Two Variables, Two Equations

Here is a story one letter cannot carry. Lena scored 53 points on a quiz made of 2-point and 5-point questions, and she does not know the split. With $a$ correct 2-point answers and $b$ correct 5-point answers, her fact reads $2 a + 5 b = 53$ . No value of $a$ can be checked on its own. The claim involves both letters at once.

Problem

Four pairs are proposed for

2 a + 5 b = 53

, each written with

a

first and

b

second.

(24, 1) (19, 3) (9, 7) (11, 6)

Substitute both values from each pair and evaluate the left side. Enter how many of the four pairs make the claim true.

Show a hint

Each pair gets the full check from 4.1. Replace $a$ and $b$ with the pair's two values, evaluate the left side, and compare it with $53$ .
The last pair gives $2 (11) + 5 (6) = 22 + 30$ . Settle whether that is $53$ , then run the other three the same way.

Show the full solution

The first three pairs pass,

48 + 5 = 53

38 + 15 = 53

, and

18 + 35 = 53

, while

(11, 6)

gives

22 + 30 = 52

. So

3

of the four pairs make the claim true. Three different pairs already satisfy the same equation, and nothing in it stops more.

Problem

The pair

x = 12, y = 2

solves

4 x + 9 y = 66

, since

48 + 18 = 66

. Now swap the two values. Substitute

x = 2, y = 12

and enter the value the left side takes.

Show a hint

A pair says which value belongs to which letter, so the swapped version is a different pair making a different substitution.
With $x = 2$ and $y = 12$ , the first term is $4 (2) = 8$ and the second is $9 (12)$ .

Show the full solution

With

x = 2

and

y = 12

, the left side is

4 (2) + 9 (12) = 8 + 108 = 116

. That is not 66, so the swapped pair fails. A pair carries its order, and

(12, 2)

and

(2, 12)

are different candidates.

Checking given pairs is one skill. Producing your own is the next. Choose a value for one variable and the equation collapses to one variable, which 4.2 solves. In Lena's equation, $a = 14$ gives $28 + 5 b = 53$ , so $5 b = 25$ and $b = 5$ , the pair $(14, 5)$ . The story allows only whole counts, but the equation is its own object and takes any number.

Problem

The equation

x + 2 y = 49

has a solution pair with

x = 17

. Substitute that value, solve the equation that remains, and enter

y

Show a hint

Substituting the given $x$ leaves an equation in $y$ alone, exactly the kind 4.2 solves.
After the substitution the claim reads $17 + 2 y = 49$ . Solve $2 y = 32$ .

Show the full solution

Substituting

x = 17

gives

17 + 2 y = 49

, so

2 y = 32

and

y = 16

. The pair

(17, 16)

checks,

17 + 32 = 49

Problem

The equation

6 x + y = 42

has a solution pair with

x = 10

. Substitute the given value, solve for the partner, and enter

y

Show a hint

Replace $x$ with $10$ and compute $6 x$ first. It is fine if that value comes out larger than $42$ .
Substituting gives $60 + y = 42$ . Since $60$ is already more than $42$ , $y$ must be negative.

Show the full solution

Substituting

x = 10

gives

60 + y = 42

, so

y = 42 - 60 = - 18

. The check passes,

60 + (- 18) = 42

. A solution pair can include a negative number.

Problem

The equation

3 x + 10 y = 58

has a solution pair with

x = 18

. Find the partner value and enter

y

as a fraction in lowest terms.

Show a hint

The collapsed equation is 4.2 work, and nothing promises a whole-number answer.
Substituting gives $54 + 10 y = 58$ . Solve $10 y = 4$ , then reduce.

Show the full solution

Substituting

x = 18

gives

54 + 10 y = 58

, so

10 y = 4

and

y = \frac{4}{10} = \frac{2}{5}

. Nothing restricts pairs to whole numbers. Every choice of

x

, whole or not, gets a partner.

