multiply both side of a equation by a variable

arman0077

New member
Joined
Apr 26, 2021
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3
hi
I think I misunderstood a part of a book
we can't multiply or divide both sides of an equation by 0 right?
in one part of that book.
we have this rational equation
\(\displaystyle \frac{2x}{(x-3)} + \frac{5}{(x+3)} = \frac{36}{(x^2-9)}\)
for solving it we multiply both side by lcd that is\(\displaystyle (x-3)(x+3)\)
and solve the equation \(\displaystyle 2x(x+3) + 5(x-3) = 36\) the answers will be \(\displaystyle x=\frac{-17}{2}\) and \(\displaystyle x = 3\)
so if \(\displaystyle x = 3\) that means \(\displaystyle (x-3) = 0 \) and we are multiplying both sides by 0
can we do that ? or I'm missing something or totally misunderstood the subject

thank you.
 

HallsofIvy

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Jan 27, 2012
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Your mistake is thinking that 0 IS a solution!

You CAN multiply both sides of an equation by zero. But there is no reason to do that since, no matter what equation you start with, the result is 0= 0!

And it's not just a matter of you cannot divide "both sides of an equation" by 0, you cannot divide anything by 0 because 0 does not have a "multiplicative inverse"- there is no number, x, such that x times 0 equals 1.

For the equation \(\displaystyle \frac{2x}{x- 3}+ \frac{5}{x+ 3}= \frac{36}{x^2- 9}\), x= 3 is NOT a solution because, if x=3 both denominators, \(\displaystyle x- 3\) and \(\displaystyle x^2- 9\) would be 0 which is impossible.

Any time you multiply an equation by something a variable, you might introduce "spurious solutions" that satisfy the new equation but not the original equation.

For an obvious example, the equation x= 7 has the single solution, x= 7.
But \(\displaystyle x^2= 7x\), multiplying each side by x, has solutions x= 7 and x= 0.
 

JeffM

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Sep 14, 2012
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6,484
My answer is just a different way to make the same substantive point as Halls of Ivy.

You may multiply anything by 0. Of course the result is zero.

You may multiply both sides of an equation by the same number. Of course multiplying both sides by zero gives 0 = 0, which is true but is of almost no practical use in solving problems.

You may divide anything by any number that is not zero. And you may divide both sides of an equation by the same number provided that number is not zero.

You cannot divide anything by zero (at least not in any number system of practical application).

So you cannot divide both sides of an equation by zero because you cannot divide either side by zero.

Now when you see an equation like

\(\displaystyle \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} = \dfrac{36}{x^2 - 9}\)

what it really means in order not to be nonsense is

\(\displaystyle \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} = \dfrac{36}{x^2 - 9},\ x \ne 3, \text { and } x \ne - 3.\)

\(\displaystyle x \ne 3 \text { and } x \ne - 3 \implies x^2 - 9 \ne 0.\)

Consequently when you multiply

\(\displaystyle (x^2 - 9) * \left ( \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} \right ) = (x^2 - 9) * \dfrac{36}{x^2 - 9}\)

you do not know yet what number you are multiplying by, but it certainly is not zero.

\(\displaystyle (x^2 - 9) * \left ( \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} \right ) = (x^2 - 9) * \dfrac{36}{x^2 - 9} \implies\)

\(\displaystyle (x + 3)(2x) + (x - 3)(5) = 36 \implies 2x^2 + 11x - 51 = 0.\)

Thus, considering only the quadratic formula, we get

\(\displaystyle x = \dfrac{-11 \pm \sqrt{11^2 - 4(2)(-51)}}{2 * 2} = \dfrac{- 11 \pm \sqrt{121 + 408}}{4} \implies \)

\(\displaystyle x = \dfrac{-11 \pm \sqrt{529}}{4} = \dfrac{-11 \pm 23}{4} = - \dfrac{17}{2} \text { or } 3.\)

But we cannot rely just on the quadratic formula. We previously said that x cannot equal 3. So the only answer that satisfies that condition and the quadratic formula is - 17/2. There is no inconsistency, no ambiguity. You just need to keep track that we previously excluded 3 and - 3 implicitly in the initial equation. It is sort of a trick question because the exclusion was not explicit.
 

lex

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Joined
Mar 3, 2021
Messages
686
hi
I think I misunderstood a part of a book
\(\displaystyle x\) is a solution of:

