Math / Algebra

Question2. Se consideră mulțimile $A=\{x \in \mathbb{R}| | x+2 \mid \leq 3\}$ și $B=\left\{x \in \mathbb{R} \left\lvert\, 1<\frac{3 x+8}{2} \leq 13\right.\right\}$ . (2p) a) Arătați că $\mathrm{A}=[-5,1]$ . (3p) b) Determinati $(A \cap B) \cap \mathbb{Z}$ .

Studdy Solution

STEP 1

What is this asking? We're finding the set A, then the integers that are in both A and B! Watch out! Don't forget to consider the endpoints when working with inequalities, and remember integers are whole numbers!

STEP 2

1. Solve for set A
2. Solve for set B
3. Find the intersection of A and B
4. Find the integers in the intersection

STEP 3

We have $|x+2| \leq 3$ .
This means the stuff inside the absolute value, which is $x+2$ , can be between $-3$ and $3$ , inclusive.
So, we can write it as $-3 \leq x+2 \leq 3$ .

STEP 4

To get $x$ by itself, let's add $-2$ to all parts of the inequality: $-3 + (-2) \leq x+2 + (-2) \leq 3 + (-2)$ .
This simplifies to $-5 \leq x \leq 1$ .

STEP 5

This means $x$ can be any number from $-5$ to $1$ , including $-5$ and $1$ .
We can write this as $A = [-5, 1]$ .
Boom!

STEP 6

We have $1 < \frac{3x+8}{2} \leq 13$ .
Let's tackle this one piece at a time!

STEP 7

First, we have $1 < \frac{3x+8}{2}$ . Multiply both sides by $2$ to get $2 \cdot 1 < 2 \cdot \frac{3x+8}{2}$ , which simplifies to $2 < 3x + 8$ .
Now, add $-8$ to both sides: $2 + (-8) < 3x + 8 + (-8)$ , which gives us $-6 < 3x$ .
Finally, divide both sides by $3$ : $\frac{-6}{3} < \frac{3x}{3}$ , so $-2 < x$ .

STEP 8

Next, we have $\frac{3x+8}{2} \leq 13$ . Multiply both sides by $2$ : $2 \cdot \frac{3x+8}{2} \leq 2 \cdot 13$ , simplifying to $3x+8 \leq 26$ . Add $-8$ to both sides: $3x+8 + (-8) \leq 26 + (-8)$ , which gives $3x \leq 18$ . Divide both sides by $3$ : $\frac{3x}{3} \leq \frac{18}{3}$ , so $x \leq 6$ .

STEP 9

Combining $-2 < x$ and $x \leq 6$ , we get $-2 < x \leq 6$ , which means $B = (-2, 6]$ .

STEP 10

We have $A = [-5, 1]$ and $B = (-2, 6]$ .
The intersection, $A \cap B$ , is where both inequalities are true.
Think of it like overlapping the two intervals on a number line!

STEP 11

The overlap starts at $-2$ (not including $-2$ because it's not in B) and goes up to $1$ (including $1$ because it's in both A and B).
So, $A \cap B = (-2, 1]$ .

STEP 12

We're looking for the integers in $(-2, 1]$ .
Remember, integers are whole numbers.
In this interval, the integers are $-1$ , $0$ , and $1$ .

STEP 13

So, $(A \cap B) \cap \mathbb{Z} = \{-1, 0, 1\}$ .
We found them!

STEP 14

The set of integers that belong to both A and B is $\{-1, 0, 1\}$ .

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Question2. Se consideră mulțimile $A=\{x \in \mathbb{R}| | x+2 \mid \leq 3\}$ și $B=\left\{x \in \mathbb{R} \left\lvert\, 1<\frac{3 x+8}{2} \leq 13\right.\right\}$ . (2p) a) Arătați că $\mathrm{A}=[-5,1]$ . (3p) b) Determinati $(A \cap B) \cap \mathbb{Z}$ .

Studdy Solution

STEP 1

What is this asking? We're finding the set A, then the integers that are in both A and B! Watch out! Don't forget to consider the endpoints when working with inequalities, and remember integers are whole numbers!

