7(3x+4)=8(2x+5)+13

how would I start a equation like this would it be 21x+28=16x+40+13 ?

Yes, that is the way to start. Now collect the x's on one side.

so it would be 21x-21x=28=
28=-5+40+13
28+13=41
x=41

7(3x+4)=8(2x+5)+13

21x+28=16x+40+13

To get all the x values on one side and the other numbers on the other side, subtract 16x and 28 from both sides.

I don't know how you got

That will leave you with a much different answer.

I hope this helps a little more. Thanks for asking.

I apologize for the confusion in my previous response. Let's go through the steps of solving the equation correctly.

Starting from the equation: 7(3x+4)=8(2x+5)+13

First, let's simplify both sides of the equation:

On the left side: 7(3x+4) = 21x + 28

On the right side: 8(2x+5) + 13 = 16x + 40 + 13 = 16x + 53

So now we have: 21x + 28 = 16x + 53

To solve for x, let's get all the terms with x on one side and the constant terms on the other side.

Subtract 16x from both sides: 21x - 16x + 28 = 16x - 16x + 53

Simplifying: 5x + 28 = 53

Next, subtract 28 from both sides: 5x + 28 - 28 = 53 - 28

Simplifying: 5x = 25

To solve for x, divide both sides by 5: 5x/5 = 25/5

Simplifying: x = 5

Therefore, the solution to the equation is x = 5.