Word Problems with Systems

Learning goals

Name two unknowns with units, one per stated fact
Match two unknowns to two independent equations
Set up value problems by counting items and totaling worth
Table amount, strength and content for a mixture
Apply $d = rt$ per leg, with a current changing the rate
Find break-even where cost equals revenue

Two unknowns, two conditions

A word problem is a situation described in words with some of its numbers left blank. Suppose two of those numbers are unknown and the problem states two separate facts about them. Then a system of two equations is the natural model: one equation for each fact. Solving it is work you already know, the substitution and elimination from the last two lessons. The new skill is the translation, now carried out twice.

The method is the same every time, whatever the story is about.

Name the two unknowns. Decide exactly which two quantities the problem asks about, and give each its own letter. Write down what each letter means, units included, as in “let $a$ be the number of adult tickets and $c$ the number of child tickets.”
Translate each condition into an equation. A well-posed problem states two facts. Turn each one into an equation in your two letters. You should finish with two equations, not one.
Solve the system. Use substitution when a variable is already isolated or easy to isolate, and elimination when both equations sit in standard form $ax + by = c$ . Either route returns the same pair.
Interpret and check. Put the pair back into the words of the problem, confirm it answers the actual question, and state it with units.

Steps 1 and 4 are the ones most often rushed. Naming both unknowns, with units, is what keeps the two equations straight. Checking against the words, not just against your own equations, is what catches a translation that came out backwards.

Why two conditions are exactly enough

A single equation in two unknowns cannot find them both. The argument below says why, and settles how many equations a two-unknown problem needs, and why the final check has to be against the words.

Why two independent conditions determine two unknowns#

Start with two unknowns, $x$ and $y$ . Before you use any information, the pair $(x, y)$ could be any point in the plane, so there are two quantities free to vary. From the two-variable-equations lesson, a single linear equation $ax + by = c$ does not fix either unknown on its own; its solutions fill an entire straight line. So one condition narrows the possibilities from the whole plane down to a single line, but no further: infinitely many pairs still fit.

A second condition, as long as it says something genuinely new, is a second line. Two lines that are not parallel cross at exactly one point. That shared point is the only pair making both statements true. So two independent conditions pin down both unknowns. The word “independent” is doing real work here. If the second equation is only the first in disguise, a multiple of it, the two describe the same line. Then nothing new is fixed, and the problem has infinitely many answers. If the second contradicts the first, the same steepness but a different intercept, the lines are parallel and no pair satisfies both. A well-posed problem avoids both traps, handing you two facts that are neither identical nor contradictory, which is exactly as many independent conditions as there are unknowns. That is the count to match: two unknowns call for two equations, and writing only one leaves a whole line of pretenders.

The final check has to go back to the words because your two equations are a translation, and a translation can be faithful or wrong. Substituting your answer into your own equations only confirms that you solved those equations correctly; it says nothing about whether the equations captured the problem. Suppose the story says the larger number exceeds the smaller by $5$ , and you translate it, with a slip, as the smaller minus the larger. You can then solve your system flawlessly and reach a pair that satisfies both of your equations perfectly while contradicting the sentence you started from. The only way to catch that is to read the answer back against the original words: does the larger number really exceed the smaller by $5$ ? Does the money add up to the stated total? Checking against the words tests the translation; checking against the equations tests only the arithmetic. A finished problem survives both.

Number and relationship problems

The plainest systems give two facts about two ordinary numbers, usually a sum or a difference together with a relationship such as “twice” or “more than.” Name the two numbers, write one fact per equation, and solve.

Worked example 1 Two numbers with a sum and a multiple relationship

The sum of two numbers is $27$ , and the larger is $3$ more than twice the smaller. Find the two numbers.

Name the two unknowns. Let $x$ be the smaller number and $y$ the larger number. The problem states two facts, so expect two equations.

The first fact is the sum:

x + y = 27.

The second fact is the relationship, “the larger is $3$ more than twice the smaller”:

y = 2x + 3.

The second equation is already solved for $y$ , so substitution is the easy route. Replace $y$ in the first equation with $2x + 3$ :

x + (2x + 3) = 27.

Solve the one-variable equation that remains:

3x + 3 = 27, \qquad 3x = 24, \qquad x = 8.

Back-substitute to recover $y$ :

y = 2(8) + 3 = 19.

Interpret and check against the words. The smaller number is $8$ and the larger is $19$ . Their sum is $8 + 19 = 27$ , and $19$ really is $3$ more than twice $8$ , since $2 \times 8 + 3 = 19$ . The two numbers are $8$ and $19$ .

