SOLUTION: Why do I have to use different steps for the same kind of problem? For x=a+b-2ab solve "a", all I have to do is flip it so it reads -2ab+b+a=x and then I can solve normally. But

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Why do I have to use different steps for the same kind of problem? For x=a+b-2ab solve "a", all I have to do is flip it so it reads -2ab+b+a=x and then I can solve normally. But      Log On


   



Question 1043899: Why do I have to use different steps for the same kind of problem?
For x=a+b-2ab solve "a", all I have to do is flip it so it reads -2ab+b+a=x and then I can solve normally.
But for x-a=a(y-b) solve "a", after I +a to both sides and end up with
x=a(y-b)-a I can't flip it to -a+a(y-b)=x because that gives a different answer than using the additive property of equality. Why can't both be flipped when a is only on 1 side?

Found 2 solutions by Boreal, josgarithmetic:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
When you add +a to both sides, you end up with x=a(y-b) PLUS a.
You don't have to move it anywhere but factor out an a and divide by whatever was left behind after factoring.
=======
x=a+b-2ab
isolate the a
x-b=a-2ab
factor an a out of the right side
x-b=a(1-2b)
divide by (1-2b) both sides
(x-b)/(1-2b)=a
==================
x-a=a(y-b)
distribute
x-a=ay-ab
add a to both sides
x=ay-ab+a
factor out an a on the right
x=a(y-b+1)
divide both sides by (y+b+1)
a=x/(y+b+1)
=================
The approach to the problem is to usually distribute everything and then take non (in this case) factors of a and move them to the other side. Then factor an a out of everything that contains it, and divide both sides by what is left over. In some ways, it is the same kind of problem, solving for a variable. In other ways, it isn't, because the a is distributed over another pair of variables or constants, and that has to be dealt with. You could in the second add the a and have x=a(y-b)+a. That would be factored to x=a(y-b+1), because what was added happened to be a itself.

Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
You want to reverse what was done to your variable, and must do the same operation to both members of the equation. Your two equations are different, and you will want to apply steps as fit each equation according to what you want to solve for.

You want to do steps to get all terms of a to one side OR the other of the equation, for your second example equation. Some expressions may be, or appear different, depending on what sequence of steps you choose. You will USUALLY find a finished answer the same regardless of what sequence of steps you choose for solving an equation for a chosen variable. Answers will NOT REALLY BE DIFFERENT.

x-a=a%28y-b%29
You chose or were told, solve for "a".
Do steps to put all terms of "a" on ONE SIDE.
x-a%2Ba=a%28y-b%29%2Ba
x%2Ba-a=ay-ab%2Ba -----(note you made a mistake in your work.)
x%2B0=ay-ab%2Ba
x=a%28y-b%2B1%29
highlight_green%28x%2F%28y-b%2B1%29=a%29, but this is NOT the only way to show the result, nor some of the previous steps.

Just as good and EQUIVALENT results,
a=x%2F%28y%2B1-b%29
or
a=x%2F%281-b%2By%29
or
a=x%2F%281%2By-b%29
or
a=-x%2F%28b-y-1%29



---
Two of my steps were unnecessary - correct, yes, but unnecessary.
The work should be more like this:
x-a=a%28y-b%29
x=a%28y-b%29%2Ba
x=a%28y-b%2B1%29
a%28y-b%2B1%29=x
a=x%2F%28y-b%2B1%29
Four steps. This is not the only sequence of steps possible, but probably most efficient/fewest steps needed.