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Basile [38]
3 years ago
11

How do I reduce each fraction to simplest form. #summer school

Mathematics
2 answers:
ra1l [238]3 years ago
7 0

So for this, you would divide the fraction by the numerator and denominator's GCF, or greatest common factor.


Starting with 18/36, the GCF would be 18 since 18 is the greatest number can go into 18 and 36. \frac{18}{36} /\frac{18}{18} =\frac{1}{2}


Next with 26/28, their GCF is 2. \frac{26}{28} /\frac{2}{2} =\frac{13}{14}


Next with 25/35, their GCF is 5. \frac{25}{35} /\frac{5}{5} =\frac{5}{7}


With 51/75, their GCF is 3. \frac{51}{75} /\frac{3}{3} =\frac{17}{25}


With 22/64, their GCF is 2. \frac{22}{64} /\frac{2}{2} =\frac{11}{32}


And lastly, 49/63 's GCF is 7. \frac{49}{63} /\frac{7}{7} =\frac{7}{9}



Also if the GCF that you know doesn't simplify all the way, you would continue reducing the fraction with more GCFs like this for 144/288:


\frac{144}{288} /\frac{12}{12} =\frac{12}{24}\\\\ \frac{12}{24}/\frac{2}{2} =\frac{6}{12}\\ \\ \frac{6}{12} /\frac{6}{6} =\frac{1}{2}

raketka [301]3 years ago
6 0
1.) you need to find your GCF in this case your GCF is 18 because the factors of 36 are:1,2,3,4,6,9,12,18,36
And the factors of 18 are: 1,2,3,6,9,18

(Factors are numbers that contribute to that number.) like 2 times 18 is 36 and 3 times 12 is 36
As you can see those are all the factors that contribute to 36...

Now 18 has those factors because 2 times 9 is 18. 3 times 6 is 18 aswell.(and so on and so fourth.) the GCF is the greatest common factor so you want to find which factor seems to appear for the both like of factors...it’s 18!! ( and if you have 2 common factors always remember to pick the highest one!)
And if there are NO factors in common, than the GCF is 1


Anyways finishing the question since we found the GCF all we have to do now is we divide 18 and 18 and you get 1 and 36 and 1 and you get 2

1/2 cannot be reduce into any simpler form so that would be your answer!!

Hope I helped!! If you’re stuck on any other question on that paper, message me and I’ll be glad to help! :)
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