# Thread: Solve equation for all real numbers a

2. ## Re: Solve equation for all real numbers a

Your first step is to put everything over a common denominator. Factor $\displaystyle x^2-3x$ and see if you can work out what the common denominator is. Then, once you've done that, remember that the only way a fraction can be equal to $\displaystyle 0$ is if the numerator is equal to $\displaystyle 0$. This will give you an equation to solve for $\displaystyle a$.

If you require further help, please show what you can do and we'll guide you from there.

3. ## Re: Solve equation for all real numbers a

Okay, so x(x-3) is the common denominator. Which gives me (x^2 + ax-3a + x-12)/ x(x-3) =0 , or (x^2 + x(a+1) - 3(a+4))/ x(x-3).

If I solve the numerator for x, I get x=3, and x=-a-4.

Should I try to solve the equation first with x=3, and then x=-a-4?

4. ## Re: Solve equation for all real numbers a

Originally Posted by cristianodoni72
Okay, so x(x-3) is the common denominator. Which gives me (x^2 + ax-3a + x-12)/ x(x-3) =0 , or (x^2 + x(a+1) - 3(a+4))/ x(x-3).

If I solve the numerator for x, I get x=3, and x=-a-4.

Should I try to solve the equation first with x=3, and then x=-a-4?
Great work so far! You should notice quickly that x=3 cannot be a solution. Why?

5. ## Re: Solve equation for all real numbers a

Thanks! Yes, with x=3, the denominator is 0. So I put x=-a-4, which gives me the numerator 2a-4 and the denominator a^2 +11a+28.
Do I solve the numerator for a now?

6. ## Re: Solve equation for all real numbers a

Originally Posted by cristianodoni72
Thanks! Yes, with x=3, the denominator is 0. So I put x=-a-4, which gives me the numerator 2a-4 and the denominator a^2 +11a+28.
Do I solve the numerator for a now?
You're overcomplicating things and getting slightly muddled. We have learned that $\displaystyle x=-a-4$. Therefore, for all real values of $\displaystyle a$,

$\displaystyle a=-4-x$

We can't simplify any further.

7. ## Re: Solve equation for all real numbers a

Okay! Thanks for guiding me!