Two secret numbers satisfy $x + y = 101$ . Finding them looks hopeless, since pairs passing that test can be listed all day and every one of them keeps the secret. Suppose a second fact about the same two numbers turns up. It is worth asking what one more fact could possibly change.

Problem

All four of these pairs satisfy

x + y = 101

(98, 3) (93, 8) (79, 22) (70, 31)

Now the second fact arrives. The same two numbers also satisfy

x - y = 39

. Enter how many of the four pairs satisfy both facts.

Show a hint

Each pair already passes the first fact, so only the new claim needs testing.
The first pair gives $98 - 3 = 95$ , not $39$ . Keep going down the list.

Show the full solution

The differences are

98 - 3 = 95

93 - 8 = 85

79 - 22 = 57

, and

70 - 31 = 39

. Only

(70, 31)

gives

39

, so

1

pair satisfies both facts. The first fact allowed an endless list, and the second fact cut it to one pair.

Four pairs pass

x + y = 126

, and the list could go on forever. Tested against

x - y = 40

, their differences are 38, 54, 40, and 34, so three pairs are struck out and

(83, 43)

alone is left. One equation allows many pairs. The second leaves one.

Lena knew one more fact all along. The quiz had 13 questions, so $a + b = 13$ must also hold. Of every pair passing $2 a + 5 b = 53$ , only $(4, 9)$ passes both, since $8 + 45 = 53$ and $4 + 9 = 13$ . She got four 2-point questions and nine 5-point questions. Checking a pair against a system means checking it against every equation in it.

Problem

For the system

2 u + v = 69

and

u + v = 37

, test the pairs

(30, 9)

(32, 5)

, and

(26, 11)

. One satisfies only the first equation, one satisfies only the second, and one satisfies both. Enter

u

for the solution of the system.

Show a hint

A solution of a system passes every equation in it. A pair that fails one check is not a solution, even if it passes every other.
$(30, 9)$ makes the first equation true, since $60 + 9 = 69$ . Check it in the second before deciding anything.

Show the full solution

(30, 9)

gives

60 + 9 = 69

but

30 + 9 = 39

, not

37

(26, 11)

gives

26 + 11 = 37

but

52 + 11 = 63

, not

69

(32, 5)

gives

64 + 5 = 69

and

32 + 5 = 37

, so it passes both and

u = 32

. Each rejected pair passed exactly one equation, and one pass is worth nothing in a system.

Problem

Both

(13, 7)

and

(29, 3)

satisfy

x + 4 y = 41

. Only one of them also satisfies

2 x - y = 19

. Enter

x + y

for the solution of the system.

Show a hint

Passing the first equation is not enough. Run each pair through the second and keep the one that passes.
For $(29, 3)$ , the left side of the second equation is $58 - 3$ . Compare that with $19$ .

Show the full solution

For

(29, 3)

2 x - y = 58 - 3 = 55

, not

19

. For

(13, 7)

2 x - y = 26 - 7 = 19

, and

13 + 28 = 41

confirms the first. The solution is

(13, 7)

, so

x + y = 13 + 7 = 20

How is that single pair found when no candidates are given? For now, by search. Take the friendlier equation, list whole-number pairs that satisfy it, and test each one in the other equation until a pair passes both. This works when the numbers are kind, and it fails the moment they are not. Lessons 5.2 and 5.3 build methods that always work.

Problem

Two numbers satisfy

x + y = 23

and

5 x - y = 25

. List whole-number pairs from the first equation and test them in the second until one passes. Enter

x

Show a hint

Every candidate from the first fact has the form $x$ and $23 - x$ , so each choice of $x$ names a full pair to test.
If a test comes out below $25$ , raise $x$ and try again. The value of $5 x - y$ climbs quickly as $x$ grows, so the search stays short.

Show the full solution

Candidates from the first equation include

(7, 16)

(8, 15)

, and

(9, 14)

. Testing

(8, 15)

in the second,

5 (8) - 15 = 40 - 15 = 25

, so both facts hold and

x = 8

. Raising

x

1

raises

5 x - y

6

, so the second equation is hit exactly once.

