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I was just now trying to solve the following simple equation:

$$(x+1)(4x-3) = 2(x+1)(2x+3)$$

In my naivety I thought dividing by $(x+1)$ would shorten the equation and lead to a quick solution, but unfortunately it lead to x disappearing from the equation. Even though I figured out that expanding both sides leads me to find the solution of $x=-1$, I don't understand why dividing by $(x+1)$ causes such problems.

Can you explain to me why that is and what I need to know to avoid such mistakes in the future?

Blue
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MarcVolkert
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2 Answers2

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When you try to divide a common factor on both sides of an equation, always ask whether this factor is $0$ or not, and with all possibilities considered, you achieve the rigor that's needed to avoid any mistake.

In this particular case,

  • If $x+1=0$, then obviously the equation holds, and hence $x=-1$ is a solution.
  • Otherwise $x+1\not=0$, so we can safely divide it on both sides, and get $4x-3=4x+6\Rightarrow -3=6$, which is impossible. (In a different scenario, this step could lead to additional solutions.)

Another approach will be to move everything into one side of the equation, and try to factor:

$$(x+1)(4x-3) = 2(x+1)(2x+3)\Leftrightarrow (x+1)(4x-3) - 2(x+1)(2x+3)=0$$

$$\Leftrightarrow (x+1)(4x-3-2(2x+3))=0\Leftrightarrow (x+1)(-9)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1$$

Another good habit for solving these problems is to keep track of the direction of arrows. If it's only $\Rightarrow$ instead of $\Leftrightarrow$, at the end, you will get a set of candidate solutions, and you will have to check whether these are really solutions, because some steps cannot be reversed to show the original equations hold.

Just a user
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  • While I agree with the fact that dividing by zero leads to ambiguous results. sometimes....I don't know why. it works well when we have a polynomial say, ($x-r_1)(x-r_2)$ and we divide by (x-r_1) to get the second root. – NoChance Apr 09 '23 at 15:39
  • But if you do it directly, you will miss the first root $x=r_1$. – Just a user Apr 10 '23 at 01:30
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More in general we have that

$$A\cdot B = C\cdot B \iff B = 0\; \lor \; A=C$$

therefore the original equation is equivalent to two cases, for the first one we consider $B=0$ and its solution, for the second one, since $B\neq 0$, we can solve the simpler equation $A=C$ and find its solution.

The final solution is the union of the solutions obtained by the two cases.

As noticed we can also proceed as follows

$$A\cdot B = C\cdot B \iff A\cdot B - C\cdot B=0 \iff B\cdot (A-C)=0 \iff B = 0\; \lor \; A-C=0$$

user
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