Let:
- Let F represent the number of frogs
- Let G represent the number of fairy godmothers
- Let D represent the number of dragons
From Tug of War #1:
4F = 5G
From Tug of War #2:
D = 2G + F
D = 4F + G
Now, we can substitute the value of F from Tug of War #1 into Tug of War #2:
D = 2(4/5G) + 4G + G
D = 8/5G + 5G
D = 33/5G
Finally, from Tug of War #3:
D = G + 3F
Substitute the value of D from Tug of War #2 into Tug of War #3:
33/5G = G + 3(4/5G)
33/5G = G + 12/5G
33G = 5G + 12G
G = 3
Now, substituting the value of G back into Tug of War #1:
4F = 5(3)
4F = 15
F = 3.75
Since we cannot have a fraction of a frog, we will round down to have a whole number. Therefore, F = 3.
Now we can find the number of dragons in Tug of War #2:
D = 2(3) + 3
D = 6 + 3
D = 9
Therefore, in Tug of War #3, there is 1 dragon and 3 fairy godmothers on one side, while the other side has 3 frogs. Since the dragon is the strongest, the team with the dragon and fairy godmothers would win the 3rd tug of war.
Read the following algebraic word problem below. Use strategies learned throughout this unit (creating "let statements", algebraic equations) to critically think/solve this problem. Hint: Figure out how to give a value to a frog, fairy godmother and dragon.
Tug of War #1: 4 frogs on one side had a tie with 5 fairy godmothers on the other side.
Tug of War #2: 1 dragon had a tie with 2 fairy godmothers and 1 frog. and 4 frogs and 1 fairy godmother on
Tug of War #3: 1 dragon and 3 fairy godmothers on one side the other side.
Who would win the 3rd tug of war? SHOW YOUR WORK FOR FULL MARKS.
1 answer
1 answer