Systems of Equations Word Problems

Real-world applications of systems

Systems Word Problems

When to Use Systems

Use a system of equations when you have:

  • Two unknowns
  • Two conditions (pieces of information)

Common Types

Mixture Problems: Combining items with different values

Rate Problems: Distance = Rate × Time

Age Problems: Current ages vs. future/past ages

Money Problems: Different coins or bills

Strategy

  1. Define two variables
  2. Write two equations from the problem
  3. Solve using substitution or elimination
  4. Check your answer in context

Setting Up Equations

Read carefully for relationships like:

  • "The sum is..."
  • "One is twice the other..."
  • "Together they..."

📚 Practice Problems

1Problem 1easy

❓ Question:

The sum of two numbers is 25. Their difference is 7. Find the numbers.

💡 Show Solution

Let xx = larger number, yy = smaller number

Equation 1 (sum): x+y=25x + y = 25

Equation 2 (difference): x−y=7x - y = 7

Add the equations: 2x=322x = 32 x=16x = 16

Substitute into equation 1: 16+y=2516 + y = 25 y=9y = 9

Check: 16+9=2516 + 9 = 25 ✓ and 16−9=716 - 9 = 7 ✓

Answer: The numbers are 16 and 9

2Problem 2medium

❓ Question:

Adult tickets cost $12 and child tickets cost $8. A family bought 7 tickets for $76. How many of each did they buy?

💡 Show Solution

Let aa = adult tickets, cc = child tickets

Equation 1 (total tickets): a+c=7a + c = 7

Equation 2 (total cost): 12a+8c=7612a + 8c = 76

From equation 1: c=7−ac = 7 - a

Substitute into equation 2: 12a+8(7−a)=7612a + 8(7 - a) = 76 12a+56−8a=7612a + 56 - 8a = 76 4a=204a = 20 a=5a = 5

Find c: c=7−5=2c = 7 - 5 = 2

Check: 5+2=75 + 2 = 7 ✓ and 12(5)+8(2)=60+16=7612(5) + 8(2) = 60 + 16 = 76 ✓

Answer: 5 adult tickets and 2 child tickets

3Problem 3hard

❓ Question:

A store sells nuts. Cashews cost $8/lb and peanuts cost $5/lb. How many pounds of each should be mixed to make 10 pounds of mix worth $62?

💡 Show Solution

Let cc = pounds of cashews, pp = pounds of peanuts

Equation 1 (total weight): c+p=10c + p = 10

Equation 2 (total value): 8c+5p=628c + 5p = 62

From equation 1: p=10−cp = 10 - c

Substitute: 8c+5(10−c)=628c + 5(10 - c) = 62 8c+50−5c=628c + 50 - 5c = 62 3c=123c = 12 c=4c = 4

Find p: p=10−4=6p = 10 - 4 = 6

Check: 4+6=104 + 6 = 10 ✓ and 8(4)+5(6)=32+30=628(4) + 5(6) = 32 + 30 = 62 ✓

Answer: 4 pounds of cashews and 6 pounds of peanuts

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