Systems of Equations Word Problems

Introduction

Many real-world problems involve two unknowns and two relationships between them. These situations require a system of two equations to solve.

Key Difference from One-Variable Problems:

• One equation: One unknown, one relationship
• System: Two unknowns, two relationships

Identifying System Problems

Look for these signs you need a system:

• Two different quantities are unknown
• Two separate pieces of information are given
• The problem involves rates, prices, or quantities of two different items
• Words like "combined," "together," "total" alongside individual descriptions

The Problem-Solving Process

Step 1: Read Carefully Understand what you're looking for (identify the two unknowns)

Step 2: Define Variables Let x = first unknown Let y = second unknown Write clearly what each represents!

Step 3: Write Two Equations Use the given information to create two different equations

Step 4: Solve the System Use substitution, elimination, or graphing

Step 5: Check and Interpret Verify solution works in both equations Answer in complete sentences with units

Number Problems

Example 1: Sum and Difference Problem: The sum of two numbers is 25. Their difference is 7. Find the numbers.

Let x = larger number Let y = smaller number

Equation 1 (sum): x + y = 25 Equation 2 (difference): x - y = 7

Using elimination (add equations): x + y = 25

• x - y = 7 2x = 32 x = 16

Substitute: 16 + y = 25 y = 9

Answer: The numbers are 16 and 9.

Check: 16 + 9 = 25 ✓ and 16 - 9 = 7 ✓

Example 2: Digit Problem Problem: A two-digit number has digits whose sum is 12. If the digits are reversed, the new number is 36 more than the original. Find the number.

Let t = tens digit Let u = units digit

Original number: 10t + u Reversed number: 10u + t

Equation 1 (sum of digits): t + u = 12 Equation 2 (reversed is 36 more): 10u + t = (10t + u) + 36

Simplify equation 2: 10u + t = 10t + u + 36 9u - 9t = 36 u - t = 4

System: t + u = 12 u - t = 4

Add equations: 2u = 16, so u = 8 Then: t + 8 = 12, so t = 4

Answer: The number is 48.

Check: 4 + 8 = 12 ✓ and 84 - 48 = 36 ✓

Money and Coin Problems

Example 1: Ticket Sales Problem: A theater sold 350 tickets. Adult tickets cost $8 and child tickets cost$5. Total sales were $2,350. How many of each type were sold?

Let a = adult tickets Let c = child tickets

Equation 1 (number of tickets): a + c = 350 Equation 2 (total money): 8a + 5c = 2350

From equation 1: c = 350 - a Substitute into equation 2: 8a + 5(350 - a) = 2350 8a + 1750 - 5a = 2350 3a = 600 a = 200

Then: c = 350 - 200 = 150

Answer: 200 adult tickets and 150 child tickets.

Check: 200 + 150 = 350 ✓ 8(200) + 5(150) = 1600 + 750 = 2350 ✓

Example 2: Coin Problem Problem: You have 27 coins in quarters and dimes worth $4.95. How many of each coin do you have?

Let q = number of quarters Let d = number of dimes

Equation 1 (number of coins): q + d = 27 Equation 2 (value in cents): 25q + 10d = 495

From equation 1: d = 27 - q Substitute: 25q + 10(27 - q) = 495 25q + 270 - 10q = 495 15q = 225 q = 15

Then: d = 27 - 15 = 12

Answer: 15 quarters and 12 dimes.

Check: 15(0.25) + 12(0.10) = 3.75 + 1.20 = 4.95 ✓

Example 3: Investment Problem: You invest $5,000 in two accounts. One earns 3$210 in interest. How much was invested in each account?

Let x = amount at 3% Let y = amount at 5%

Equation 1 (total investment): x + y = 5000 Equation 2 (total interest): 0.03x + 0.05y = 210

From equation 1: y = 5000 - x Substitute: 0.03x + 0.05(5000 - x) = 210 0.03x + 250 - 0.05x = 210 -0.02x = -40 x = 2000

Then: y = 5000 - 2000 = 3000

Answer: $2,000 at 3$3,000 at 5%.

