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Systems of two equations
Tickets at two prices, two numbers with a sum and a difference, a jar of coins: two things you don’t know, and two clues. One clue alone is not enough. Learn two ways to use both clues together, how to check your answer, and the shortcut of testing the choices.
Two unknowns need two clues
A sandwich and a drink cost $9 together. The sandwich costs $5 more than the drink. How much is each one?
There are two things you don’t know, so there are two unknowns. Give each one a letter: s for the sandwich and d for the drink. Each clue becomes one equation: s + d = 9 and s − d = 5. Two equations that share the same two unknowns are called a system of equations.
| s | d | s − d | Second clue? |
|---|---|---|---|
| 5 | 4 | 1 | No |
| 6 | 3 | 3 | No |
| 7 | 2 | 5 | Yes |
| 8 | 1 | 7 | No |
The answer is a pair
The first clue alone has many answers: every row of the table adds up to 9. The second clue keeps only one row. The solution of a system is the pair of numbers that makes both equations true at the same time: here s = 7 and d = 2, a $7 sandwich and a $2 drink.
How to solve them
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1
Turn the words into two equations
Give each unknown a letter. Then read the problem one clue at a time: each clue becomes one equation. These words tell you which sign to write.
sum, total, together+x + y = 20difference, how much more−x − y = 6x is 6 more than y− or +x − y = 6, same as x = y + 6each, per×4 tickets at t dollars each: 4tNext step: SubstitutionExample45 tickets were sold. Adult tickets cost $10 and kids’ tickets cost $6. They brought in $370.
- Letters: a is the number of adult tickets, k is the number of kids’ tickets.
- Clue 1, how many tickets: a + k = 45.
- Clue 2, how much money: each adult ticket brings 10 dollars, so the adult tickets bring 10a. The kids’ tickets bring 6k. Together: 10a + 6k = 370.
Answer a + k = 45 and 10a + 6k = 370
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2
Substitution: put one equation inside the other
Use it when one letter is already alone, like y = x + 3. Wherever you see y in the other equation, write (x + 3) instead. Now there is only one letter, and you solve it like any equation.
When you have x, go back to the equation where y was alone to find y.
Next step: Elimination- x + y = 17, with y = x + 3
- x + (x + 3) = 17
- 2x + 3 = 17
- 2x = 14
- x = 7
- y = 7 + 3 = 10
The pair is x = 7 and y = 10. -
3
Elimination: add the equations so one letter cancels
Write one equation under the other and add them: left side with left side, right side with right side. If a letter has the same number in front of it, with + in one equation and − in the other, it cancels and disappears.
If the same letter has the same number and the same sign in both, subtract the equations instead. If the numbers are different, see the next step: multiply first.
Next step: Multiply first- x + y = 15 and x − y = 7
- (x + y) + (x − y) = 15 + 7
- 2x = 22
- x = 11
- y = 15 − 11 = 4
The +y and the −y cancel. ShortcutSum and difference
When a problem gives the sum and the difference of two numbers, the bigger number is (sum + difference) ÷ 2 and the smaller is (sum − difference) ÷ 2. Here: (15 + 7) ÷ 2 is 11 and (15 − 7) ÷ 2 is 4.
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4
When nothing cancels, multiply one equation first
In 2x + 3y = 19 and x + y = 8, adding or subtracting removes nothing. Multiply every term of the second equation by 2. Now both equations have 2x, and subtracting cancels it.
Multiply every term, the number on the right side too. Forgetting it is the most common slip here.
Next step: Check both- 2x + 3y = 19 and x + y = 8
- 2x + 3y = 19 and 2x + 2y = 16
- (2x + 3y) − (2x + 2y) = 19 − 16
- y = 3
- x + 3 = 8, so x = 5
The 2x cancels. Then put y = 3 into the easier equation. -
5
Check the pair in both equations
Put both numbers into both original equations. A pair that works in only one equation is the most common wrong choice on the test: it was put there to fool you.
Next step: Test choicesExampleIs (5, 3) right for 2x + 3y = 19 and x + y = 8?
- First equation: 2 × 5 + 3 × 3 = 10 + 9 = 19. It works.
- Second equation: 5 + 3 = 8. It works.
