(y+3)(y+1)=2y(y+3)

Solution for (y+3)(y+1)=2y(y+3) equation:

Simplifying
(y + 3)(y + 1) = 2y(y + 3)

Reorder the terms:
(3 + y)(y + 1) = 2y(y + 3)

Reorder the terms:
(3 + y)(1 + y) = 2y(y + 3)

Multiply (3 + y) * (1 + y)
(3(1 + y) + y(1 + y)) = 2y(y + 3)
((1 * 3 + y * 3) + y(1 + y)) = 2y(y + 3)
((3 + 3y) + y(1 + y)) = 2y(y + 3)
(3 + 3y + (1 * y + y * y)) = 2y(y + 3)
(3 + 3y + (1y + y2)) = 2y(y + 3)

Combine like terms: 3y + 1y = 4y
(3 + 4y + y2) = 2y(y + 3)

Reorder the terms:
3 + 4y + y2 = 2y(3 + y)
3 + 4y + y2 = (3 * 2y + y * 2y)
3 + 4y + y2 = (6y + 2y2)

Solving
3 + 4y + y2 = 6y + 2y2

Solving for variable 'y'.

Reorder the terms:
3 + 4y + -6y + y2 + -2y2 = 6y + 2y2 + -6y + -2y2

Combine like terms: 4y + -6y = -2y
3 + -2y + y2 + -2y2 = 6y + 2y2 + -6y + -2y2

Combine like terms: y2 + -2y2 = -1y2
3 + -2y + -1y2 = 6y + 2y2 + -6y + -2y2

Reorder the terms:
3 + -2y + -1y2 = 6y + -6y + 2y2 + -2y2

Combine like terms: 6y + -6y = 0
3 + -2y + -1y2 = 0 + 2y2 + -2y2
3 + -2y + -1y2 = 2y2 + -2y2

Combine like terms: 2y2 + -2y2 = 0
3 + -2y + -1y2 = 0

Factor a trinomial.
(1 + -1y)(3 + y) = 0

Subproblem 1
Set the factor '(1 + -1y)' equal to zero and attempt to solve:

Simplifying
1 + -1y = 0

Solving
1 + -1y = 0

Move all terms containing y to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + -1y = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -1y = 0 + -1
-1y = 0 + -1

Combine like terms: 0 + -1 = -1
-1y = -1

Divide each side by '-1'.
y = 1

Simplifying
y = 1
Subproblem 2
Set the factor '(3 + y)' equal to zero and attempt to solve:

Simplifying
3 + y = 0

Solving
3 + y = 0

Move all terms containing y to the left, all other terms to the right.

Add '-3' to each side of the equation.
3 + -3 + y = 0 + -3

Combine like terms: 3 + -3 = 0
0 + y = 0 + -3
y = 0 + -3

Combine like terms: 0 + -3 = -3
y = -3

Simplifying
y = -3
Solution
y = {1, -3}