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(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)

I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.

However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?

I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".

I have a theory:

  • Invert both sides and then multiply both sides by 2;

Is this correct?

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  • $\begingroup$ After all the good answers posted, one point worth emphasizing is that the "move numbers from side to side" mnemonic relies on the implicit assumption that none of those you move are $0$. For example, $5 = \frac{0}{x}$ is not equivalent to $\frac{0}{5}=x$. $\endgroup$ – dxiv Dec 1 '16 at 6:49
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You are allowed to invert both sides, given you invert the entire side, like such: $$2 + x = \frac1y \rightarrow \frac{1}{2 + x} = y$$ A common mistake is to invert only one term. Note that "inverting" happens because we can multiply both sides of the equation by the product of both sides. Take for example: $$\frac{1}{2 + x} = \frac{1}{y} \rightarrow \frac{y(2 + x)}{2 + x} = \frac{y(2 + x)}{y}\rightarrow y = 2 + x$$ We did the above by multiplying both sides by y(2 + x), which is allowed, as long as it is done to both sides. You can solve your example in the same way.

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    $\begingroup$ +1 for answering the OPs question about method rather than just solving the equation. $\endgroup$ – Ethan Bolker Dec 1 '16 at 0:55
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Multiply by $x$: $$5x=2$$

Divide by $5$: $$\frac{2}{5} = x$$

More generally, $$\frac{x}{b} = a \iff x=ab \iff \frac{x}{a} = b$$

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We have that $$5 = \frac2x.$$ Now multiply by $x$ on each side, and get $$5x=2. $$ Next, divide by $5$ on each side, and get $$x=\frac25. $$

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  • $\begingroup$ Beat me to it... :( $\endgroup$ – Frank Dec 1 '16 at 0:51
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From $5=\frac 2x$ to get $x=\frac 25$, you're multiplying both sides by $x$ to get $$5x=2$$ And dividing by $5$, we get $x=\frac 25$.

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Other way for solving it is by the reciprocal or inverse process:

$5 = \frac{2}{x} \Rightarrow 5^{-1} = (\frac{2}{x})^{-1}$

$\frac{1}{5} = \frac{x}{2} \Rightarrow \frac{2}{5} = x$

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You can do it by means the second equivalence principle: "multiplying or dividing both sides of an equation by a non-zero constant" we obtain an equivalent equation. This is the basis of your calculations in the example $5=\frac{2}{x}$.

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Well... $$\begin{equation}\begin{aligned}5 &= \frac{2}{x} &&\text{From question}\\5x&=2 &&\text{Multiply by x on each side}\\ x&=\frac{2}{5} &&\text{Divide by 5 on each side} \end{aligned}\end{equation}$$ Or $$\begin{equation}\begin{aligned}\frac{2}{5} &= x &&\text{From question}\\2&=5x &&\text{Multiply by 5 on each side}\\ \frac{2}{x}&=5 &&\text{Divide by x on each side} \end{aligned}\end{equation}$$

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