Question

Solve $$ax-a^2=bx-b^2$$ for $$x$$.

Answer 1

you need to isolate the x
so the first thing you do is
combine the x's as follows:
ax - bx = a^2 -b^2

Answer 2

now take the x out and you have
x(a-b) = a^2 =- b^2

Answer 3

then divide by a-b
and you have
x = a^2-b^2
---------
a - b

Answer 4

x= (a-b) (a+b) divded by a-b
X= a+b

Answer 5

right you can simplify

Answer 6

x= a+b ..is the answer

Answer 7

correct

Answer 8

Is that the "complete" answer?

Answer 9

yes u cant simplify further

Answer 10

Are you sure that is the ENTIRE answer???

Answer 11

what else do you want on the answer

Answer 12

My book gives a restriction on x=a+b.

Answer 13

then don't simplify it, use the one I gave you before simplifying.

Answer 14

But I will lose marks for not simplifying?

Answer 15

then that makes no sense, if you simplify that is the answer x= a+b, there is nothing else to it.

Answer 16

$$a \neq b$$

Answer 17

\[a \neq b\] is condition?

Answer 18

Yes, a restriction on the simplified answer x=a+b.

Answer 19

nope, my question is a not equal b is a restriction, too? I solved and got a=b then for all x you have that form

Answer 20

I mean you have infinite solution except x =a +b

Answer 21

x=a+b is the solution; except when a=b, then there is no solution.

Answer 22

how can it be. ok, correct me if I am wrong, please

Answer 23

\[ax-a^2 =bx-b^2\\ax-a^2-bx+b^2=0 \]
only one way to move terms without making any condition.

Answer 24

factor out to get (a-b) (x-(a+b)) =0

Answer 25

if you restrict the second part, so, the product =0 if and only if a=b

Answer 26

So , when a= b replace to original problem
\[ax -a^2 = ax -a^2\] for all x

Answer 27

Point out my mistake, please

Answer 28

$$ (a-b) (x-(a+b)) =0$$becomes a true statement when
$$a-b=0$$or $$x-(a+b)=0$$
If $$a-b=0,$$then the solution set is {real numbers}

If $$x-(a+b)=0,$$ then the solution set is all the values of x for which x =a+b