How do I solve this algebraic equation?

1. Start with distributing...

distribute the x to the 2x + 1

the -2 to the x - 3

and then the 2x to the x + 1

2. Subtract 2x squared from both sides

Leaves you wish 1x - 2x + 6 = 2x

3. Add/ subtract like terms

1x - 2x = -1x

Leaves -1x + 6 = 2x

4. Add 1x to both sides

Leaves you with 6 = 3x

5. Divide each side by three and get what x equals

First remove the brackets

Gives

2x^2 + x - 2x + 6 = 2x^2 + 2x

Now gather like terms together

Remember that when a number crosses the equals sign then minus becomes plus and plus becomes minus

All that is left

3x = 6

x = 6/3

So x = 2

Hey,

Multiply everything, then move all the x terms to one side and simplify.

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

2x^2 + x - 2x +6 = 2x^2 + 2x

2x^2 -2x^2 + x - 2x - 2x = -6

x - 4x = -6

-3x = -6

x = -6/-3

x = 2

You're welcome!

2x^2+x-2x+6=2x^2+2x

x=-6

distribute the x then solve for its value.