Free Systems of Equations Word Problems Worksheet Collection - Free Printable
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Step-by-step solution for: Free Systems of Equations Word Problems Worksheet Collection
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Step-by-step solution for: Free Systems of Equations Word Problems Worksheet Collection
Let's solve each of these systems of equations word problems step by step.
---
- Adult ticket: $5
- Student ticket: $3
- Total tickets: 18
- Total cost: $82
Let:
- $ a $ = number of adult tickets
- $ s $ = number of student tickets
Equations:
$$
a + s = 18 \quad \text{(1)}
$$
$$
5a + 3s = 82 \quad \text{(2)}
$$
From (1): $ s = 18 - a $
Substitute into (2):
$$
5a + 3(18 - a) = 82 \\
5a + 54 - 3a = 82 \\
2a = 28 \\
a = 14
$$
✔ Answer: 14 adult tickets
---
Let:
- $ x $ = price of one shirt
- $ y $ = price of one hat
Eva: 2 shirts, 5 hats → $ 2x + 5y = 154 $
Nicole: 3 shirts, 4 hats → $ 3x + 4y = 168 $
System:
$$
2x + 5y = 154 \quad \text{(1)} \\
3x + 4y = 168 \quad \text{(2)}
$$
Multiply (1) by 3: $ 6x + 15y = 462 $
Multiply (2) by 2: $ 6x + 8y = 336 $
Subtract:
$$
(6x + 15y) - (6x + 8y) = 462 - 336 \\
7y = 126 \Rightarrow y = 18
$$
Plug back into (1):
$$
2x + 5(18) = 154 \\
2x + 90 = 154 \\
2x = 64 \Rightarrow x = 32
$$
✔ Answer: $32 per shirt
---
- Total animals: 10
- Total legs: 36
- Pigs have 4 legs, ducks have 2 legs
Let:
- $ p $ = number of pigs
- $ d $ = number of ducks
$$
p + d = 10 \quad \text{(1)} \\
4p + 2d = 36 \quad \text{(2)}
$$
From (1): $ d = 10 - p $
Substitute into (2):
$$
4p + 2(10 - p) = 36 \\
4p + 20 - 2p = 36 \\
2p = 16 \Rightarrow p = 8
$$
✔ Answer: 8 pigs
---
- Total students: 195
- Total vehicles: 19
- Car holds 5 students
- Bus holds 25 students
Let:
- $ c $ = number of cars
- $ b $ = number of buses
$$
c + b = 19 \quad \text{(1)} \\
5c + 25b = 195 \quad \text{(2)}
$$
Simplify (2): divide by 5 → $ c + 5b = 39 $
Now subtract (1) from this:
$$
(c + 5b) - (c + b) = 39 - 19 \\
4b = 20 \Rightarrow b = 5
$$
✔ Answer: 5 buses
---
Let:
- $ s $ = price of senior ticket
- $ c $ = price of child ticket
Mr. Smith: $ 8s + 5c = 136 $
Mr. Jackson: $ 4s + 6c = 96 $
System:
$$
8s + 5c = 136 \quad \text{(1)} \\
4s + 6c = 96 \quad \text{(2)}
$$
Multiply (2) by 2: $ 8s + 12c = 192 $
Subtract (1):
$$
(8s + 12c) - (8s + 5c) = 192 - 136 \\
7c = 56 \Rightarrow c = 8
$$
Plug into (2):
$$
4s + 6(8) = 96 \\
4s + 48 = 96 \\
4s = 48 \Rightarrow s = 12
$$
✔ Answer: $12 per senior ticket
---
Let:
- $ x $ = bigger number
- $ y $ = smaller number
$$
x - y = 6 \quad \text{(1)} \\
x + y = 14 \quad \text{(2)}
$$
Add equations:
$$
2x = 20 \Rightarrow x = 10
$$
✔ Answer: 10
---
Let the number be $ 10a + b $, where:
- $ a $ = tens digit
- $ b $ = units digit
Given:
