# Is my proof valid? Prove that if a, b, and c are real numbers and a ≠ 0, then there is a unique solution for x of the equation ax + b = c.

I know this is painfully basic for some of you, but please do help me, proof writing is very new to me. thanks

We start with $$ax+b=c.$$

We obtain $$ax=c-b$$ by subtracting $$b$$ from both sizes of the equation.

Then, we get the following solution to the equation. $$x=(c-b)/a$$

Now we must prove that $$x=(c-b)/a$$ is a unique solution.

let $$z$$ be a solution to our equation.

We have $$az+b=c$$.

Because $$ax+b=c$$ and $$az+b=c$$ we get $$ax+b=az+b$$

Now we obtain $$ax=az$$ by subtracting $$b$$ from both sides.

Then, we divide by $$a$$ and obtain $$x=z$$.

Therefore, $$x=(c-b)/a$$ is a unique solution to our equation.

• Yeah... sounds good. A little more thorough then necessary but that's not a bad thing. Feb 8 '20 at 23:44
• Welcome to Mathematics Stack Exchange. You might note that we can divide by $a$ since it is given that $a\ne0$ Feb 9 '20 at 0:05

Perfect!

But it might help to note that the solution must be unique because every step must be true. That is.

If $$ax + b = c$$ then it MUST be true that

$$ax = c-b$$.

And if $$ax =c-b$$ and $$a\ne 0$$ then it MUST be true that

$$x =\frac {c-b}a$$.

And if it must be true that $$x = \frac {c-b}a$$ then that is the only possible solution.

.....

Note a less direct way would be:

Claim: $$x = \frac {c-b}a$$ is a solution.

Verify $$a(\frac {c-b}a) + b=$$

$$(c-b) + b = c+(-b+b) =c + 0 =c$$.

So it is a solution....

If you had done that (but why would you) then you would have to verify it is unique.

But by SOLVING well, solving is solving. It finds all and only all solutions. So it would be unique.