Check your understanding

Two numbers have a sum of $30$ and a difference of $8$ . If $x$ is the larger and $y$ is the smaller, which system matches the words?

x + y = 30

and

x - y = 8

x + y = 30

and

y - x = 8

x + y = 8

and

x - y = 30

x - y = 30

and

x + y = 8

Money and value problems

Coins, bills, tickets, and stamps all share a shape. There are two facts: a count (how many items) and a total value (what they are worth together). The count equation just adds the numbers of items; the value equation multiplies each count by its unit price before adding. Work in a single unit, all cents or all dollars, so nothing is mixed.

Worked example 2 A ticket problem: counts and total value

A theater sold $200$ tickets one night and collected $2440$ dollars. Adult tickets cost $15$ dollars and child tickets cost $8$ dollars. How many of each were sold?

Name the two unknowns. Let $a$ be the number of adult tickets and $c$ the number of child tickets.

The first fact counts the tickets:

a + c = 200.

The second fact is the money, each ticket contributing its price:

15a + 8c = 2440.

Both equations are in standard form, so eliminate a variable. Multiply the count equation by $8$ so the $c$ terms match:

8a + 8c = 1600.

Subtract this from the value equation to cancel $c$ :

(15a + 8c) - (8a + 8c) = 2440 - 1600, \qquad 7a = 840, \qquad a = 120.

Back-substitute into the count equation:

120 + c = 200, \qquad c = 80.

Interpret and check. The theater sold $120$ adult tickets and $80$ child tickets. The counts add to $120 + 80 = 200$ , and the money comes to $15(120) + 8(80) = 1800 + 640 = 2440$ dollars, matching both facts. So $120$ adult tickets and $80$ child tickets.

Check your understanding

A piggy bank holds $n$ nickels and $d$ dimes, $32$ coins in all, worth $250$ cents. A nickel is $5$ cents and a dime is $10$ cents. Which system models this?

n + d = 250

and

5n + 10d = 32

5n + 10d = 32

and

n + d = 250

n + d = 32

and

5n + 10d = 250

n + d = 32

and

10n + 5d = 250

Mixture and blend problems

A mixture problem combines two ingredients of different strengths or prices into one batch. Again there are two facts: the total amount and the total content (of acid, salt, or value). A small table keeps the bookkeeping straight, with a row for each ingredient and a row for the mixture.

Solution	Liters	Acid fraction	Liters of acid
$20\%$ acid	$x$	$0.20$	$0.20x$
$50\%$ acid	$y$	$0.50$	$0.50y$
Mixture	$12$	$0.30$	$3.6$

The two columns that must balance give the system. The Liters column says $x + y = 12$ . The Liters of acid column says $0.20x + 0.50y = 3.6$ , because the finished $12$ liters is $30\%$ acid and $0.30 \times 12 = 3.6$ .

Worked example 3 A mixture problem: blending two acid solutions

A chemist blends a $20\%$ acid solution with a $50\%$ acid solution to make $12$ liters of a $30\%$ acid solution. How many liters of each go into the blend?

Name the two unknowns. Let $x$ be the liters of $20\%$ solution and $y$ the liters of $50\%$ solution.

The first fact is the total volume:

x + y = 12.

The second fact is the acid content. The $20\%$ solution contributes $0.20x$ liters of acid, the $50\%$ solution contributes $0.50y$ , and the finished $12$ liters holds $0.30 \times 12 = 3.6$ liters of acid:

0.20x + 0.50y = 3.6.

Solve the volume equation for $x$ and substitute. From $x = 12 - y$ :

0.20(12 - y) + 0.50y = 3.6.

Distribute and solve:

2.4 - 0.20y + 0.50y = 3.6, \qquad 0.30y = 1.2, \qquad y = 4.

Then $x = 12 - 4 = 8$ .

Interpret and check. Use $8$ liters of the $20\%$ solution and $4$ liters of the $50\%$ solution. The volumes add to $8 + 4 = 12$ liters, and the acid comes to $0.20(8) + 0.50(4) = 1.6 + 2.0 = 3.6$ liters, which is $30\%$ of $12$ . So $8$ liters of $20\%$ and $4$ liters of $50\%$ .

Distance problems: rate times time

Distance problems rest on one relationship, distance equals rate times time, written $d = rt$ . Two unknowns appear when a moving thing has both its own speed and the help or hindrance of a current or wind. Going downstream a current adds to the boat’s speed; going upstream it subtracts. That single difference turns one trip into two equations.

Worked example 4 A current problem: boat speed and stream speed

A motorboat travels $36$ miles downstream in $2$ hours, then returns the same $36$ miles upstream in $3$ hours. Find the speed of the boat in still water and the speed of the current.

Name the two unknowns. Let $b$ be the boat’s speed in still water and $c$ the speed of the current, both in miles per hour.

