15
$\begingroup$

$$ \frac{x-4}{(x-1)} = \frac{1-4}{(x-1)} $$

Can someone tell me how many solutions are there for the above equation?

MY APPROACH:

I cross multiplied the equations and re-arranged to get a quadratic eqaution which gave me $x=1$ repeated twice. So I answered the question by telling there are two solutions for this equations where one equals the other.

TRAINER'S APPROACH:

He cancelled out the denominators on both sides and he imposed a condition that $x \ne 1 $ but still ended up with $x = 1$ after solving the remaining parts of the equation. Since the solution itself contradicts the condition imposed upon the cancellation, he told that the equation has no solutions.

Which approach is right? And why the other one is wrong?

$\endgroup$
7
  • $\begingroup$ trainer's one is right. I could not give more explanations than the trainer. $\endgroup$
    – user126154
    Mar 17 '14 at 11:27
  • $\begingroup$ The problem should be interpreted as "find the extension of the set $\left\{x\in \mathbb R\colon \frac{x-4}{(x-1)} = \frac{1-4}{(x-1)}\right\}$". Certainly $1$ isn't in the set. In fact it is empty. Multiplicity of solutions has no bearing here. $\endgroup$
    – Git Gud
    Mar 17 '14 at 11:29
  • $\begingroup$ Why is my solution wrong? $\endgroup$
    – valyrian
    Mar 17 '14 at 11:29
  • $\begingroup$ @GitGud So the trainer's right? $\endgroup$
    – valyrian
    Mar 17 '14 at 11:30
  • 2
    $\begingroup$ For $x=1$ you are multiplying both sides of your equation by $0$. $\endgroup$ Mar 17 '14 at 11:30
10
$\begingroup$

The trainer is right, there is no solution. Your approach of crossing the equations has an implicit demand that the denominators are nonzero, so your approach should show there are no solutions as well.

$\endgroup$
23
$\begingroup$

Here's another way to look at it. You can also recast the equation as follows, without cancelling anything or multiplying or dividing by anything which might be zero: $$0=\frac {x-4}{x-1}-\left(\frac {1-4}{x-1}\right)=\frac {x-1}{x-1} $$Now do you see what is going on?

$\endgroup$
2
  • 4
    $\begingroup$ Very well. That is one bogus way of proving $1 = 0$ !! :P $\endgroup$
    – valyrian
    Mar 17 '14 at 11:46
  • 1
    $\begingroup$ I like this approach; clean simple contradiction $\endgroup$
    – Dan Bryant
    Mar 17 '14 at 12:25
3
$\begingroup$

If you use $ x=1 $ you get $ 1-1=0 $ on the denominator, which gives you a division by zero. And since that's the only value that gets an equality on both sides of the equation, it shows that there's no solution.

$\endgroup$
1
  • $\begingroup$ Thank you. That's what I was thinking since the comments came on. $\endgroup$
    – valyrian
    Mar 17 '14 at 11:31
1
$\begingroup$

We may never divide by 0, so we require for the solution that the denominator $(x-1) \not =0$, meaning $x \not= 1$.

However, since the only possible value for $x$ (given by both you and the trainer) is $x=1$. Hence we are forced to conclude that no solutions exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.