example 1: this is basically reducing WHILE multiplying. since 5 and 10 have a greatest common factor and is able to be reduced by each other, we do so. hence the reason 5 went to 1 and 10 went to 2 because 5 goes into 5 once and 5 goes into 10 twice. it was a different case for 3 and 8. 3 obviously cant go into 8 and 3 is prime so nothing can be done with those. ONLY REDUCE WHILE MULTIPLYING DIAGONALLY you CANNOT do up and down or side to side. so our new fractions are [tex]\frac{3}{2}[/tex] and [tex]\frac{1}{8}[/tex] so multiply across and get [tex]\frac{3}{16}[/tex] which cant be reduced more.
example 2: this one in my opinion is pretty easy. you take the two fractions and multiply regularly across. you get [tex]\frac{27}{80}[/tex]. that fraction can be reduced. so we find the prime factorization of each number. 3 is a prime factor. 3x3x3=27 which is our top number. 2 and 5 are all prime numbers used to reach 80. 2x2x2x2x5=80. that's the prime factorization method its an easier way to check your work also.
example 3: this one is practically a combination of the two previous ones. [tex]\frac{16}{21}[/tex] and [tex]\frac{7}{8}[/tex] can be diagonally reduced on both diagonals. 7 goes into 7 once and 7 goes into 21 three times 8 goes into 8 once and 8 goes into 16 twice. now we have [tex]\frac{1}{1}[/tex] and [tex]\frac{2}{3}[/tex] multiply across and get just [tex]\frac{2}{3}[/tex]. prime factorization is the best way to do this but, more effective with larger numbers. [tex]\frac{2*2*2*2*7}{3*7*2*2*2}[/tex] now cross out the numbers that are the same on the top and bottom. three 2's cancel out, one 7 cancels out, and now we're left with [tex]\frac{2}{3}[/tex].
I really hope this helps you!!