Mixture Problems

Example 1: Solution Concentration Problem: A chemist needs 10 liters of 30% acid solution. She has 20% and 50% solutions available. How much of each should she mix?

Let x = liters of 20% solution Let y = liters of 50% solution

Equation 1 (total volume): x + y = 10 Equation 2 (amount of acid): 0.20x + 0.50y = 0.30(10)

Simplify equation 2: 0.20x + 0.50y = 3

Multiply by 10: 2x + 5y = 30

System: x + y = 10 2x + 5y = 30

From first: x = 10 - y Substitute: 2(10 - y) + 5y = 30 20 - 2y + 5y = 30 3y = 10 y = 10/3 ≈ 3.33 liters

x = 10 - 10/3 = 20/3 ≈ 6.67 liters

Answer: About 6.67 liters of 20% solution and 3.33 liters of 50% solution.

Example 2: Nut Mixture Problem: A store mixes peanuts ($3/lb) with cashews ($8/lb) to make 20 pounds of mixed nuts worth $5/lb. How many pounds of each should be used?

Let p = pounds of peanuts Let c = pounds of cashews

Equation 1 (total weight): p + c = 20 Equation 2 (total value): 3p + 8c = 5(20)

Simplify equation 2: 3p + 8c = 100

From equation 1: c = 20 - p Substitute: 3p + 8(20 - p) = 100 3p + 160 - 8p = 100 -5p = -60 p = 12

Then: c = 20 - 12 = 8

Answer: 12 pounds of peanuts and 8 pounds of cashews.

Distance, Rate, and Time Problems

Formula: Distance = Rate × Time (d = rt)

Example 1: Traveling Toward Each Other Problem: Two cars are 400 miles apart and drive toward each other. One travels at 50 mph, the other at 60 mph. How long until they meet?

Let t = time until they meet (same for both) Let d₁ = distance first car travels Let d₂ = distance second car travels

Equation 1 (total distance): d₁ + d₂ = 400 Equation 2 (using d = rt): First car: d₁ = 50t Second car: d₂ = 60t

Substitute into equation 1: 50t + 60t = 400 110t = 400 t = 400/110 ≈ 3.64 hours

Answer: They meet in about 3.64 hours (or 3 hours 38 minutes).

Example 2: Boat in Current Problem: A boat travels 30 miles downstream in 2 hours. The return trip upstream takes 3 hours. Find the boat's speed in still water and the current's speed.

Let b = boat speed in still water Let c = current speed

Downstream speed: b + c Upstream speed: b - c

Using d = rt: Equation 1 (downstream): (b + c) × 2 = 30 Equation 2 (upstream): (b - c) × 3 = 30

Simplify: 2b + 2c = 30 → b + c = 15 3b - 3c = 30 → b - c = 10

Add equations: 2b = 25 b = 12.5

Then: 12.5 + c = 15 c = 2.5

Answer: Boat speed is 12.5 mph, current is 2.5 mph.

Example 3: Round Trip Problem: You drive to a destination at 60 mph and return at 40 mph. The total trip takes 5 hours. How far is the destination?

Let d = distance to destination (same both ways) Let t₁ = time going Let t₂ = time returning

Using d = rt: Equation 1: d = 60t₁ Equation 2: d = 40t₂ Equation 3: t₁ + t₂ = 5

From equations 1 and 2: 60t₁ = 40t₂ Simplify: 3t₁ = 2t₂ So: t₂ = (3/2)t₁

Substitute into equation 3: t₁ + (3/2)t₁ = 5 (5/2)t₁ = 5 t₁ = 2

Then: d = 60(2) = 120

Answer: The destination is 120 miles away.

Work Rate Problems

Formula: Work = Rate × Time

Example 1: Working Together Problem: One pipe can fill a pool in 6 hours. Another can fill it in 4 hours. How long does it take both pipes together?

Let t = time to fill together

Pipe 1 rate: 1/6 pool per hour Pipe 2 rate: 1/4 pool per hour Combined rate: 1/6 + 1/4 = 2/12 + 3/12 = 5/12 pool per hour

Equation: (5/12)t = 1 t = 12/5 = 2.4 hours

Answer: 2.4 hours (or 2 hours 24 minutes).