- Compare (6, 2): 6 + 2 = 8 works, but 2 × 6 + 3 × 2 is 18, not 19. One equation is not enough.
Answer (5, 3) works in both
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6
Shortcut: test the answer choices
The answer is already in front of you. When solving looks long, take a choice, find the other letter from the easier equation, and check it in the harder one.
Number choices are usually listed from small to big. If they are, start with a middle one: if it comes out too small, try a bigger choice next. If the choices are pairs, check each pair in both equations.
Next step: Three examplesExampleIf 3x + y = 26 and x − y = 2, what is x? (A) 5 (B) 6 (C) 7 (D) 8
- Try (B) 6: x − y = 2 gives y = 4. Then 3 × 6 + 4 is 22, not 26. Too small.
- Try (C) 7: x − y = 2 gives y = 5. Then 3 × 7 + 5 is 26. It works.
Answer x = 7, choice (C)
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7
Three worked examples, from easy to test level
Each one starts with the first try many people make, crossed out, and then the way that works. Read the wrong try too: it is the kind of answer the test puts in the choices.
The test-level one is about coins. Before you write anything, put every amount in the same unit: cents.
Example · EasyTwo numbers add up to 31. Their difference is 9. What are the two numbers?
First try
31 − 9 = 22, so the numbers are 22 and 9- Check the first try: 22 + 9 = 31, but 22 − 9 = 13, not 9. It fits only one clue.
- Write both clues: x + y = 31 and x − y = 9.
- Add the two equations: the y terms cancel, and 2x = 40.
- x = 40 ÷ 2 = 20. Then y = 31 − 20 = 11.
- Check both: 20 + 11 = 31 and 20 − 11 = 9. The numbers are 20 and 11.
Answer 20 and 11
Example · MediumA school play sold 120 tickets. Adult tickets cost $8 and student tickets cost $5. The play took in $750. How many adult tickets were sold?
First try
120 ÷ 2 = 60 adult tickets- Check the first try: 60 adult and 60 student tickets bring 480 + 300 = 780 dollars, not 750. Half and half does not fit the money.
- Letters: a adult tickets, s student tickets. Count: a + s = 120. Money: 8a + 5s = 750.
- Multiply every term of the count by 5: 5a + 5s = 600. Now both equations have 5s.
- Subtract: (8a + 5s) − (5a + 5s) is 3a, and 750 − 600 = 150. So 3a = 150.
- a = 150 ÷ 3 = 50 adult tickets, and s = 120 − 50 = 70. Check: 400 + 350 = 750 dollars.
Answer 50 adult tickets
Example · Test levelA jar holds 30 coins, all dimes and quarters. Together they are worth $5.10. How many quarters are in the jar?
First try
10d + 25q = 5.10- The first try mixes units: a dime is 10 cents and a quarter is 25 cents, but $5.10 is in dollars. Write the total in cents too: $5.10 is 510 cents.
- Count: d + q = 30. Value: 10d + 25q = 510.
- Multiply every term of the count by 10: 10d + 10q = 300.
- Subtract: 25q − 10q is 15q, and 510 − 300 = 210. So 15q = 210.
- Without a calculator: 15 × 14 = 210, so q = 14 quarters. Then d = 30 − 14 = 16 dimes.
- Check: 16 dimes are 160 cents and 14 quarters are 350 cents, and 160 + 350 = 510.
Answer 14 quarters
4 traps in the answer choices
The test writes its wrong answers on purpose: each one is what you get if you make one of these mistakes.
Checking only one equation
x + y = 11 and 3x − y = 9. Someone tries the pair (4, 7):
Wrong4 + 7 = 11, so (4, 7) is the answer
Right(5, 6): 5 + 6 = 11 and 15 − 6 = 9
Subtracting only part of an equation
3x + 2y = 16 minus x + 2y = 8. The numbers on the right are subtracted too:
Wrong2x = 16 + 8 = 24, so x = 12
Right2x = 16 − 8 = 8, so x = 4
Answering the wrong letter
x + y = 14 and x − y = 2. The question asks for y:
Wrong2x = 16, so the answer is 8
Rightx = 8, then y = 14 − 8 = 6
Mixing dollars and cents
12 coins, nickels and dimes, worth $0.85. The value equation:
Wrong5n + 10d = 0.85
Right5n + 10d = 85 (in cents)
15 questions in three levels
Warm-up first, then core, then test level. Pick an answer, then read why.
