A system is appropriate when two unknown quantities are connected by two genuinely different relationships. One equation narrows the possibilities; the second relationship identifies the pair that satisfies the entire situation.
From two unknowns to an interpreted pair
System-modeling workflow
- Identify two unknowns
Choose symbols and state their meanings and units.
- Find relationship one
Translate a total count, total amount, or difference into the first equation.
- Find relationship two
Translate an independent cost, value, or comparison statement.
- Solve
Use substitution when isolation is convenient or elimination when coefficients align.
- Interpret both values
Attach each value to the correct variable and answer the requested combination or quantity.
- Verify
Substitute into both original relationships and check contextual restrictions.
| Stage | Question to answer | Mathematical output |
|---|---|---|
| Unknown A and Unknown B | What two quantities vary independently? | Define \(x\) and \(y\) |
| Relationship 1 | What total or comparison links their counts? | Equation 1 |
| Relationship 2 | What different value or amount links them? | Equation 2 |
| Solve and interpret | Which pair satisfies both? | \((x,y)\) with contextual labels |
Count and value systems
| Pass type | Quantity | Price each | Revenue contribution |
|---|---|---|---|
| Standard | \(s\) | \(\$8\) | \(8s\) |
| Premium | \(p\) | \(\$13\) | \(13p\) |
| Total | \(52\) | — | \(\$526\) |
Two pass types
An event sells \(52\) passes. Standard passes cost \(\$8\), premium passes cost \(\$13\), and total revenue is \(\$526\). How many of each type are sold?
- Define
Let \(s\) be standard passes and \(p\) be premium passes.
- Count relationship
\(s+p=52\).
- Revenue relationship
\(8s+13p=526\).
- Substitute
From \(s=52-p\), write \(8(52-p)+13p=526\).
- Solve
\(416+5p=526\), so \(p=22\) and \(s=30\).
- Check
\(30+22=52\) and \(8(30)+13(22)=526\).
Difference and comparison systems
Two inventory quantities
Two bins contain \(94\) components. The larger bin holds \(18\) more components than the smaller bin. Find both counts using a system.
- Define
Let \(L\) be the larger count and \(S\) be the smaller count.
- Total
\(L+S=94\).
- Difference
\(L-S=18\), equivalently \(L=S+18\).
- Eliminate
Add the equations: \(2L=112\), so \(L=56\).
- Find the other count
\(S=94-56=38\).
Choose substitution or elimination
Method selection
Substitution
Efficient when one equation already states \(x=\cdots\), \(y=\cdots\), or a total equation makes isolation immediate.
Elimination
Efficient when a variable's coefficients are equal, opposite, or become opposite after a simple multiplication.
Application patterns
- Ticket or product problems often pair total quantity with total revenue.
- Coin problems pair total coin count with total monetary value; express values consistently in cents or dollars.
- Vehicle or seating problems pair total objects with a feature count such as wheels or seats.
- Age and inventory comparisons pair a total or future relationship with a difference relationship.
- Mixture problems require two distinct component contributions and should remain within the information supplied.
Wheels and vehicles
A recreation center stores \(26\) vehicles consisting only of bicycles and tricycles. Together they have \(62\) wheels. Find each vehicle count.
- Define
Let \(b\) be bicycles and \(t\) be tricycles.
- Vehicle count
\(b+t=26\).
- Wheel count
\(2b+3t=62\).
- Substitute
\(b=26-t\), so \(2(26-t)+3t=62\).
- Solve
\(52+t=62\), so \(t=10\) and \(b=16\).
Contextual feasibility
Counts of tickets, coins, vehicles, or people must normally be nonnegative whole numbers. Prices and measured quantities may be decimals. A negative or fractional count signals incorrect equations or incompatible data rather than permission to round.
Check your understanding
A store sells \(x\) small packages and \(y\) large packages. It sells \(40\) packages for \(\$312\); prices are \(\$6\) and \(\$10\). Which system is correct?
- \(x+y=40\), \(6x+10y=312\)
- \(6x+10y=40\), \(x+y=312\)
- \(x+y=40\), \(16x+16y=312\)
- \(x-y=40\), \(6x-10y=312\)
Show answer and explanation
Answer: \(x+y=40\), \(6x+10y=312\)
The first equation counts packages. The second multiplies each package count by its own price to total dollars.
What to remember
- Two unknowns require two independent relationships to determine a unique pair.
- Define variables once, then keep quantity, unit value, and total value distinct.
- Choose substitution or elimination based on equation structure, not habit.
- Verify both equations and contextual whole-number or nonnegative restrictions.
Put these notes into practice
Apply the ideas with SAT-style questions, then reinforce key details with flashcards.