Writing a Pair of Equations from a Word Problem

Name both unknowns, then say each fact once.

Name both unknowns first

A family buys 9 tickets for a show. An adult ticket costs $8 and a child ticket costs $5, and the bill is $57. How many of each did they buy?

There are two numbers we do not know, so give each one a letter before writing anything else. Let a be the number of adult tickets and c the number of child tickets. Then a adult tickets cost 8a dollars, and c child tickets cost 5c dollars.

eachhow manycostadult8a8achild5c5c

The price of one ticket times the number of tickets gives the cost of that kind of ticket.

One equation for each fact

The question gives two facts. The number of tickets is 9, so a + c = 9. The bill is $57, and it is the adult cost plus the child cost, so 8a + 5c = 57.

Two unknowns need two facts. The first fact alone allows many answers: 1 adult and 8 children, 2 adults and 7 children, and so on. Each of those makes a different bill, and only one of them makes $57.

childrenbill2 adults7513 adults6544 adults5575 adults460

Every row is 9 tickets. Each adult in place of a child adds $3 to the bill, and only 4 adults and 5 children cost $57.

Each fact is a straight line of possible answers. The answer to the problem is the point on both lines.

aca + c = 98a + 5c = 57

The line for the number of tickets and the line for the bill meet once, where a = 4 and c = 5.

Solve the pair

Solve the equations as before. Multiply a + c = 9 by 5 so that its c term matches the 5c: 5a + 5c = 45. Subtract this from 8a + 5c = 57: 8a − 5a = 3a and 57 − 45 = 12, so 3a = 12 and a = 4. Then a + c = 9 gives c = 5.

Read the answer back in words

The letters stood for tickets, so the answer is given as tickets: 4 adult tickets and 5 child tickets. A bare 4 does not answer the question, because it does not say what the 4 counts.

Check the answer against the words of the question, not only against the equations: 4 + 5 = 9 tickets, and the bill is 8 × 4 + 5 × 5 = 32 + 25 = 57 dollars.

The usual mistakes

Writing a + c = 57, which adds a number of tickets to a number of dollars. Each equation must count one kind of thing: tickets in one, dollars in the other.

Giving the answer to the wrong letter. The letter c was chosen for child tickets, so c = 5 means 5 child tickets. Writing down what each letter stands for at the start prevents this.

Worked example: Adult and Child Tickets with a Total Count and a Total Value

Question For a school concert an adult ticket costs $12 and a child ticket costs $7. Altogether 150 tickets are sold for a total of $1360. (a) How many tickets of each kind are sold? (b) How much more money comes from the adult tickets than from the child tickets?

  1. 1.Let a be the number of adult tickets and c the number of child tickets. The count gives a + c = 150, and the money gives 12a + 7c = 1360.

    a + c=150(1) the tickets12a + 7c=1360(2) the dollars
    (1) the ticketsa + c=150(2) the dollars12a + 7c=1360
    With a adult tickets and c child tickets, a + c = 150 and 12a + 7c = 1360.

  2. 2.From the first equation, a = 150 − c. Substitute this for a in the second equation: 12(150 − c) + 7c = 1360.

    a + c=150(1) the tickets12a + 7c=1360(2) the dollars12(150 − c) + 7c=1360put a = 150 − c in (2)
    (1) the ticketsa + c=150(2) the dollars12a + 7c=1360put a = 150 − c in (2)12(150 − c) + 7c=1360
    From the first equation a = 150 − c. Substitute it into the second equation.

  3. 3.Expand the bracket: 1800 − 12c + 7c = 1360. Collect the c terms: 1800 − 5c = 1360.

    a + c=150(1) the tickets12a + 7c=1360(2) the dollars12(150 − c) + 7c=1360put a = 150 − c in (2)1800 − 12c + 7c=1360expand the bracket1800 − 5c=1360collect the c terms
    (1) the ticketsa + c=150(2) the dollars12a + 7c=1360put a = 150 − c in (2)12(150 − c) + 7c=1360expand the bracket1800 − 12c + 7c=1360collect the c terms1800 − 5c=1360
    Expand the bracket and collect the c terms: 1800 − 5c = 1360.

  4. 4.Add 5c to both sides: 1800 = 1360 + 5c. Subtract 1360 from both sides: 440 = 5c. Divide both sides by 5: c = 88.

    a + c=150(1) the tickets12a + 7c=1360(2) the dollars12(150 − c) + 7c=1360put a = 150 − c in (2)1800 − 12c + 7c=1360expand the bracket1800 − 5c=1360collect the c terms1800=1360 + 5cadd 5c to both sides440=5csubtract 1360 from both sidesc=88divide both sides by 5
    (1) the ticketsa + c=150(2) the dollars12a + 7c=1360put a = 150 − c in (2)12(150 − c) + 7c=1360expand the bracket1800 − 12c + 7c=1360collect the c terms1800 − 5c=1360add 5c to both sides1800=1360 + 5csubtract 1360 from both sides440=5cdivide both sides by 5c=88
    Add 5c, subtract 1360, then divide by 5, each on both sides: c = 88.

  5. 5.Then a = 150 − 88 = 62. (a) 62 adult tickets and 88 child tickets are sold. Check in the second equation: 12 × 62 + 7 × 88 = 744 + 616 = 1360.

    a = 150 − 88 = 62 adult tickets, and c = 88 child ticketsAdult62 × $12 = $744$744Child88 × $7 = $616$616check in (2): 744 + 616 = 1360
    a = 150 − 88 = 62 adult tickets, and c = 88 childticketsAdult62 × $12 = $744Child88 × $7 = $616check in (2): 744 + 616 = 1360
    (a) a = 150 − 88 = 62: there are 62 adult tickets and 88 child tickets, and 744 + 616 = 1360.