On the computer test, Mathematics Knowledge has 15 questions in 31 minutes: about 2 min 4 s each. On paper it is 25 questions in 24 minutes, a little under 1 minute each. A system can come as two bare equations in Mathematics Knowledge, or as a word problem about tickets or coins in Arithmetic Reasoning (15 questions in 55 minutes on the computer test).
No calculator. Pick the method that fits: substitution when a letter is already alone, elimination when a letter has the same number in both equations. Then read the question again: it may ask for x, for y, or for x + y.
A tip: if solving looks long, test the answer choices in both equations. On the computer test you cannot go back to a question, so check your pair before you choose.
If y = 4 and x − y = 19, what is x?
Show the answer
D. Put 4 in place of y: x − 4 = 19. Undo the subtraction by adding 4 to both sides: x = 19 + 4 = 23. Check: 23 − 4 = 19. 15 subtracts the 4 instead of adding it, 4.75 divides, and 19 forgets y. x is 23.
Which pair (x, y) makes both x + y = 12 and x − y = 6 true?
Show the answer
D. Test each pair in both equations. (9, 3): 9 + 3 = 12 and 9 − 3 = 6, so both work. (3, 9) has the numbers in the wrong order: 3 − 9 is −6. (6, 6) and (8, 4) add up to 12, but their differences are 0 and 4, not 6. The pair is (9, 3).
If y = 2x and x + y = 21, what is x?
Show the answer
A. y is already alone, so put 2x in place of y: x + 2x = 21, which is 3x = 21. Divide by 3: x = 21 ÷ 3 = 7. Check: y = 14, and 7 + 14 = 21. 14 is y, not x; 10.5 treats y as if it were x; and 21 stops at 3x. x is 7.
Add the equations x + y = 17 and x − y = 5. What equation do you get?
Show the answer
B. Add left side to left side: (x + y) + (x − y). The +y and the −y cancel, and x + x is 2x. Add right side to right side: 17 + 5 = 22. So 2x = 22. 2x = 12 adds the left sides but subtracts the right sides (17 − 5). 2y = 12 is what you get when you subtract the equations instead. 2x + 2y = 22 forgets that +y and −y cancel. The sum is 2x = 22.
Is (2, 5) a solution of 3x + y = 11 and x + 2y = 13?
Show the answer
C. Put x = 2 and y = 5 into each equation. First: 3 × 2 is 6, and 6 + 5 = 11. It works. Second: 2 × 5 is 10, and 2 + 10 = 12, not 13. It fails. A pair must work in both, so (2, 5) is not a solution. The answer is No: it works only in the first.
If y = x − 3 and 2x + y = 30, what is x?
Show the answer
D. y is alone, so put (x − 3) in place of y: 2x + x − 3 = 30, which is 3x − 3 = 30. Add 3 to both sides: 3x = 33, and 33 ÷ 3 = 11. Check: y = 11 − 3 = 8, and 2 × 11 + 8 is 30. 8 is y, 9 subtracts the 3 instead of adding it (3x = 27), and 10 forgets the 3. x is 11.
If 3x + 2y = 18 and x + y = 7, what is x?
Show the answer
B. Nothing cancels yet. Multiply every term of x + y = 7 by 2: 2x + 2y = 14. Now subtract it from 3x + 2y = 18: the 2y cancels, and x = 18 − 14 = 4. Then y = 7 − 4 = 3. 3 is y, 6 drops the 2y (3x = 18), and 11 subtracts the equations before multiplying. x is 4.
A movie theater sold 90 tickets. Adult tickets cost $12 and child tickets cost $7. Ticket sales were $880. How many child tickets were sold?
Show the answer
A. Letters: a adult, c child. Count: a + c = 90. Money: 12a + 7c = 880. Multiply every term of the count by 7: 7a + 7c = 630. Subtract: 5a = 880 − 630 = 250, so a = 250 ÷ 5 = 50 adults, and c = 90 − 50 = 40. Check: 12 × 50 is 600 and 7 × 40 is 280, and 600 + 280 = 880. 50 is the adult tickets, 45 is half of 90, and 250 forgets to divide by 5. Child tickets: 40.
