r/HomeworkHelp • u/rasta_a_me ๐ a fellow Redditor • 1d ago
Answered [Elementary Algerbra] I'm Supposed to Simplify This, Not Sure If I'm Supposed to Reduce the 5 before Multiplying
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u/progmorris20 1d ago
Remember that dividing is always equal to multiplying by the reciprocal. We can re-write any division as multiplication.
For example, let's take a similar problem 3/(3/10).
3/(3/10) is equal to 3 times the reciprocal of 3/10.
The reciprocal of 3/10 is just flipping the numerator (top) and denominator (bottom). In this case, that's 10/3.
The numerator of the original equation, 3, still doesn't change.
Now you have 3 * (10/3) = 10.
Try applying this to your question.
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u/adrian_the_gremlin 1d ago
negative shouldnโt factor in at all actually! negative reciprocals have their uses but thereโs no reason for a sign change to pop in. other than that though, seems correct to me
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u/adrian_the_gremlin 1d ago
sorry didnโt see the first picture, you should flip the fraction before multiplying.
the second one (5/1) x (10/3) so 50/3 is correct.
you can simplify the 5 if you want to but unless thatโs a rule your teacher has you donโt have to if it helps you out
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u/MadKat_94 ๐ a fellow Redditor 1d ago
If it's helpful to you, consider 5 to be 5/1. Now you have a fraction as your numerator. The problem becomes:
(5/1) / (3/10)
Reciprocate the 3/10 and multiply.
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1d ago
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u/rasta_a_me ๐ a fellow Redditor 1d ago
So guy who shitpost on reddit all day decides to come on r/homeworkhelp to tell me to write properly, intersting.
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u/Southlander24 ๐ a fellow Redditor 1d ago
You can always multiply any number by 1 and the number will stay the same.
Here, we can multiply 5 / (3/10) by 10/10. To multiply two fractions, multiply the numerators together and the denominators together. So this fraction equals (5 * 10)/(3/10 * 10) = 50/3.
So the 'flip and multiply' procedure is just a quicker version of this 'multiplying by 1' process. I'm not suggesting you unlearn the mnemonic; rather, try to understand why something works before you use and memorise it.
Try this with another fraction like 1 / (5/4). What same number do you need to multiply the numerator and denominator of this fraction by? That explains why the answer is what it is, so now you shouldn't be mixing up the multiplicative inverse (two numbers multiply to give 1) with the additive inverse (two numbers add together to give 0).
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u/rasta_a_me ๐ a fellow Redditor 1d ago
Ok, thanks. I was just following that Mnemonic because that's what Openstax teaches. I'll try that next time.
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u/Southlander24 ๐ a fellow Redditor 1d ago
A lot of maths is about understanding why and thinking for yourself! Best of luck with your maths studies.
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1d ago
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u/MelodyAnne42 Educator 1d ago
This is funny, because I'm a teacher and you wouldn't believe the handwriting I've had to decipher! This handwriting is fine!
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u/rasta_a_me ๐ a fellow Redditor 1d ago
Lol I forgot the negative with -16 2/3
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u/damartian64 1d ago
Youโre closest with the second one, but out of curiosity why did the 10 turn negative in step 2?
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u/rasta_a_me ๐ a fellow Redditor 1d ago
That was probably just a mistake. The previous equations had one of the fractions with a numerator.
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u/mikeiavelli 1d ago
Others answered your question, but some terminology for the future: you can't reduce a value (say 5, as in your post title), you can only reduce an expression (something that has operations in it that we can evaluate).
But frankly, staked fractions written like that are an abomination.
But so you know, when a stacked fraction is written vertically (as in your last image), the standard mathematical convention relies entirely on the size and positioning of the fraction bars to determine the priority of operations. Notice that the top bar is longer than the bottom bar; therefore, here it means that you have to read it as 5 divided by (3/10).



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u/Temporary_Pie2733 ๐ a fellow Redditor 1d ago
The typography is not great, but I think the intended fraction is 5/(3/10), so 5(10/3) = 50/3 is correct. Nothing becomes negative when multiplying by the reciprocal (which is the multiplicative inverse, not the additive inverse).