  6. 6.(b) The adult tickets bring in $744 and the child tickets $616, so the adult tickets bring in 744 − 616 = $128 more.

    a = 150 − 88 = 62 adult tickets, and c = 88 child ticketsAdult62 × $12 = $744$744Child88 × $7 = $616$616744 − 616 = $128 morecheck in (2): 744 + 616 = 1360
    a = 150 − 88 = 62 adult tickets, and c = 88 childticketsAdult62 × $12 = $744Child88 × $7 = $616744 − 616 = $128 morecheck in (2): 744 + 616 = 1360
    (b) The adult tickets bring in 744 − 616 = $128 more.

Answer: (a) 62 adult tickets and 88 child tickets; (b) $128

Common mistakes

  • Expanding 12(150 − c) as 1800 − c. The 12 multiplies both terms inside the bracket, so the expansion is 1800 − 12c.
  • Stopping at c = 88 and giving it as the number of adult tickets. The letter c was chosen for the child tickets, and a = 150 − c still has to be worked out.

More equations and inequalities problems, worked step by step →

Worked example: Two Orders at a Cafe with Two Unknown Prices

Question At a cafe every coffee costs the same and every tea costs the same. One table orders 3 coffees and 2 teas and pays $21. Another table orders 2 coffees and 5 teas and pays $25. (a) Find the price of a coffee and the price of a tea. (b) Find the cost of an order of 5 coffees and 4 teas.

  1. 1.Let a coffee cost x dollars and a tea cost y dollars. The first order gives equation (1), 3x + 2y = 21. The second order gives equation (2), 2x + 5y = 25.

    3x + 2y=21(1)2x + 5y=25(2)
    (1)3x + 2y=21(2)2x + 5y=25
    A coffee costs x dollars and a tea costs y dollars. Each order is one equation.

  2. 2.Multiply equation (1) by 2 and equation (2) by 3, so that both have 6x: 6x + 4y = 42 and 6x + 15y = 75.

    3x + 2y=21(1)2x + 5y=25(2)6x + 4y=42(1) × 26x + 15y=75(2) × 3
    (1)3x + 2y=21(2)2x + 5y=25(1) × 26x + 4y=42(2) × 36x + 15y=75
    Multiply equation (1) by 2 and equation (2) by 3, so that both have 6x.

  3. 3.Subtract the first of these from the second. The 6x terms cancel: 15y − 4y = 75 − 42, so 11y = 33 and y = 3.

    3x + 2y=21(1)2x + 5y=25(2)6x + 4y=42(1) × 26x + 15y=75(2) × 311y=33subtract: the 6x terms cancely=3divide both sides by 11
    (1)3x + 2y=21(2)2x + 5y=25(1) × 26x + 4y=42(2) × 36x + 15y=75subtract: the 6x terms cancel11y=33divide both sides by 11y=3
    Subtract one from the other. The 6x terms cancel: 11y = 33, so y = 3.

  4. 4.Substitute y = 3 into equation (1): 3x + 6 = 21, so 3x = 15 and x = 5.

    3x + 2y=21(1)2x + 5y=25(2)6x + 4y=42(1) × 26x + 15y=75(2) × 311y=33subtract: the 6x terms cancely=3divide both sides by 113x + 6=21put y = 3 in (1)3x=15subtract 6 from both sidesx=5divide both sides by 3
    (1)3x + 2y=21(2)2x + 5y=25(1) × 26x + 4y=42(2) × 36x + 15y=75subtract: the 6x terms cancel11y=33divide both sides by 11y=3put y = 3 in (1)3x + 6=21subtract 6 from both sides3x=15divide both sides by 3x=5
    Substitute y = 3 into equation (1): 3x + 6 = 21, so x = 5.

  5. 5.(a) A coffee costs $5 and a tea costs $3. Check in equation (2): 2 × 5 + 5 × 3 = 10 + 15 = 25.

    A coffee costs $5 and a tea costs $3.Table 1$5$5$5$3$3$21Table 2$5$5$3$3$3$3$3$25check in (2): 2 × 5 + 5 × 3 = 10 + 15 = 25
    A coffee costs $5 and a tea costs $3.Table 1$5$5$5$3$3$21Table 2$5$5$3$3$3$3$3$25check in (2): 2 × 5 + 5 × 3 = 10 + 15 = 25
    (a) A coffee costs $5 and a tea costs $3. Equation (2) agrees: 10 + 15 = 25.

  6. 6.(b) 5 coffees and 4 teas cost 5 × 5 + 4 × 3 = 25 + 12 = $37.

    A coffee costs $5 and a tea costs $3.Table 1$5$5$5$3$3$21Table 2$5$5$3$3$3$3$3$25New order$5$5$5$5$5$3$3$3$3$375 × $5 + 4 × $3 = $25 + $12 = $37
    A coffee costs $5 and a tea costs $3.Table 1$5$5$5$3$3$21Table 2$5$5$3$3$3$3$3$25New order$5$5$5$5$5$3$3$3$3$375 × $5 + 4 × $3 = $25 + $12 = $37
    (b) 5 coffees and 4 teas cost 25 + 12 = $37.

Answer: (a) a coffee costs $5 and a tea costs $3; (b) $37

Common mistakes

  • Multiplying only the left-hand side and writing 6x + 4y = 21. Every term on both sides is multiplied, so equation (1) becomes 6x + 4y = 42.
  • Checking the prices in equation (1) only, the equation they were found from. A slip made earlier can still satisfy that equation, so the check uses the other equation.

More equations and inequalities problems, worked step by step →

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