\(\displaystyle \frac{2x}{(x-3)} + \frac{5}{(x+3)} = \frac{36}{(x^2-9)} \\
\text{ }\\
\rightarrow \text{ 3 things:}\\\\
\begin{align*}\\
&\text{(1) } x≠3\\
\text{and } &\text{(2) } x≠-3\\
\text{and } &\text{(3) } 2x(x+3) + 5(x-3) = 36 \hspace5ex ^*\\
\end{align*}\\
\text{ }\\
\text{(3) } \rightarrow x=\frac{-17}{2} \text{ or } x = 3\\
\text{ }\\
\therefore (1), (2), (3) \text{ together } \rightarrow x=\frac{-17}{2}
\text{ }\\
\text{ }\\
\)
(* the manipulation to get (3) was OK, as we know (1) and (2))
'→' stands for 'implies'
 
Last edited:

nasi112

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Aug 23, 2020
Messages
425
You are allowed to multiply and divide by \(\displaystyle (x + 3)\) or \(\displaystyle (x - 3)\) as long as \(\displaystyle x \neq 3\) and \(\displaystyle x \neq - 3\)
 

JeffM

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Joined
Sep 14, 2012
Messages
6,484
You are allowed to multiply and divide by \(\displaystyle (x + 3)\) or \(\displaystyle (x - 3)\) as long as \(\displaystyle x \neq 3\) and \(\displaystyle x \neq - 3\)
Although I prefer to say that you are allowed to multiply by (x + 3) or (x - 3) even if x = 3 or x = - 3 although doing so is never (almost never?) useful. But I am being ridiculously fussy at this point.
 

Jomo

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Dec 30, 2014
Messages
10,568
When you have say for example \(\displaystyle \dfrac {(2x)}{(x-3)}(x-3)\) and say that it equals 2x you are making a mistake if x=3. Of course if you make a mistake you should not expect to get the correct answer. Since \(\displaystyle \dfrac {(2x)}{(x-3)}(x-3) = 2x, x\neq3\), you must exclude x=3 as a solution.
 

arman0077

New member
Joined
Apr 26, 2021
Messages
3
My answer is just a different way to make the same substantive point as Halls of Ivy.

You may multiply anything by 0. Of course the result is zero.

You may multiply both sides of an equation by the same number. Of course multiplying both sides by zero gives 0 = 0, which is true but is of almost no practical use in solving problems.

You may divide anything by any number that is not zero. And you may divide both sides of an equation by the same number provided that number is not zero.

You cannot divide anything by zero (at least not in any number system of practical application).

So you cannot divide both sides of an equation by zero because you cannot divide either side by zero.

Now when you see an equation like

\(\displaystyle \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} = \dfrac{36}{x^2 - 9}\)

what it really means in order not to be nonsense is

\(\displaystyle \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} = \dfrac{36}{x^2 - 9},\ x \ne 3, \text { and } x \ne - 3.\)

\(\displaystyle x \ne 3 \text { and } x \ne - 3 \implies x^2 - 9 \ne 0.\)

Consequently when you multiply

\(\displaystyle (x^2 - 9) * \left ( \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} \right ) = (x^2 - 9) * \dfrac{36}{x^2 - 9}\)

you do not know yet what number you are multiplying by, but it certainly is not zero.

\(\displaystyle (x^2 - 9) * \left ( \dfrac{2x}{x - 3} + \dfrac{5}{x + 3} \right ) = (x^2 - 9) * \dfrac{36}{x^2 - 9} \implies\)

\(\displaystyle (x + 3)(2x) + (x - 3)(5) = 36 \implies 2x^2 + 11x - 51 = 0.\)

Thus, considering only the quadratic formula, we get

\(\displaystyle x = \dfrac{-11 \pm \sqrt{11^2 - 4(2)(-51)}}{2 * 2} = \dfrac{- 11 \pm \sqrt{121 + 408}}{4} \implies \)

\(\displaystyle x = \dfrac{-11 \pm \sqrt{529}}{4} = \dfrac{-11 \pm 23}{4} = - \dfrac{17}{2} \text { or } 3.\)

But we cannot rely just on the quadratic formula. We previously said that x cannot equal 3. So the only answer that satisfies that condition and the quadratic formula is - 17/2. There is no inconsistency, no ambiguity. You just need to keep track that we previously excluded 3 and - 3 implicitly in the initial equation. It is sort of a trick question because the exclusion was not explicit.
oooooooowww i get it now thank you
and thank you all
 
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