# Algebra equations - how to solve?

I have two equations that I want to solve but I can solve the first but not the second, here's an example: \begin{align*} 100 &= 120 \times x\\ 0.83 &= 123/ x \end{align*}

The first one I know how to solve, basically I am doing on both sides of the equation with the same inverse operation of the 120 which is division, In order to separate the variable.

But when I'am try to do the same method on the seconed equation (by multiply both sides of the equation with the same number 120 in this case, it doesn't work as I am expect). I can solve it by dividing the 120 by 0.83, but it isn't the same method as above.

The main thing for me here is to acquire some kind of method on how to solve this kind of equation, and I have learned to separate the variable by doing the inverse operation on both sides, so if you can please point me what I am doing wrong here? I am prefer to stay with the method above. Thanks.

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does it have solution?if x is simple variable then set of solution is nothing,it has not solution – dato datuashvili Nov 13 '11 at 8:39
That is: are you looking for an $x$ that satisfies both equations, or are you giving two separate examples to explain what your issue is? – Arturo Magidin Nov 13 '11 at 8:53
I need just to know how to solve the second example, the first one is just to show you what kind of method I am using to solve this kind of problems. – Hanan N. Nov 13 '11 at 10:17
@HananN.: basically to use the first method you need $x$ to be in the numerator, not on the denominator. So you do a first step where you put $x$ on the numerator (by multiplying both sides by $x$), and then apply your usual method. – essay May 15 '14 at 18:08

For the second equation, multiply both sides by $x$; that will give you $$0.83x = 123.$$ Now you are in the same situation as the first equation, which you know how to solve.

Alternatively, you can take reciprocals on both sides, and go from $$0.83 = 123/x$$ to $$\frac{1}{0.83} = \frac{x}{123}.$$ Since $$\frac{x}{123} = \frac{1}{123}\times x,$$ you are again in the same situation as the first equation.

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I would like to know how to do with the same method as i ahve wrote. – Hanan N. Nov 13 '11 at 13:48
@HananN. I told you how: you multiply both sides by $x$ and divide both sides by $0.83$. This is exactly the same as the method you use in the first equation, where you divide both sides by $120$, just with one extra step (the first step of multiplying both sides by $x$. – Arturo Magidin Nov 13 '11 at 21:59

In 2nd equation,

0.83 = 123/x

0.83 = 123 X 1/x

0.83/123 = 1/x

Taking the reciprocal of the above equation, we get

123/0.83 = x

By finding the value of x in the above equation, you will get the solution in the similar way as you got in equation 1 of your question.

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Provided $x\neq0$, we can multiply both sides of our equation by $x$ to get: \begin{align*} 0.83 &= 123/ x \iff 0.83x=123. \end{align*} We then divide both sides by $0.83$, so that the LHS contains only the variable we're looking for: \begin{align*} \dfrac{0.83x}{0.83}=\dfrac{123}{0.83}\Rightarrow x=\dfrac{123}{0.83}\approx 148.19 \end{align*}

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