Problem

The system

m + n = 67

and

m + n = 73

makes two claims about the same two numbers. Take any pair satisfying the first equation, test it in the second, and enter the number of solutions of the system.

Show a hint

Read the two claims side by side. Both are about the sum of the same two numbers.
A pair with $m + n = 67$ makes the second left side $67$ , and $67$ is not $73$ .

Show the full solution

Any pair passing the first equation has

m + n = 67

, so in the second equation its left side is

67

, never

73

. No pair passes both, and the system has

0

solutions. Two claims about the same numbers can simply contradict each other. Chapter 7 will draw what that looks like.

Problem

Two numbers satisfy

x + y = 6

and

x - y = 14

. Search pairs from the first fact, and do not stop at positive candidates. Enter

y

Show a hint

The difference $14$ is larger than the sum $6$ . Two positive numbers always differ by less than their sum, so one member must be negative.
Pairs summing to 6 include ones with a negative member, like $9$ and $- 3$ .

Show the full solution

Pairs from the first fact run

(7, - 1)

(8, - 2)

(9, - 3)

(10, - 4)

. Testing

(10, - 4)

in the second,

10 - (- 4) = 14

, so both facts hold and

y = - 4

. Positive pairs summing to

6

differ by less than

6

, so the pair that works has to include a negative member.

A system is two claims sharing one pair. Testing confirms a pair, and searching finds one while the numbers stay kind, but they will not stay kind. Lesson 5.2, Substitution, turns two equations in two variables into one equation in one variable, chapter 4's home ground.

Practice these ideas

Practice

Substitute the pair

x = 11, y = 2

into

8 x + 3 y = 94

and evaluate the left side. It matches the right side. Enter the number both sides equal.

Show the solution

The left side is

8 (11) + 3 (2) = 88 + 6 = 94

. One check settles a pair, both values go in together and the two sides are compared once.

Practice

The pair

(7, 6)

is claimed to solve

9 x + 4 y = 91

. Substitute

x = 7, y = 6

and enter the value the left side actually takes.

Show the solution

9 (7) + 4 (6) = 63 + 24 = 87

. That is not

91

, so

(7, 6)

is not a solution. A check either matches the right side or it does not, and this one misses by

4

Practice

The equation

7 x + y = 86

has a solution pair with

x = 11

. Substitute that value and solve for the letter that remains. Enter

y

Show the solution

Substituting

x = 11

gives

77 + y = 86

, so

y = 9

. Check,

7 (11) + 9 = 77 + 9 = 86

, and the pair

(11, 9)

passes.

Practice

The equation

x - 2 y = 11

has a solution pair with

y = 12

. Enter

x

Show the solution

Substituting

y = 12

gives

x - 24 = 11

, so

x = 35

. Check,

35 - 2 (12) = 35 - 24 = 11

. Either member of the pair can be the given one.

Practice

Check the pairs

(6, 0)

(4, 6)

(3, 8)

(5, 2)

against

3 x + y = 18

, substituting

x

first and

y

second. Enter how many of the four are solutions.

Show the solution

The left sides are

18 + 0 = 18

12 + 6 = 18

9 + 8 = 17

15 + 2 = 17

, so

2

of the pairs are solutions. Zero is a legal value for a member of a pair, and

(6, 0)

passes like any other.

Practice

Substitute the pair

(24, 9)

into

x + 9 y = 4 x + y

, evaluating each side separately. Both sides come out to the same number. Enter it.

Show the solution

The left side is

24 + 81 = 105

and the right side is

96 + 9 = 105

, so both sides equal

105

. The pair checks against the whole claim at once, whichever side each letter appears on.

Practice

The equation

8 x + y = 50

has a solution pair with

x = 9

. Enter

y

Show the solution

Substituting

x = 9

gives

72 + y = 50

, so

y = 50 - 72 = - 22

. Check,

8 (9) + (- 22) = 72 - 22 = 50

. The pair has a negative member, and the check passes all the same.

