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I am solving the one-step equation: $$-\frac{3}{4}\alpha = \frac{5}{4}$$ In the answer sheet it says to multiply both sides by the "reciprocal" $-\dfrac{4}{3}$. In one step equations you are supposed to do the opposite on both sides and in this wouldn't the opposite be divide both sides by $-\dfrac{3}{4}$ and not multiply by $-\dfrac{4}{3}$ ? Am I making a mistake or is this just another way of doing it?

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    $\begingroup$ Dividing by $x$ is equivalent to multiplying by $1/x$ (also written $x^{-1}$), called the reciprocal of $x$. $\endgroup$ – molarmass Aug 20 '18 at 18:45
  • $\begingroup$ It helps to remember that division $a\div b$ is rigorously defined as multiplication by the multiplicative inverse $a\times (b)^{-1}$, just like how subtraction $a-b$ is rigorously defined as addition by the additive inverse $a+(-b)$ $\endgroup$ – JMoravitz Aug 20 '18 at 19:38
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Multiplying by the "reciprocal" is exactly the same as dividing by the number in front of $\alpha$. In your example it's equivalent to notice that :

$$ \frac{1}{\frac{a}{b}} = \frac{b}{a} $$

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  • $\begingroup$ So I could do it either way? $\endgroup$ – Redline Aug 20 '18 at 18:46
  • $\begingroup$ Yes, to solve $p\alpha = q$ you can multiply each side by $\frac{1}{p}$ or divide each side by $p$ ! $\endgroup$ – tmaths Aug 20 '18 at 18:47
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You like to get rid of $$ -\frac{3}{4}$$ in$$ -\frac{3}{4} \alpha = \frac{5}{4}$$

Dividing by $$ -\frac{3}{4}$$ means multiplying by $$ -\frac{4}{3}$$ because dividing by $x$ means multiplying by $1/x$

So it is fine to multiply by reciprocal to make our life easier.

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If you try it out, you will notice that this will free the $\alpha$ on the LHS. So this operation is the inverse of multiplication with $-3/4$.

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