Systems Word Problems (Mixture, Cost, Rate, Work)

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Translate real-world scenarios into systems of linear equations.
  • Recognize common structures: mixture, ticket/cost, distance-rate-time, and work problems.
  • Solve systems using substitution, elimination, or reasoning.
  • Interpret solutions in meaningful real-world context.

Key Ideas

Systems word problems often look complicated, but most of them fit into a small set of repeatable patterns. Once you recognize the pattern, you can set up a system almost on autopilot.

Common structures:

  • mixture / concentration
  • ticket / cost / revenue
  • distance–rate–time (D = RT)
  • work / rate problems

Each type has a standard way to define variables and a standard way to form equations.

Common Problem Types

Mixture Problems

Used when combining solutions, concentrations, or substances.

Core idea:

\[ \text{amount} \times \text{concentration} = \text{pure amount} \]

Typical situations:

  • mixing percent solutions
  • combining items with different prices into one mixture
  • weighted averages

Ticket / Price / Cost Problems

Combine a count equation with a money equation.

Standard structure:

\[ \begin{split} \text{items:} \quad & x + y = \text{total items} \\ \text{cost:} \quad & ax + by = \text{total revenue} \end{split} \]

Distance / Rate / Time (D = RT)

Motion problems with two moving objects.

Basic relationship:

\[ d = rt \]

Typical setups:

  • traveling toward each other
  • traveling in the same direction (one catching up)
  • round-trip problems
  • delayed starts

Work Problems

Focus on rates, not times. If one person takes \(a\) hours and another takes \(b\) hours:

\[ r_1 = \frac{1}{a}, \quad r_2 = \frac{1}{b} \]

Combined rate:

\[ r = r_1 + r_2 \]

Time together:

\[ t = \frac{1}{r_1 + r_2} \]

Systems appear when you have extra information, like “one works twice as fast as the other.”

Strategies

  • Start by defining variables in words (e.g., “Let \(x\) be the number of adult tickets.”).
  • Use the structure of the problem (mixture, cost, D = RT, work) to build your equations.
  • Keep units consistent everywhere (all hours, all miles, all dollars).
  • Choose the solving method (substitution or elimination) based on which gives cleaner algebra.
  • After solving, interpret the answer: does it make sense in the story?

Worked Examples

Example 1 — Mixture

A chemist mixes 10 L of 20% solution with \(x\) L of 50% solution to produce a 30% solution. Find \(x\).

Amount of pure substance:

\[ 10(0.20) + 0.50x = (10 + x)(0.30) \]

Simplify:

\[ \begin{split} 2 + 0.5x &= 3 + 0.3x \\ 0.2x &= 1 \\ x &= 5 \end{split} \]

So the chemist needs 5 L of the 50% solution.

Example 2 — Ticket/Cost Problem

Adult tickets cost $12 and student tickets cost $8.
If 150 tickets were sold for a total of $1520, how many of each were sold?

Let:
\(x\) = number of adult tickets
\(y\) = number of student tickets

System:

\[ \begin{split} x + y &= 150 \\ 12x + 8y &= 1520 \end{split} \]

Multiply the first equation by 8:

\[ \begin{split} 8x + 8y &= 1200 \\ 12x + 8y &= 1520 \end{split} \]

Subtract:

\[ 4x = 320 \Rightarrow x = 80 \]

Then:

\[ y = 150 - 80 = 70 \]

So there were 80 adult and 70 student tickets.

Example 3 — Distance/Rate/Time

Two cars drive toward each other from 300 miles apart.
Car A: 50 mph
Car B: 70 mph
When do they meet?

Let \(t\) = time in hours until they meet.

Combined distance:

\[ 50t + 70t = 300 \]

\[ \begin{split} 120t &= 300 \\ t &= 2.5 \text{ hours} \end{split} \]

They meet after 2.5 hours.

Example 4 — Overtaking Problem

A cyclist leaves 2 hours after a runner.
Runner speed: 6 mph
Cyclist speed: 12 mph
When will the cyclist catch up?

Runner’s head start distance:

\[ 6(2) = 12 \text{ miles} \]

Let \(t\) = cyclist’s time (in hours) until catch-up.

Distances when they meet:

\[ \begin{split} \text{Runner:} &\quad 6(t + 2) \\ \text{Cyclist:} &\quad 12t \end{split} \]

Set equal:

\[ \begin{split} 12t &= 6(t + 2) \\ 12t &= 6t + 12 \\ 6t &= 12 \\ t &= 2 \end{split} \]

The cyclist catches up 2 hours after starting.

