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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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    @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.2014-05-15

4 Answers 4

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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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    @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 $x$.2011-11-13
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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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Basically,

for the first one: $ 100 = 120\times x. $ Dividing both sides by $120$ gets: $ 100/120 =(120×x)/120 \\ 100/120 = x. $ And for the second one: $ 0.83 = 123/x. $ Multiplying both sides by $x$ gets: $ 0.83\times x = (123/x)\times x \\ 0.83 x = 123. $ Note: $0.83\approx100/120$. Hence we can say $0.83$ is also same as $100/120$ (dividing this gives us $0.83333333\dots$ and so on).