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.
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.