- $ a + b = 7 $
- Reversed number: $ 10b + a $
- $ 10b + a = 10a + b + 9 $
Simplify second equation:
$$
10b + a - 10a - b = 9 \\
9b - 9a = 9 \Rightarrow b - a = 1
$$
Now:
$$
a + b = 7 \\
b - a = 1
$$
Add:
$$
2b = 8 \Rightarrow b = 4 \Rightarrow a = 3
$$
Number: $ 10a + b = 30 + 4 = 34 $
Check: reverse is 43; $ 43 - 34 = 9 $ ✔
✔ Answer: 34
---
Let:
- $ x $ = larger
- $ y $ = smaller
$$
x - y = 18 \\
x + y = 66
$$
Add:
$$
2x = 84 \Rightarrow x = 42 \Rightarrow y = 24
$$
✔ Answer: 42 and 24
---
Let:
- $ w $ = width
- $ l $ = length
Given:
- $ l = 2w + 3 $
- Perimeter = 30 → $ 2(l + w) = 30 \Rightarrow l + w = 15 $
Substitute:
$$
(2w + 3) + w = 15 \\
3w + 3 = 15 \\
3w = 12 \Rightarrow w = 4 \\
l = 2(4) + 3 = 11
$$
✔ Answer: Length is 11 meters
---
- Nickels = $0.05
- Dimes = $0.10
- Total coins: 44
- Total value: $2.95
Let:
- $ n $ = number of nickels
- $ d $ = number of dimes
$$
n + d = 44 \quad \text{(1)} \\
0.05n + 0.10d = 2.95 \quad \text{(2)}
$$
Multiply (2) by 100 to eliminate decimals:
$$
5n + 10d = 295
$$
Divide by 5:
$$
n + 2d = 59 \quad \text{(2')}
$$
Now subtract (1):
$$
(n + 2d) - (n + d) = 59 - 44 \\
d = 15
\Rightarrow n = 44 - 15 = 29
$$
✔ Answer: 29 nickels
---
1) 14
2) $32
3) 8 pigs
4) 5 buses
5) $12
6) 10
7) 34
8) 42 and 24
9) 11 meters
10) 29 nickels
Let me know if you'd like explanations in a different format or want to see the steps graphically!
---
1) Movie Tickets
- Adult ticket: $5
- Student ticket: $3
- Total tickets: 18
- Total cost: $82
Let:
- $ a $ = number of adult tickets
- $ s $ = number of student tickets
Equations:
$$
a + s = 18 \quad \text{(1)}
$$
$$
5a + 3s = 82 \quad \text{(2)}
$$
From (1): $ s = 18 - a $
Substitute into (2):
$$
5a + 3(18 - a) = 82 \\
5a + 54 - 3a = 82 \\
2a = 28 \\
a = 14
$$
✔ Answer: 14 adult tickets
---
2) Eva and Nicole’s Purchase
Let:
- $ x $ = price of one shirt
- $ y $ = price of one hat
Eva: 2 shirts, 5 hats → $ 2x + 5y = 154 $
Nicole: 3 shirts, 4 hats → $ 3x + 4y = 168 $
System:
$$
2x + 5y = 154 \quad \text{(1)} \\
3x + 4y = 168 \quad \text{(2)}
$$
Multiply (1) by 3: $ 6x + 15y = 462 $
Multiply (2) by 2: $ 6x + 8y = 336 $
Subtract:
$$
(6x + 15y) - (6x + 8y) = 462 - 336 \\
7y = 126 \Rightarrow y = 18
$$
Plug back into (1):
$$
2x + 5(18) = 154 \\
2x + 90 = 154 \\
2x = 64 \Rightarrow x = 32
$$
✔ Answer: $32 per shirt
---
3) Pigs and Ducks
- Total animals: 10
- Total legs: 36
- Pigs have 4 legs, ducks have 2 legs
Let:
- $ p $ = number of pigs
- $ d $ = number of ducks
$$
p + d = 10 \quad \text{(1)} \\
4p + 2d = 36 \quad \text{(2)}
$$
From (1): $ d = 10 - p $
Substitute into (2):
$$
4p + 2(10 - p) = 36 \\
4p + 20 - 2p = 36 \\
2p = 16 \Rightarrow p = 8
$$
✔ Answer: 8 pigs
---
4) Field Trip Vehicles
- Total students: 195
- Total vehicles: 19
- Car holds 5 students
