Answer:
[tex]x=\frac{10}{a+b}[/tex] with [tex]a \neq -b[/tex].
Step-by-step explanation:
So we have:
[tex]ax+bx=10[/tex] and we are solving for [tex]x[/tex].
We want to get all terms that have x on one side and all terms without x on the opposing side. This is actually already done.
Now we are going to factor the side that has the x's. We are going to factor x out.
[tex]x(a+b)=10[/tex]
Now you have a number times x and you want x by itself. The inverse operation of multiplication is division. So we are going to divide both sides by [tex](a+b)[/tex]:
[tex]x=\frac{10}{a+b}[/tex]
You could probably stop here but just in case....
[tex]a+b[/tex] can't be 0 because then you would have 10/0 which is not a number.
So if we solve [tex]a+b=0[/tex] we can find out what we don't want and b to be:
[tex]a+b=0[/tex]
Subtract b on both sides:
[tex]a=-b[/tex]
So we we don't want [tex]a[/tex] and [tex]-b[/tex] to be the same value.
