Mathematical-Reasoning · Unit 18 · Video 1 · Interactive Practice

Ordered Pairs as Solutions: The Truth Set of 2x + 3y = 12

IKey Formulas

Formula	Name	What it says
$\{x \in U \mid P(x)\}$	Truth set, one variable	The elements of $U$ that make $P(x)$ true
$(a, b)$ solves $2x + 3y = 12$ $\iff$ $2a + 3b = 12$	A solution is an ordered pair	First number for $x$ , second number for $y$
$y = 4 - \tfrac{2}{3}x$	Solving for $y$	One $y$ for each real $x$ : infinitely many solutions
$S = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid 2x + 3y = 12\}$	Set builder notation	The pair $(x, y)$ goes before the bar

Key Insight: $\{3, 2\} = \{2, 3\}$ , but $(3, 2) \neq (2, 3)$ : $(3, 2) \in S$ while $(2, 3) \notin S$ . The line is a picture of $S$ , since a point $(a, b)$ lies on it exactly when $2a + 3b = 12$ .

IIIs the Pair a Solution?

A pair belongs to $S$ exactly when its point lies on the line; swapping the pair's numbers usually moves it off.

Pair (a, b)

IIIPick $x$ , Solve for $y$

Each real $x$ gives exactly one $y = 4 - \tfrac{2}{3}x$ , so the solution set $S$ is infinite.

x 0

x	y	2x + 3y

IVQuiz Questions

Problem 1 · Test a Pair

Given: the open sentence $2x + 3y = 12$ with $x, y \in \mathbb{R}$ — which pair is in its truth set?

✅ Correct!

2(-3) + 3(6) = -6 + 18 = 12

, so the equation is true for

(-3, 6)

❌ Not quite. With

x = 2

and

y = 3

2 \cdot 2 + 3 \cdot 3 = 4 + 9 = 13 \neq 12

. It is

(3, 2)

, the pair in the other order, that is a solution.

❌ Not quite.

2 \cdot 4 + 3 \cdot 1 = 8 + 3 = 11 \neq 12

, so the equation is false for

(4, 1)

❌ Not quite.

2 \cdot 0 + 3 \cdot 6 = 18 \neq 12

. Setting

x = 0

gives

3y = 12

, so

y = 4

, not

6

Show solution

Substitute the first number for $x$ and the second for $y$ in each pair:

(2, 3):\ 4 + 9 = 13 \qquad (4, 1):\ 8 + 3 = 11

(-3, 6):\ -6 + 18 = 12\ \checkmark \qquad (0, 6):\ 0 + 18 = 18

Only $(-3, 6)$ makes the sentence a true statement, so only $(-3, 6)$ is in the truth set.

Problem 2 · Sets Versus Ordered Pairs

Given: $S$ is the truth set of $2x + 3y = 12$ , and $(3, 2) \in S$ — which statement is true?

✅ Correct! A set ignores order, so

\{2, 3\} = \{3, 2\}

; an ordered pair does not, and

2 \cdot 2 + 3 \cdot 3 = 13 \neq 12

❌ Not quite. The set equation is true, but membership in

S

is decided by the ordered pair:

x = 2

y = 3

gives

4 + 9 = 13 \neq 12

❌ Not quite.

(2, 3) \notin S

is right, but

\{2, 3\}

and

\{3, 2\}

have exactly the same elements, so they are the same set.

❌ Not quite.

(2, 3)

and

(3, 2)

have different first numbers, so they are different pairs, and

(2, 3)

gives

13 \neq 12

Show solution

Two sets are equal when they have the same elements, and $\{2, 3\}$ and $\{3, 2\}$ both contain exactly $2$ and $3$ . So $\{2, 3\} = \{3, 2\}$ .

An ordered pair records which number comes first. In $(2, 3)$ the $2$ is substituted for $x$ and the $3$ for $y$ :

2 \cdot 2 + 3 \cdot 3 = 4 + 9 = 13 \neq 12

So $(2, 3) \notin S$ , even though $(3, 2) \in S$ because $2 \cdot 3 + 3 \cdot 2 = 6 + 6 = 12$ . Statement (b) is the true one.