Going downstream the current helps, so the speed over the ground is $b + c$ ; going upstream the current fights the boat, so the speed is $b - c$ . Each leg gives a distance-equals-rate-times-time equation. Downstream, $36$ miles in $2$ hours:

2(b + c) = 36, \qquad b + c = 18.

Upstream, $36$ miles in $3$ hours:

3(b - c) = 36, \qquad b - c = 12.

The $c$ terms are already opposites, so add the two equations to eliminate $c$ :

(b + c) + (b - c) = 18 + 12, \qquad 2b = 30, \qquad b = 15.

Back-substitute into $b + c = 18$ :

15 + c = 18, \qquad c = 3.

Interpret and check. The boat makes $15$ mph in still water and the current runs at $3$ mph. Downstream the speed is $15 + 3 = 18$ mph, covering $18 \times 2 = 36$ miles; upstream it is $15 - 3 = 12$ mph, covering $12 \times 3 = 36$ miles. Both legs check, so the boat is $15$ mph and the current is $3$ mph.

Check your understanding

Solving a boat-and-current problem you reach $b + c = 20$ and $b - c = 14$ , where $b$ is the boat's still-water speed and $c$ is the current. What is the current's speed?

3

mph B

6

mph C

17

mph D

34

mph

Break-even: where cost meets revenue

A business has a cost that grows with how many items it makes and a revenue that grows with how many it sells. Break-even is the number of items where the two are equal, so the company neither loses nor gains. Because cost and revenue are two lines, break-even is exactly the point where they cross, the picture that opened this chapter.

Worked example 5 A break-even problem: cost meets revenue

A small studio makes phone cases. It spends $1200$ dollars in fixed costs plus $4$ dollars in materials per case, and sells each case for $10$ dollars. How many cases must it sell to break even, and what are the cost and revenue there?

Name the two quantities. Let $x$ be the number of cases and $y$ the number of dollars.

The cost is the fixed amount plus $4$ dollars per case:

y = 4x + 1200.

The revenue is $10$ dollars per case:

y = 10x.

Break-even is the point that lies on both lines, where cost and revenue are the same $y$ . Both equations are solved for $y$ , so set the two expressions equal:

10x = 4x + 1200.

Solve:

6x = 1200, \qquad x = 200.

Then $y = 10(200) = 2000$ .

Interpret and check. The studio breaks even at $200$ cases, where cost and revenue are both $2000$ dollars. The cost is $4(200) + 1200 = 2000$ dollars and the revenue is $10(200) = 2000$ dollars, equal as required. Selling fewer than $200$ cases loses money, and selling more turns a profit.

Break-even as two lines meeting at one point. The revenue line y = 10x and the cost line y = 4x + 1200 cross at (200, 2000): selling 200 cases makes revenue equal cost, so that crossing point is the break-even the system solves for.

The break-even point is nothing more exotic than the crossing point of two lines, the same picture that opened this chapter. Every system you solve, whether it wears a story or not, is a search for where two lines meet.

Common mistakes

Writing only one equation for two unknowns. Two unknowns need two conditions. A single equation leaves an entire line of pairs still fitting, so read the problem again and find the second fact before you start solving.
Mislabeling which quantity is which. Decide before you translate whether $x$ is the larger or the smaller, the adults or the children, and hold it fixed. Swapping midway flips a difference or a total and quietly wrecks the answer.
Blurring the count and the value. In a coin or ticket problem, the count equation and the value equation are different facts. Do not add coins to cents. Multiply each count by its unit price, and keep every term in one unit.
Mixing amount with content in a mixture. The volumes add to the batch volume; the acid, salt, or value adds to the batch content. Keep the two columns of the table separate, or the two equations collapse into one.
Checking only against your own equations. Substituting the pair into your equations confirms the arithmetic, not the translation. Read the answer back into the sentences, and reject a count of people or coins that comes out negative or fractional.

Key takeaways

Model a two-unknown situation with a system: name each unknown with its own letter and units, and turn each stated fact into one equation. Solve by substitution or elimination, then interpret and check against the words.
Match the counts: two unknowns need two independent equations. One equation alone leaves a whole line of possible pairs; a second, genuinely new condition is the line that crosses it at the single answer.
Value problems: one equation counts the items, one totals their worth (count times unit price); work in a single unit.
Mixture problems: one equation adds the amounts, one adds the content; a table of amount, strength, and content keeps them apart.
Distance problems: apply $d = rt$ to each leg, and let a current or wind make the rate $b + c$ one way and $b - c$ the other. Break-even: set cost equal to revenue, the point where the two lines cross.
Always check against the original words, not just your equations, and report the quantity the problem actually asked for, with units.

Word Problems with Systems

Learning goals

Two unknowns, two conditions

Why two conditions are exactly enough

Number and relationship problems

Money and value problems

Mixture and blend problems

Distance problems: rate times time

Break-even: where cost meets revenue

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)