Word Problems with Systems

Learning goals

Name two unknowns with units and turn each independent fact into an equation
Set up value problems by counting items and totaling worth
Organize a mixture's amount, strength and content in a table, then write its two equations
Apply $d = rt$ per leg, with a current changing the rate
Find break-even where cost equals revenue

Number and relationship problems

Here is a situation with two numbers unknown, and two separate facts connecting them. Name the two numbers, write one equation per fact, 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

Two unknowns, two conditions

You just named two unknowns and wrote one equation per fact. That pattern is the general idea: a system of two equations, one equation for each independent fact. Solving it is work you already know, the substitution and elimination from the last two lessons; the new skill was the translation you just 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

Worked Example 1 used two equations, not one. A single equation in two unknowns cannot find them both. The picture below shows why, and the proof below it says why the final check has to be against the words.

One equation, x + y = 6, is true for many pairs, three of them marked: (1, 5), (2, 4), (3, 3). A second, different equation is a second line. The two cross at exactly one point, (2, 4): the only pair making both equations true.

Why two independent conditions determine two unknowns#

One equation, $x + y = 6$ , is true for many pairs, the picture above marks three of them: $(1, 5)$ , $(2, 4)$ , and $(3, 3)$ . A second, different equation is a second line, and two non-parallel lines cross at exactly one point, here $(2, 4)$ : the only pair making both equations true at once.

The word “independent” is doing real work. A fact that only restates the first gives the very same line, so nothing new is pinned down. A fact that contradicts the first, same steepness but a different total, gives a distinct but parallel line that never crosses the first at all.

The final check has to go back to the words, not just your own two equations. Solving a system only tests whether the algebra was done correctly; it says nothing about whether the equations were a faithful translation of the problem. Worked Example 1’s check reread the original sentences, the sum and the relationship, rather than stopping at the equations $x + y = 27$ and $y = 2x + 3$ .

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, one row per ingredient plus a row for the mixture, keeps the bookkeeping straight.

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. A table keeps the amount and the content straight:

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 Liters column gives the first fact, the total volume:

x + y = 12.

The Liters of acid column gives the second, the total acid content. The finished $12$ liters is $30\%$ acid, so $0.30 \times 12 = 3.6$ :

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 $50\%$ .

Check your understanding

A mixer blends a $10\%$ juice concentrate with a $40\%$ juice concentrate to make $9$ liters of a $30\%$ mixture. If $x$ is the liters of the $10\%$ concentrate and $y$ the liters of the $40\%$ concentrate, which system models this?

x + y = 9

and

0.10x + 0.40y = 0.30(9)

x + y = 9

and

0.10x + 0.40y = 9

0.10x + 0.40y = 9

and

x + y = 0.30(9)

x + y = 30

and

0.10x + 0.40y = 9

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.

Check your understanding

A vendor spends $60$ dollars on a stand plus $3$ dollars per item, and sells each item for $8$ dollars. Which equation gives the break-even number of items $x$ ?

3x + 60 = 8x

3x = 8x + 60

3x + 60 = 8

60x + 3 = 8x

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 can reverse a subtraction or attach the wrong price, percentage, or rate to a variable, quietly wrecking 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 every term in one equation in the same unit, liters in one equation and liters of acid in the other. Otherwise the equation stops matching the fact it was meant to state.
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. Give each unknown its own letter and units, translate each independent fact into an equation, and solve by substitution or elimination. Two unknowns need two independent equations. One 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 count the items and total their worth, count times unit price. Mixture problems total the amount and the content separately; a table keeps them apart. Distance problems apply $d = rt$ to each leg, letting a current or wind make the rate $b + c$ one way and $b - c$ the other. Break-even sets 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

Number and relationship problems

Two unknowns, two conditions

Why two conditions are exactly enough

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)

Core practice