Problem 3 · Complete the Pair

Given: $S = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid 2x + 3y = 12\}$ — find the missing coordinate in each pair so that the pair is in $S$ .

$(a, -4) \in S$ . What is $a$ ?

$\left(\tfrac{5}{2}, b\right) \in S$ . What is $b$ ?

✅ Correct!

(12, -4)

and

\left(\tfrac{5}{2}, \tfrac{7}{3}\right)

both satisfy

2x + 3y = 12

❌ Check the sign.

3 \cdot (-4) = -12

, not

+12

: the equation is

2a - 12 = 12

❌ Check the order. In

(a, -4)

the first number

a

x

and

-4

y

; you substituted

-4

for

x

❌ One step short.

2a = 24

is right, but

a

itself is

24 / 2

❌ Check the sign.

2 \cdot \tfrac{5}{2} = 5

is moved to the other side by subtracting:

3b = 12 - 5

❌ Check the order.

\tfrac{5}{2}

is the first number, so it is substituted for

x

, not

y

❌ One step short.

3b = 7

is right, but

b

itself is

7 / 3

Show solution

First pair. Substitute $x = a$ and $y = -4$ :

2a + 3(-4) = 12 \;\Longrightarrow\; 2a - 12 = 12 \;\Longrightarrow\; 2a = 24 \;\Longrightarrow\; a = 12

Check: $2 \cdot 12 + 3(-4) = 24 - 12 = 12$ ✓.

Second pair. Substitute $x = \tfrac{5}{2}$ and $y = b$ :

2 \cdot \tfrac{5}{2} + 3b = 12 \;\Longrightarrow\; 5 + 3b = 12 \;\Longrightarrow\; 3b = 7 \;\Longrightarrow\; b = \tfrac{7}{3}

Check: $5 + 3 \cdot \tfrac{7}{3} = 5 + 7 = 12$ ✓.

Problem 4 · A Pair and Its Swap

Given: $S$ is the truth set of $2x + 3y = 12$ over $\mathbb{R}$ — find all real pairs $(a, b)$ with both $(a, b) \in S$ and $(b, a) \in S$ .

✅ Correct! The two conditions force

a = b

, and then

5a = 12

❌ Not quite. Try

a = b = \tfrac{12}{5}

2 \cdot \tfrac{12}{5} + 3 \cdot \tfrac{12}{5} = \tfrac{60}{5} = 12

, and its swap is the same pair.

❌ Not quite. A pair

(a, a)

is its own swap, but it still has to be in

S

(1, 1)

gives

2 + 3 = 5 \neq 12

❌ Not quite.

(3, 2) \in S

, but its swap

(2, 3)

gives

4 + 9 = 13 \neq 12

, so the pair

(3, 2)

fails the second condition.

Show solution

$(a, b) \in S$ and $(b, a) \in S$ mean

2a + 3b = 12 \qquad\text{and}\qquad 2b + 3a = 12.

Subtract the second equation from the first: $(2a + 3b) - (3a + 2b) = 0$ , so $b - a = 0$ and $a = b$ .

Substituting $b = a$ into the first equation gives $5a = 12$ , so $a = b = \tfrac{12}{5}$ .

Check: $2 \cdot \tfrac{12}{5} + 3 \cdot \tfrac{12}{5} = \tfrac{24}{5} + \tfrac{36}{5} = 12$ ✓. On the graph, $\left(\tfrac{12}{5}, \tfrac{12}{5}\right)$ is where the line $S$ crosses the line $y = x$ , the line across which a pair and its swap are mirror images.

Solved: 0 / 4

Ordered Pairs as Solutions: The Truth Set of 2x + 3y = 12

IKey Formulas

IIIs the Pair a Solution?

IIIPick xxx, Solve for yyy

IVQuiz Questions

IIIPick $x$ , Solve for $y$