Solving 2a $\sum (2x_i)^2 + 2b \sum x_i - 2 \sum x_i y_i$ for a and b

[ISU20CpreE]

2a \sum (2x_i)^2 + 2b \sum x_i - 2 \sum x_i y_i I need to set it equal to zero and solve for a.

2a \sum x_i + 2nb - 2\sum y_i

and solve for b.

I need a hint so i can start doing it, it confuses me all those adding symbols.

The Answer the book came up with is:

a= \frac{(n\sum x_i y_i - \sum x_i \sum y_i)} {(n \sum (x_i)^2 - (\sum x_i)^2}

b= \frac{1 / n} (\sum y_i - a \sum x_i)

 

[berkeman]

First of all, these look like simple algebraic manipulations. a and b are not inside any of the summations, so solving for them is straightforward. The solution for b looks fine (simple re-ordering of the equation), but the answer for a looks wrong. like, where did the n come from? Could that be the answer to a different question or something?

Do the 2nd one first to see how easy it is. Then re-check the first problem statement and answer to be sure you have them synchronized.

 

[ISU20CpreE]

First of all, these look like simple algebraic manipulations. a and b are not inside any of the summations, so solving for them is straightforward. The solution for b looks fine (simple re-ordering of the equation), but the answer for a looks wrong. like, where did the n come from? Could that be the answer to a different question or something?

Do the 2nd one first to see how easy it is. Then re-check the first problem statement and answer to be sure you have them synchronized.

Thank you for your help!