- Bus holds 25 students
Let:
- $ c $ = number of cars
- $ b $ = number of buses
$$
c + b = 19 \quad \text{(1)} \\
5c + 25b = 195 \quad \text{(2)}
$$
Simplify (2): divide by 5 → $ c + 5b = 39 $
Now subtract (1) from this:
$$
(c + 5b) - (c + b) = 39 - 19 \\
4b = 20 \Rightarrow b = 5
$$
✔ Answer: 5 buses
---
5) Theater Tickets
Let:
- $ s $ = price of senior ticket
- $ c $ = price of child ticket
Mr. Smith: $ 8s + 5c = 136 $
Mr. Jackson: $ 4s + 6c = 96 $
System:
$$
8s + 5c = 136 \quad \text{(1)} \\
4s + 6c = 96 \quad \text{(2)}
$$
Multiply (2) by 2: $ 8s + 12c = 192 $
Subtract (1):
$$
(8s + 12c) - (8s + 5c) = 192 - 136 \\
7c = 56 \Rightarrow c = 8
$$
Plug into (2):
$$
4s + 6(8) = 96 \\
4s + 48 = 96 \\
4s = 48 \Rightarrow s = 12
$$
✔ Answer: $12 per senior ticket
---
6) Two Numbers – Difference 6, Sum 14
Let:
- $ x $ = bigger number
- $ y $ = smaller number
$$
x - y = 6 \quad \text{(1)} \\
x + y = 14 \quad \text{(2)}
$$
Add equations:
$$
2x = 20 \Rightarrow x = 10
$$
✔ Answer: 10
---
7) Two-Digit Number – Digit Sum is 7, Reverse Increases by 9
Let the number be $ 10a + b $, where:
- $ a $ = tens digit
- $ b $ = units digit
Given:
- $ a + b = 7 $
- Reversed number: $ 10b + a $
- $ 10b + a = 10a + b + 9 $
Simplify second equation:
$$
10b + a - 10a - b = 9 \\
9b - 9a = 9 \Rightarrow b - a = 1
$$
Now:
$$
a + b = 7 \\
b - a = 1
$$
Add:
$$
2b = 8 \Rightarrow b = 4 \Rightarrow a = 3
$$
Number: $ 10a + b = 30 + 4 = 34 $
Check: reverse is 43; $ 43 - 34 = 9 $ ✔
✔ Answer: 34
---
8) Two Numbers – Difference 18, Sum 66
Let:
- $ x $ = larger
- $ y $ = smaller
$$
x - y = 18 \\
x + y = 66
$$
Add:
$$
2x = 84 \Rightarrow x = 42 \Rightarrow y = 24
$$
✔ Answer: 42 and 24
---
9) Rectangle Dimensions
Let:
- $ w $ = width
- $ l $ = length
Given:
- $ l = 2w + 3 $
- Perimeter = 30 → $ 2(l + w) = 30 \Rightarrow l + w = 15 $
Substitute:
$$
(2w + 3) + w = 15 \\
3w + 3 = 15 \\
3w = 12 \Rightarrow w = 4 \\
l = 2(4) + 3 = 11
$$
✔ Answer: Length is 11 meters
---
10) Nickels and Dimes
- Nickels = $0.05
- Dimes = $0.10
- Total coins: 44
- Total value: $2.95
Let:
- $ n $ = number of nickels
- $ d $ = number of dimes
$$
n + d = 44 \quad \text{(1)} \\
0.05n + 0.10d = 2.95 \quad \text{(2)}
$$
Multiply (2) by 100 to eliminate decimals:
$$
5n + 10d = 295
$$
Divide by 5:
$$
n + 2d = 59 \quad \text{(2')}
$$
Now subtract (1):
$$
(n + 2d) - (n + d) = 59 - 44 \\
d = 15
\Rightarrow n = 44 - 15 = 29
$$
✔ Answer: 29 nickels
---
✔ Final Answers:
1) 14
2) $32
3) 8 pigs
4) 5 buses
5) $12
6) 10
7) 34
8) 42 and 24
9) 11 meters
10) 29 nickels
Let me know if you'd like explanations in a different format or want to see the steps graphically!
Parent Tip: Review the logic above to help your child master the concept of system of equation word problem worksheet.