Activity W 3.5 Fun riddle
Students may be encouraged to solve this riddle and should be asked to explain their strategy to solve it. This can give them an idea about solving linear equations. If the following equations are true: Then solve these
1οΈβ£ π± + πΈ = π¦
2οΈβ£ π¦ - π± = πΈ
3οΈβ£ π¦ - πΈ = π±
4οΈβ£ πΈ+ π± - πΈ =π±
Letβs assign:
-
π± = p
-
πΈ = f
-
π¦ = b
Step 1: Solve the first two equations.
From Equation 1: p + f = b
From Equation 2: b - p = f
Step 2: Substitute Equation 1 into Equation 2.
Substitute b = p + f into Equation 2:
(p + f) - p = f
f = f
This is always true, so the values depend on the third equation.
Step 3: Use the third equation.
From Equation 3:
b + f - p = p
Substitute b = p + f:
(p + f) + f - p = p
p + f + f - p = p
2f = p
Now substitute back:
If p = 2f,
then from Equation 1:
2f + f = b
so, b = 3f
So the values are:
-
π± (p) = 2f
-
πΈ (f) = f
-
π¦ (b) = 3f
Now solve the bottom part!
1οΈβ£ π± + π± + πΈ = ?
= 2p + f
= 2(2f) + f = 4f + f = 5f
2οΈβ£ π± - πΈ = ?
= p - f = 2f - f = f
3οΈβ£ π± + πΈ - π¦ = ?
= p + f - b
= 2f + f - 3f = 0
4οΈβ£ π¦ - π± = ?
= b - p = 3f - 2f = f
5οΈβ£ πΈ + π± - π± = ?
= f + p - p = f
Final Answers:
-
π± + π± + πΈ = π¦ 5f (depends on f)
-
π± - πΈ = πΈ f
-
π± + πΈ - π¦ = 0
-
π¦ - π± = πΈ
-
πΈ + π± - π± = πΈ f
We can also solve by assume values, like if πΈ = 1, then:
π± = 2, π¦ = 3 β and substitute!
TRY THIS!
π§ Puzzle 1: Fruit Equation
π + π + π = 12
π + π = 8
π + π = 10
Questions:
a) What is the value of each fruit?
b) π + π - π = ?
π§ Puzzle 2: Animal Equation
πΆ + π± = 14
π± + π° = 10
πΆ - π° = 6
Questions:
a) What is the value of πΆ, π±, and π°?
b) πΆ + π± + π° = ?
π§ Puzzle 3: Shape Equation
πΊ + π΅ = 15
π΅ + πͺ = 20
πΊ + πͺ = 25
Questions:
a) What is the value of each shape?
b) πΊ + π΅ + πͺ = ?
SOLUTIONS
π Puzzle 1: Fruit Equation
Given:
1οΈβ£ π + π + π = 12
2οΈβ£ π + π = 8
3οΈβ£ π + π = 10
Solution:
From equation 1οΈβ£:
2π + π = 12
So, π = 12 β 2π
Substitute into 2οΈβ£:
(12 β 2π) + π = 8
π = 8 β (12 β 2π)
π = 2π β 4
Now substitute into 3οΈβ£:
π + (2π β 4) = 10
3π = 14
π = 14 Γ· 3 β 4.67
But since the puzzle suggests whole numbers, let's double-check.
Actually, this puzzle seems set up for decimals or might have a typo β usually these are whole numbers.
π If you want, I can adjust the numbers slightly so they fit perfectly.
πΆ Puzzle 2: Animal Equation
Given:
1οΈβ£ πΆ + π± = 14
2οΈβ£ π± + π° = 10
3οΈβ£ πΆ β π° = 6
Solution:
From 1οΈβ£: πΆ = 14 β π±
From 2οΈβ£: π° = 10 β π±
Substitute into 3οΈβ£:
(14 β π±) β (10 β π±) = 6
14 β π± β 10 + π± = 6
4 = 6 β β this is impossible!
So this puzzle needs correction β the values don't balance.
πΊ Puzzle 3: Shape Equation
Given:
1οΈβ£ πΊ + π΅ = 15
2οΈβ£ π΅ + πͺ = 20
3οΈβ£ πΊ + πͺ = 25
Solution:
From 1οΈβ£: πΊ = 15 β π΅
Substitute into 3οΈβ£:
(15 β π΅) + πͺ = 25
πͺ = 10 + π΅
Now substitute πͺ into 2οΈβ£:
π΅ + (10 + π΅) = 20
2π΅ = 10
π΅ = 5
Now substitute π΅ = 5 into:
πΊ = 15 β 5 = 10
πͺ = 10 + 5 = 15
β
Final values:
πΊ = 10
π΅ = 5
πͺ = 15
Extra Question: πΊ + π΅ + πͺ = 10 + 5 + 15 = 30
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