Example 5 — Work Problem

One machine finishes a job in 6 hours; another in 4 hours.
How long to finish working together?

Rates:

\[ r_1 = \frac{1}{6}, \quad r_2 = \frac{1}{4} \]

Combined rate:

\[ \begin{split} r &= \frac{1}{6} + \frac{1}{4} \\ &= \frac{2}{12} + \frac{3}{12} \\ &= \frac{5}{12} \end{split} \]

Time:

\[ t = \frac{1}{5/12} = \frac{12}{5} = 2.4 \text{ hours} \]

So they complete the job together in 2.4 hours.

WarningCommon Mistakes
  • Forgetting to multiply concentration by total amount in mixture problems.
  • Mixing counts and costs in the same equation for ticket problems.
  • Using the wrong form of the distance formula (it should be \(d = rt\)).
  • Ignoring unit consistency (e.g., mixing minutes and hours).
  • Adding times for work problems instead of rates.

Practice Problems

  1. A chemist mixes 5 L of 10% solution with \(x\) L of 40% solution to obtain a 25% solution. Find \(x\).

  2. Adult tickets are $15 and student tickets are $10. A total of 140 tickets bring in $1900. How many adult tickets were sold?

  3. A train going 80 mph leaves 1 hour after a car going 60 mph on the same road in the same direction. How long will it take the train to catch the car after it leaves?

  4. Two workers together complete a job in 3 hours. One works twice as fast as the other. How long would the slower worker take alone?

  5. A cyclist has already traveled 20 miles and a runner has already traveled 10 miles. After \(t\) additional hours, their combined distance must be 50 miles. Speeds: cyclist 10 mph, runner 5 mph. Find \(t\).

1. Mixture

Let \(x\) = liters of 40% solution.

\[ \begin{split} 5(0.10) + 0.40x &= (5 + x)(0.25) \\ 0.5 + 0.4x &= 1.25 + 0.25x \\ 0.15x &= 0.75 \\ x &= 5 \end{split} \]

2. Tickets

Let \(x\) = adult tickets, \(y\) = student tickets.

\[ \begin{split} x + y &= 140 \\ 15x + 10y &= 1900 \end{split} \]

Multiply the first equation by 10:

\[ 10x + 10y = 1400 \]

Subtract:

\[ 5x = 500 \Rightarrow x = 100 \]

So 100 adult tickets and 40 student tickets.

3. Overtaking

Car’s head start distance:

\[ 60(1) = 60 \]

Let \(t\) = hours after the train leaves.

Distances:

\[ \begin{split} \text{Car:} &\quad 60(t + 1) \\ \text{Train:} &\quad 80t \end{split} \]

Set equal:

\[ \begin{split} 80t &= 60(t + 1) \\ 80t &= 60t + 60 \\ 20t &= 60 \\ t &= 3 \end{split} \]

4. Work

Let \(x\) = time (in hours) for the slower worker alone.
Then faster worker time = \(x/2\).

Rates:

\[ \frac{1}{x} + \frac{2}{x} = \frac{1}{3} \]

\[ \begin{split} \frac{3}{x} &= \frac{1}{3} \\ x &= 9 \end{split} \]

So the slower worker would take 9 hours alone.

5. Distance

Initial distance: \(20 + 10 = 30\) miles.
They need a total of 50 miles, so 20 more miles together.

Let \(t\) = additional hours.

Total distance:

\[ \begin{split} 20 + 10t + 10 + 5t &= 50 \\ 30 + 15t &= 50 \\ 15t &= 20 \\ t &= \frac{20}{15} = \frac{4}{3} \end{split} \]

So \(t = \tfrac{4}{3}\) hours (1 hour 20 minutes).

Summary

  • Many word problems map cleanly onto a small set of system structures: mixture, ticket/cost, distance-rate-time, and work.
  • For each type, defining variables clearly and respecting units makes equation-building much easier.
  • Systems can then be solved with substitution, elimination, or sometimes direct reasoning.
  • Always interpret your final answer back in the real-world context.
  • Before writing equations, ask: “Which type is this—mixture, cost, rate, or work?”
  • Keep a mental template for each type (e.g., \(d = rt\) or items + cost equations).
  • Check that your solution is reasonable (no negative tickets, negative time, etc.).
  • If the algebra looks messy, re-check your setup—most word-problem errors come from the translation step.