1)\({2 \over 3} \ \times \ {17 \over 15} = \)\( \ \color{red}{{2 \times 17 \over 3\times15} = } \) \(\color{red}{{34 \over 45}}\)
Solution:
Multiply the top numbers, and then multiply the bottom numbers.
\({2 \over 3} \ \times \ {17 \over 15} = \)\({2 \times 17 \over 3\times15} = \) \({34 \over 45}\)
2)\({3 \over 8} \ \times \ {10 \over 16} = \)\( \ \color{red}{{3 \times 10 \over 8\times16} = } \) \(\color{red}{{30 \over 128}}\)\(\color{red}{ = {30 \div 2 \over 128 \div 2} = {15 \over 64}} \)
GCF(30,128) = 2
Solution:
Step1: Multiply the top numbers, and then multiply the bottom numbers.
\({3 \over 8} \ \times \ {10 \over 16} = \)\( {3 \times 10 \over 8\times16} = \) \({30 \over 128}\)
Step 2: Simplify your answer. \({30 \over 128}\)\( = {30 \div 2 \over 128 \div 2} = {15 \over 64} \)
3)\({4 \over 6} \ \times \ {18 \over 17} = \)\( \ \color{red}{{4 \times 18 \over 6\times17} = } \) \(\color{red}{{72 \over 102}}\)\(\color{red}{ = {72 \div 6 \over 102 \div 6} = {12 \over 17}} \)
GCF(72,102) = 6
Solution:
Step1: Multiply the top numbers, and then multiply the bottom numbers.
\({3 \over 8} \ \times \ {10 \over 16} = \)\({3 \times 10 \over 8\times16} = \) \({30 \over 128}\)
Step 2: Simplify your answer. \({72 \over 102}\)\( = {72 \div 6 \over 102 \div 6} = {12 \over 17}\)
4)\({7 \over 10} \ \times \ {7 \over 8} = \)\( \ \color{red}{{7 \times 7 \over 10\times8} = } \) \(\color{red}{{49 \over 80}}\)
Solution:
Step1: Multiply the top numbers, and then multiply the bottom numbers.
\({7 \over 10} \ \times \ {7 \over 8} = \)\({7 \times 7 \over 10\times8} = \) \({49 \over 80}\)
5)\({7 \over 3} \ \times \ {1 \over 5} = \)\( \ \color{red}{{7 \times 1 \over 3\times5} = } \) \(\color{red}{{7 \over 5}}\)
Solution:
Step1: Multiply the top numbers, and then multiply the bottom numbers.
\({7 \over 3} \ \times \ {1 \over 5} = \)\({7 \times 1 \over 3\times5} =\) \({7 \over 5}\)
6)\({9 \over 7} \ \times \ {6 \over 5} = \)\( \ \color{red}{{9 \times 6 \over 7\times5} = } \) \(\color{red}{{54 \over 35}}\)
Solution:
Step1: Multiply the top numbers, and then multiply the bottom numbers.
\({9 \over 7} \ \times \ {6 \over 5} = \)\({9 \times 6 \over 7\times5} = \) \({54 \over 35}\)
7)\({10 \over 5} \ \div \ {8 \over 3} = \)\({10 \over 5} \ \times \ {3 \over 8} = \)\( \ \color{red}{{10 \times 3 \over 5\times8} = } \) \(\color{red}{{30 \over 40}}\)\(\color{red}{ = {30 \div 10 \over 40 \div 10} = {3 \over 4}} \)
GCF(30,40) = 10
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then multiply them.
\({10 \over 5} \ \div \ {8 \over 3} = \)\({10 \over 5} \ \times \ {3 \over 8} = \)\({10 \times 3 \over 5\times8} =\) \({30 \over 40}\)
Step 2: Simplify your answer. \({30 \over 40}\)\( = {30 \div 10 \over 40 \div 10} = {3 \over 4} \)
8)\({6 \over 3} \ \div \ {14 \over 18} = \)\({6 \over 3} \ \times \ {18 \over 14} = \)\( \ \color{red}{{6 \times 18 \over 3\times14} = } \) \(\color{red}{{108 \over 42}}\)\(\color{red}{ = {108 \div 6 \over 42 \div 6} = {18 \over 7}}\)
GCF(108,42) = 6
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then Multiply them:
\({6 \over 3} \ \div \ {14 \over 18} = \)\({6 \over 3} \ \times \ {18 \over 14} = \)\({6 \times 18 \over 3\times14} = \) \({108 \over 42}\)
Step 2: Simplify your answer. \({108 \over 42}\)\( = {108 \div 6 \over 42 \div 6} = {18 \over 7} \)
9)\({6\over 10} \ \div \ {16 \over 17} = \)\({6 \over 10} \ \times \ {17 \over 16} = \)\( \ \color{red}{{6 \times 17 \over 10\times 16} = } \)\(\color{red}{{102 \over 160}}\)\(\color{red}{ = {102 \div 2\over 160\div 2} = {51 \over 80}}\)
GCF(102,160) = 2
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then Multiply them:
\({6\over 10} \ \div \ {16 \over 17} = \)\({6 \over 10} \ \times \ {17 \over 16} = \)\({6 \times 17 \over 10\times 16} = \)\({102 \over 160}\)
Step 2: Simplify your answer. \({102 \over 160}\)\( = {102 \div 2\over 160\div 2} = {51 \over 80}\)
10)\({8\over 7} \ \div \ {15 \over 20} = \)\({8 \over 7} \ \times \ {20 \over 15} = \)\( \ \color{red}{{8 \times 20 \over 7\times 15} = } \)\(\color{red}{{160 \over 105}}\)\(\color{red}{ = {160 \div 5\over 105\div 5} = {32 \over 21}} \)
GCF(160,105) = 5
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then Multiply them:
\({8\over 7} \ \div \ {15 \over 20} = \)\({8 \over 7} \ \times \ {20 \over 15} = \)\({8 \times 20 \over 7\times 15= } \)\({160 \over 105}\)
Step 2: Simplify your answer. \({160 \over 105}\)\( = {160 \div 5\over 105\div 5} = {32 \over 21}\)
11)\({2\over 5} \ \div \ {3 \over 2} = \)\({2 \over 5} \ \times \ {2 \over 3} = \)\( \ \color{red}{{2 \times 2 \over 5\times3} = } \)\(\color{red}{{4 \over 15}}\)
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then Multiply them:
\({2\over 5} \ \div \ {3 \over 2} = \)\({2 \over 5} \ \times \ {2 \over 3} = \)\({2 \times 2 \over 5\times3} = \)\({4 \over 15}\)
12)\({1\over 6} \ \div \ {3 \over 6} = \)\({1 \over 6} \ \times \ {6 \over 3} = \)\( \ \color{red}{{1 \times 6 \over 6\times 3} = } \)\(\color{red}{{6 \over 18}}\)\(\color{red}{ = {6 \div 6\over 18\div 6} = {1 \over 3}} \)
GCF(6,18) = 6
Solution:
Step1: Keep first fraction, change division sign to multiplication, and flip the numerator and denominator of the second fraction. Then Multiply them:
\({1\over 6} \ \div \ {3 \over 6} = \)\({1 \over 6} \ \times \ {6 \over 3} = \)\({1 \times 6 \over 6\times 3} = \)\({6 \over 18}\)
Step 2: Simplify your answer. \({6 \over 18}\)\( = {6 \div 6\over 18\div 6} = {1 \over 3}\)