The sum of two numbers is 56 and their difference is 18. What is the smaller number?
Show the answer
B. Call them x and y: x + y = 56 and x − y = 18. Subtract the equations: 2y = 56 − 18 = 38, so y = 19. The bigger number is 56 − 19 = 37. Check: 37 − 19 = 18. 9 halves the difference, 28 halves the sum, and 37 is the bigger number. The smaller number is 19.
Eli has 20 coins, all nickels and dimes. They are worth $1.40. How many nickels does he have?
Show the answer
C. Work in cents: $1.40 is 140 cents. Count: n + d = 20. Value: 5n + 10d = 140. Multiply every term of the count by 5: 5n + 5d = 100. Subtract: 5d = 40, so d = 8 dimes, and n = 20 − 8 = 12 nickels. Check: 12 nickels are 60 cents and 8 dimes are 80 cents, and 60 + 80 = 140. 8 is the dimes, 10 is half of 20, and 14 treats every coin as a dime. Nickels: 12.
If 4x − y = 13 and 2x + 3y = 31, what is the value of x + y?
Show the answer
D. Multiply every term of 4x − y = 13 by 3: 12x − 3y = 39. Add it to 2x + 3y = 31: the y terms cancel, and 14x = 70, so x = 5. Then y = 4 × 5 − 13 = 7. Check the second equation: 2 × 5 is 10 and 3 × 7 is 21, and 10 + 21 = 31. The question asks for x + y: 5 + 7 = 12. 5 and 7 answer only half the question, and 2 is y − x. x + y is 12.
A unit fundraiser sold 150 plates: chicken plates at $9 and veggie plates at $7. They raised $1,210. How many chicken plates did they sell?
Show the answer
C. Letters: c chicken, v veggie. Count: c + v = 150. Money: 9c + 7v = 1,210. Multiply every term of the count by 7: 7c + 7v = 1,050. Subtract: 2c = 1,210 − 1,050 = 160, so c = 80. Then v = 150 − 80 = 70. Check: 9 × 80 is 720 and 7 × 70 is 490, and 720 + 490 = 1,210. 70 is the veggie plates, 75 is half of 150, and 160 forgets to divide by 2. Chicken plates: 80.
A jar has 42 coins, all dimes and quarters, worth $7.80 in all. How many dimes are in the jar?
Show the answer
A. Work in cents: $7.80 is 780 cents. Count: d + q = 42. Value: 10d + 25q = 780. Multiply every term of the count by 10: 10d + 10q = 420. Subtract: 15q = 780 − 420 = 360. Without a calculator: 15 × 20 is 300 and 15 × 4 is 60, so 15 × 24 = 360 and q = 24 quarters. Then d = 42 − 24 = 18 dimes. Check: 18 dimes are 180 cents, 24 quarters are 600 cents, and 180 + 600 = 780. 24 is the quarters, 21 is half of 42, and 78 treats every coin as a dime. Dimes: 18.
At the PX (the store on base), 2 shirts and 3 hats cost $57. 1 shirt and 2 hats cost $33. Every shirt costs the same and every hat costs the same. How much does one hat cost?
Show the answer
A. Double the second clue: 2 shirts and 4 hats cost $66. Compare it with the first clue, 2 shirts and 3 hats for $57: the shirts are the same, so the extra hat costs 66 − 57 = 9 dollars. Then one shirt is 33 − 2 × 9 = 15 dollars. Check: 2 × 15 is 30 and 3 × 9 is 27, and 30 + 27 = 57. $15 is the shirt, $24 subtracts the clues without doubling, and $11.40 splits $57 as if every item cost the same. One hat costs $9.
If y = 3x − 5 and x + 2y = 39, what is y?
Show the answer
B. Put (3x − 5) in place of y: x + 2(3x − 5) = 39. Multiply both terms in the parentheses by 2: x + 6x − 10 = 39, so 7x = 49 and x = 7. Then y = 3 × 7 − 5 = 16. Faster: test the choices. If y = 16, then 3x − 5 = 16, so x = 7, and 7 + 2 × 16 is 39. It works. 7 is x, 19.5 ignores x, and 28 forgets the 2 in front of y. y is 16.
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