0
$\begingroup$

Someone is asked to solve the following equation: $(x-5)/(x+1)= (x-5)/(x+3)$. This person respond "There is no solution. Cross multiply to get $(x-5)(x+3)=(x-5)(x+1)$. Divide both sides by $x-5$ and I get $x+3=x+1$. Subtracting $x$ from both sides, I get $1=3$ which is impossible. So there is no solution." Is this person right?

I mean the person is right but instead of subtracting $x$, wouldn't you subtract the constants. So I would subtract 1 on both sides and get $x+2=x$ and subtract $x$ and you get $0=2$ which is not true.

Any ideas?

$\endgroup$
  • 7
    $\begingroup$ Hint: you may not be able to divide by $(x-5)$. $\endgroup$ – lulu Feb 14 '16 at 0:26
  • 2
    $\begingroup$ Notice that $x=5$ is a solution, so there's something wrong with your argument. $\endgroup$ – Gregory Grant Feb 14 '16 at 0:26
7
$\begingroup$

First of all $x$ must be different from $-1$ and $-3$.

Then, $x=5$ is obviously a solution, since $0=0$.

Supposing moreover $x\neq5$, we can divide both sides by $(x-5)$ and get $$ \frac1{x+1}=\frac1{x+3} $$ which is equivalent to $$ x+1=x+3 $$ but this last one is clearly always false.

Thus the only solution of your initial equation is $x=5$.

$\endgroup$
3
$\begingroup$

Cross multiplying is ok, but first you must specify that $x \neq -1,-3$, or the values for which the denominators are null, and you cannot divide by $(x-5)$ without first stating $x \neq 5$ (infact one/the only solution is $x = 5$)

Rather, multiply out to get a polynomial, like the following (after setting $x \neq -1,-3$): $$ (x-5)(x+3)=(x-5)(x+1) $$ Then: $$ x^2-2x-15 = x^2-4x-5 $$ Finally $2x=10$, so $x=5$

$\endgroup$
2
$\begingroup$

By the same logic of that person, the identity $$ \frac{0}{1}=\frac{0}{2} $$ is false because you can factor out $0$ and get the false identity $$ \frac{1}{1}=\frac{1}{2} $$

Or you can deduce the well known fact that $2=1$: set $x=y=1$; then $x^2-y^2=x^2-xy$ and you can factor out $x-y$ from both sides, getting $$ x+y=x $$ that is, $2=1$.


If $x\ne 5$, you can factor out $x-5$ from both sides, getting $$ \frac{1}{x+3}=\frac{1}{x+1} $$ that's clearly false for every $x$ (of course we cannot consider $-3$ or $-1$ to begin with).

However, for $x=5$, the left-hand side is equal to the right-hand side. So $5$ is a solution.

$\endgroup$
1
$\begingroup$

You cannot divide by $(x-5)$ when $x=5$. So, $x=5$ is a solution.

$\endgroup$
1
$\begingroup$

If you cross multiply and simplify $ (x-5)= 0, x = 5 $. Do not divide by a zero on either side of equation.

$\endgroup$
0
$\begingroup$

After cross multiplying, expand each side of the equation $ (x-5)(x+3) = (x-5)(x+1) $ as a quadratic expression.

On the left side: $ (x-5)(x+3) = x^{2}+3x-5x-15 $ $$= x^{2}-2x-15 $$

On the right side: $(x-5)(x+1) = x^{2}+x-5x-5 $ $$= x^{2}-4x-5$$

Thus, $$ (x-5)(x+3) = (x-5)(x+1) $$ $$<=> x^{2}-2x-15 = x^{2}-4x-5 $$

The $ x^{2}$ terms on each side cancel each other out, so we are left with $$ -2x - 15 = -4x - 5 $$

Add $ 2x $ to each side:

$$ -2x - 15 +2x = -4x - 5 +2x $$ $$<=> -15 = -2x-5$$

Add $ 5 $ to each side: $$ -15 + 5 = -2x-5 + 5$$ $$<=> -10 = -2x $$

Finally divide a $-2$ to each side: $$ \frac{-10}{-2} = \frac{-2x}{-2} $$ $$<=> 5 = x $$

Thus $ x = 5 $ is a solution that is not extraneous.

$\endgroup$

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.