Example 2: Different Work Rates Problem: Working alone, John can paint a fence in 8 hours and Mary can paint it in 6 hours. They work together for 2 hours, then Mary leaves. How much longer does John need to finish?

Let x = additional time John works alone

Work done together in 2 hours: 2(1/8 + 1/6) = 2(3/24 + 4/24) = 2(7/24) = 7/12

Work remaining: 1 - 7/12 = 5/12

John finishes at rate 1/8: (1/8)x = 5/12 x = (5/12) × 8 = 40/12 = 10/3 ≈ 3.33 hours

Answer: About 3.33 hours (or 3 hours 20 minutes).

Age Problems

Example 1: Current and Future Ages Problem: Sarah is 4 years older than Tom. In 6 years, the sum of their ages will be 36. How old are they now?

Let s = Sarah's current age Let t = Tom's current age

Equation 1 (current): s = t + 4 Equation 2 (in 6 years): (s + 6) + (t + 6) = 36

Simplify equation 2: s + t + 12 = 36 s + t = 24

Substitute s = t + 4: (t + 4) + t = 24 2t = 20 t = 10

Then: s = 10 + 4 = 14

Answer: Tom is 10, Sarah is 14.

Geometry Problems

Example 1: Rectangle Dimensions Problem: A rectangle's perimeter is 60 cm. The length is twice the width. Find the dimensions.

Let l = length Let w = width

Equation 1 (relationship): l = 2w Equation 2 (perimeter): 2l + 2w = 60

Substitute: 2(2w) + 2w = 60 4w + 2w = 60 6w = 60 w = 10

Then: l = 2(10) = 20

Answer: Width is 10 cm, length is 20 cm.

Example 2: Angles Problem: Two angles are complementary (sum to 90°). One angle is 15° more than twice the other. Find both angles.

Let x = smaller angle Let y = larger angle

Equation 1 (complementary): x + y = 90 Equation 2 (relationship): y = 2x + 15

Substitute: x + (2x + 15) = 90 3x + 15 = 90 3x = 75 x = 25

Then: y = 2(25) + 15 = 65

Answer: The angles are 25° and 65°.

Common Mistakes to Avoid

1. Not defining variables clearly Always write what x and y represent!

2. Writing only one equation Need TWO equations for TWO unknowns

3. Mixing up units Keep dollars vs cents, hours vs minutes consistent

4. Forgetting to answer the question "Find how many of each" means find BOTH values

5. Not checking the solution Verify in both original equations AND in context

6. Setting up equations incorrectly "Total" usually means addition "Combined rate" means add rates

Problem-Solving Strategy

Before You Start:

1. Read the entire problem twice
2. Identify what you're looking for (the two unknowns)
3. Look for two separate pieces of information

Setting Up:

1. Define variables with words, not just letters
2. Write one equation for each piece of information
3. Make sure equations are independent (not the same info)

Solving:

1. Choose substitution or elimination
2. Solve for one variable
3. Substitute to find the other
4. Show all work

Finishing:

1. Check in both original equations
2. Does answer make sense in context?
3. Answer in complete sentence with units
4. Did you answer what was asked?

Quick Reference - Problem Types

| Type | Variables | Common Equations | |------|-----------|------------------| | Coin/Ticket | Number of each | Total count, Total value | | Mixture | Amount of each | Total volume, Total concentration | | Age | Current ages | Current relationship, Future/past | | Distance | Distances or speeds | d = rt for each, Total distance | | Work | Time or rates | Combined work = 1 | | Geometry | Dimensions | Perimeter/area, Relationships |

Practice Tips

• Start with simpler two-variable problems
• Draw diagrams when helpful (especially for distance/geometry)
• Make tables to organize information
• Label which equation represents which information
• Practice both substitution and elimination
• Check reasonableness (negative ages? fractional tickets?)
• Read carefully - word problems test reading as much as math
• Write clear, complete solutions showing all steps