Simplify each term.

Multiply y by y.

y2-1⋅y+2y⋅y=0⋅y

Rewrite -1y as -y.

y2-y+2y⋅y=0⋅y

Cancel the common factor of y.

Cancel the common factor.

y2-y+2y⋅y=0⋅y

Rewrite the expression.

y2-y+2=0⋅y

y2-y+2=0⋅y

y2-y+2=0⋅y

Multiply 0 by y.

y2-y+2=0

y2-y+2=0

Solve the equation.

Use the quadratic formula to find the solutions.

-b±b2-4(ac)2a

Substitute the values a=1, b=-1, and c=2 into the quadratic formula and solve for y.

1±(-1)2-4⋅(1⋅2)2⋅1

Simplify.

Simplify the numerator.

Raise -1 to the power of 2.

y=1±1-4⋅(1⋅2)2⋅1

Multiply 2 by 1.

y=1±1-4⋅22⋅1

Multiply -4 by 2.

y=1±1-82⋅1

Subtract 8 from 1.

y=1±-72⋅1

Rewrite -7 as -1(7).

y=1±-1⋅72⋅1

Rewrite -1(7) as -1⋅7.

y=1±-1⋅72⋅1

Rewrite -1 as i.

y=1±i72⋅1

y=1±i72⋅1

Multiply 2 by 1.

y=1±i72

y=1±i72

The final answer is the combination of both solutions.

y=1+i72,1-i72

y=1+i72,1-i72

y=1+i72,1-i72

The solution to the equation includes both the positive and negative portions of the solution.

y=2,-1,1+i72,1-i72

Solve for y |y-1|=2/y