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Why are these simplifications wrong? $$\begin{align} (2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt] x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3 \end{align}$$

Blue
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user551850
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  • You can divide the first equation by $2x-3$ assuming that $x\ne 1.5$ – Peter Apr 13 '18 at 13:15
  • SImilar in the second example, as long as $x\ne 0$, you can didive the equation by $x$ – Peter Apr 13 '18 at 13:16
  • Cancelletions are not wrong in general is the quantity is $\neq 0$ thus we need to consider 2 cases. In the first set the quantity $=0$ and verify on the original equation if that leads to some solution. Then set the quantity $\neq0$ and simplify. – user Apr 13 '18 at 14:38
  • @user551815 Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 14 '18 at 14:17

4 Answers4

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Remember the one rule when it comes to division:

Never divide by zero

If you want to remove $(2x-3)$ you must check that it is not equal to $0$ that is, you have to assume that $x\neq \frac{3}{2}$. Same goes with division by $x$.

In your first example, if you are looking for $x$, dividing by $(2x-3)$ without notice means that you lost a solution $\left(x=\frac{3}{2}\right)$.

gebruiker
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Bill O'Haran
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$$(2x-3)(x+5)=9(2x-3).$$ Since you are multiplying both sides by the same constant, namely $2x-3$, then they cancel out, thus bringing forward the following equation: $$x+5=9.\tag1$$

$$x(x+2)=x(-x+3).$$ In the same fashion as before, the $x$'s cancel out, i.e. $$x+2=-x+3.$$ Now, adding $x$ to both sides yeilds the following equation: $$2x + 2 = 3$$ $$\Downarrow$$ $$2x = 1.\tag2$$

Finally, solve for $x$ in Eq. $(1)$ and Eq. $(2)$. Also note that each of the common factors, $(2x-3)$ and $x$, can be equal to $0$, because for all values $n$, one always has that $0\cdot n = 0$. Thus, you have a bonus two other solutions.

Mr Pie
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From here

$$(2x-3)(x+5)=9(2x-3)$$

we can observe that $2x-3=0$ is a solution and for $2x-3\neq 0$ we can cancel out and obtain

$$(2x-3)(x+5)=9(2x-3)\iff x+5=9\iff x=4$$

thus the solutions for the original equation are $x=\frac32$ and $x=4$.

As an alternative note that

$$(2x-3)(x+5)=9(2x-3)\iff 2x^2+7x-15=18x-27 \iff2x^2-11x+12=0$$

and then

$$x_{1,2}=\frac{11\pm\sqrt{121-96}}{4}=\frac{11\pm5}{4}=\frac32,4$$

user
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Observe that the solution set of $x^2=x$ is $\{0,1\}$ whereas the solution set of $x=1$ is $\{1\}$. Hence dividing by $x$ is not an equivalence transformation. Instead of dividing by $x$ factor it out: $x^2-x=0\iff x(x-1)=0$ to get $x=0$ or $x=1$.

In your first case we will get $(2x-3)(x+5-9)=0\iff 2x-3=0$ or $x-4=0$.

Michael Hoppe
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