Solve $(x-5)/(x+1)= (x-5)/(x+3)$.

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?

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

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 $x=5$.

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 $x=5$

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.

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

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

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.