Practice

The equation

8 x + 6 y = 85

has a solution pair with

x = 8

. Enter

y

as a fraction in lowest terms.

Show the solution

Substituting

x = 8

gives

64 + 6 y = 85

, so

6 y = 21

and

y = \frac{21}{6} = \frac{7}{2}

. Check,

64 + 6 \cdot \frac{7}{2} = 64 + 21 = 85

. Nothing in the equation forces

y

to be a whole number.

Practice

The equation

4 y - 3 x = 21

has a solution pair with

x = 9

. Enter

y

Show the solution

Substituting

x = 9

gives

4 y - 27 = 21

, so

4 y = 48

and

y = 12

. Check,

4 (12) - 3 (9) = 48 - 27 = 21

Practice

Every pair solving

x + 5 y = 88

also solves

3 x + 15 y = 264

, since the second equation is the first with both sides multiplied by

3

. The second fact adds nothing new. Enter

y

for the pair with

x = 13

Show the solution

Using the first equation,

13 + 5 y = 88

, so

5 y = 75

and

y = 15

. Check in the second,

3 (13) + 15 (15) = 39 + 225 = 264

. A second equation only pins the pair when it says something the first did not.

Practice

All four of

(35, 11)

(37, 7)

(39, 3)

(28, 25)

satisfy

2 x + y = 81

. Exactly one also satisfies

3 x - y = 104

. Enter

x

for that pair.

Show the solution

Testing the second equation,

105 - 11 = 94

111 - 7 = 104

117 - 3 = 114

84 - 25 = 59

. Only

(37, 7)

passes both, so

x = 37

. A solution of a system must satisfy every equation, and failing even one rules a pair out.

Practice

Two numbers add to

87

and differ by

33

. Enter the larger number.

Show the solution

The facts are

x + y = 87

and

x - y = 33

. Searching pairs that sum to

87

, the pair

(60, 27)

gives

60 - 27 = 33

, so the larger number is

60

. Check,

60 + 27 = 87

Practice

Two numbers satisfy

x + y = 29

and

4 x + y = 71

. Test whole-number pairs from the first equation in the second until one passes, then enter the product

x y

Show the solution

The pair

(14, 15)

sums to

29

, and

4 (14) + 15 = 56 + 15 = 71

passes the second equation. So

x y = 14 \cdot 15 = 210

. No other whole pair works, since raising

x

1

raises

4 x + y

3

, so

71

is hit exactly once.

Practice

For the system

x + y = 102

and

x + y = 89

, any pair satisfying the first equation misses the second by one fixed amount. Enter that amount.

Show the solution

Any pair with

x + y = 102

makes the second equation read

102 = 89

, which misses by

102 - 89 = 13

. The two claims conflict for every pair at once, so this system has no solution.

Practice

The system

x + y = 112

and

x - y = 112

looks like the impossible kind, but the left sides differ, and it has exactly one solution. Enter

x

Show the solution

If adding

y

and subtracting

y

leave the same value, then

y = 0

, and the first equation gives

x = 112

. Check,

112 + 0 = 112

and

112 - 0 = 112

. Zero is a legal member of a pair, so the two claims agree instead of conflicting.

Practice

Two numbers satisfy

m + n = 5

and

m - n = 17

. Search pairs from the first fact, and let the second number go negative. Enter

n

Show the solution

The pair

(11, - 6)

gives

11 + (- 6) = 5

and

11 - (- 6) = 11 + 6 = 17

, so

n = - 6

. A search that stops at positive candidates never finds this pair, since positive pairs summing to

5

differ by at most

5

Practice

For one particular number

k

, the pair

x = 16, y = - 9

is the solution of the system

x + y = 7

and

2 x + 5 y = k

. Enter

k

Show the solution

The first equation checks,

16 + (- 9) = 7

. Substituting the pair into the second,

k = 2 (16) + 5 (- 9) = 32 - 45 = - 13

. Keep the parentheses,

5 (- 9)

- 45

, and the term joins the sum with its sign.