1. กำหนดให้ \( (8 x, 15)=(15+3 x,-5 y) \) จงหาค่าของ \( x-y, x-y= \) \( \qquad \) 2. กำหนดให้ \( A=\{1,3,5\}, B=\{3,6\} \) จงหาค่าของ 2.1 \( \mathrm{A} \times \mathrm{B}= \) \( \qquad \) \[ 2.2 \mathrm{~B} \times \mathrm{A}= \] \( \qquad \) \( 2.3 \mathrm{~B} \times \mathrm{B}= \) \( \qquad \) \( 2.3 \mathrm{~A} \times \mathrm{A}= \) \( \qquad \) 3. กำหนดให้ \( A=\{1,3,5\}, B=\{3,6\} \) จงเขียนความสัมพันธ์ \( r \) แบบแจกแจงสมาชิก \( 3.1 r_{1}=\{(x, y) \in A \times B \mid y=x-2\} \) \[ , \mathrm{r}_{1} \doteq \] \( \qquad \) \( 3.2 r_{2}=\{(x, y) \in A \times A \mid y

Ask by Bates Johnston. in Thailand

Jan 26,2025

Alright, I need to solve these mathematical problems step by step. Let's start with the first one. **Problem 1:** We have the equation \( (8x, 15) = (15 + 3x, -5y) \). I need to find the value of \( x - y \). First, let's break down the equation into two separate equations since they are equal component-wise. 1. \( 8x = 15 + 3x \) 2. \( 15 = -5y \) Starting with the first equation: \( 8x = 15 + 3x \) Subtract \( 3x \) from both sides: \( 8x - 3x = 15 \) \( 5x = 15 \) Divide both sides by 5: \( x = 3 \) Now, let's solve the second equation: \( 15 = -5y \) Divide both sides by -5: \( y = -3 \) Now that I have \( x = 3 \) and \( y = -3 \), I can find \( x - y \): \( x - y = 3 - (-3) = 3 + 3 = 6 \) So, \( x - y = 6 \). **Problem 2:** Given sets \( A = \{1, 3, 5\} \) and \( B = \{3, 6\} \), I need to find the Cartesian products and their intersections. 1. \( A \times B \) 2. \( B \times A \) 3. \( B \times B \) 4. \( A \times A \) Let's compute each one: 1. \( A \times B = \{(1,3), (1,6), (3,3), (3,6), (5,3), (5,6)\} \) 2. \( B \times A = \{(3,1), (3,3), (3,5), (6,1), (6,3), (6,5)\} \) 3. \( B \times B = \{(3,3), (3,6), (6,3), (6,6)\} \) 4. \( A \times A = \{(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)\} \) **Problem 3:** Given sets \( A = \{1, 3, 5\} \) and \( B = \{3, 6\} \), I need to define the relations \( r_1 \) and \( r_2 \) as specified. 1. \( r_1 = \{(x, y) \in A \times B \mid y = x - 2\} \) 2. \( r_2 = \{(x, y) \in A \times A \mid y < x\} \) Let's find each relation: 1. For \( r_1 \): - \( x = 1 \): \( y = 1 - 2 = -1 \) → Not in \( B \) - \( x = 3 \): \( y = 3 - 2 = 1 \) → Not in \( B \) - \( x = 5 \): \( y = 5 - 2 = 3 \) → In \( B \) So, \( r_1 = \{(5, 3)\} \) 2. For \( r_2 \): - \( x = 1 \): \( y < 1 \) → \( y = 1 \) (since \( y \) must be in \( A \)) - \( x = 3 \): \( y < 3 \) → \( y = 1 \) - \( x = 5 \): \( y < 5 \) → \( y = 1 \) or \( y = 3 \) So, \( r_2 = \{(1,1), (3,1), (5,1), (5,3)\} \)