STEP 2

1. Solve for set A
2. Solve for set B
3. Find the intersection of A and B
4. Find the integers in the intersection

STEP 3

We have $|x+2| \leq 3$ .
This means the stuff inside the absolute value, which is $x+2$ , can be between $-3$ and $3$ , inclusive.
So, we can write it as $-3 \leq x+2 \leq 3$ .

STEP 4

To get $x$ by itself, let's add $-2$ to all parts of the inequality: $-3 + (-2) \leq x+2 + (-2) \leq 3 + (-2)$ .
This simplifies to $-5 \leq x \leq 1$ .

STEP 5

This means $x$ can be any number from $-5$ to $1$ , including $-5$ and $1$ .
We can write this as $A = [-5, 1]$ .
Boom!

STEP 6

We have $1 < \frac{3x+8}{2} \leq 13$ .
Let's tackle this one piece at a time!

STEP 7

STEP 8

STEP 9

Combining $-2 < x$ and $x \leq 6$ , we get $-2 < x \leq 6$ , which means $B = (-2, 6]$ .

STEP 10

We have $A = [-5, 1]$ and $B = (-2, 6]$ .
The intersection, $A \cap B$ , is where both inequalities are true.
Think of it like overlapping the two intervals on a number line!

STEP 11

The overlap starts at $-2$ (not including $-2$ because it's not in B) and goes up to $1$ (including $1$ because it's in both A and B).
So, $A \cap B = (-2, 1]$ .

STEP 12

We're looking for the integers in $(-2, 1]$ .
Remember, integers are whole numbers.
In this interval, the integers are $-1$ , $0$ , and $1$ .

STEP 13

So, $(A \cap B) \cap \mathbb{Z} = \{-1, 0, 1\}$ .
We found them!

STEP 14

The set of integers that belong to both A and B is $\{-1, 0, 1\}$ .

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Studdy Solution

STEP 1

What is this asking? We're finding the set A, then the integers that are in both A and B! Watch out! Don't forget to consider the endpoints when working with inequalities, and remember integers are whole numbers!

STEP 2

1. Solve for set A2. Solve for set B3. Find the intersection of A and B4. Find the integers in the intersection

STEP 3

We have ∣x+2∣≤3|x+2| \leq 3∣x+2∣≤3.This means the *stuff* inside the absolute value, which is x+2x+2x+2, can be between −3-3−3 and 333, inclusive.So, we can write it as −3≤x+2≤3-3 \leq x+2 \leq 3−3≤x+2≤3.

STEP 4

To get xxx by itself, let's **add −2-2−2** to all parts of the inequality: −3+(−2)≤x+2+(−2)≤3+(−2)-3 + (-2) \leq x+2 + (-2) \leq 3 + (-2)−3+(−2)≤x+2+(−2)≤3+(−2).This simplifies to −5≤x≤1-5 \leq x \leq 1−5≤x≤1.

STEP 5

This means xxx can be any number from −5-5−5 to 111, including −5-5−5 and 111.We can write this as A=[−5,1]A = [-5, 1]A=[−5,1].Boom!

STEP 6

We have 1<3x+82≤131 < \frac{3x+8}{2} \leq 131<23x+8​≤13.Let's tackle this one piece at a time!

STEP 7

STEP 8

STEP 9

Combining −2<x-2 < x−2<x and x≤6x \leq 6x≤6, we get −2<x≤6-2 < x \leq 6−2<x≤6, which means B=(−2,6]B = (-2, 6]B=(−2,6].

STEP 10

We have A=[−5,1]A = [-5, 1]A=[−5,1] and B=(−2,6]B = (-2, 6]B=(−2,6].The intersection, A∩BA \cap BA∩B, is where *both* inequalities are true.Think of it like overlapping the two intervals on a number line!

STEP 11

The overlap starts at −2-2−2 (not including −2-2−2 because it's not in B) and goes up to 111 (including 111 because it's in both A and B).So, A∩B=(−2,1]A \cap B = (-2, 1]A∩B=(−2,1].

STEP 12

We're looking for the *integers* in (−2,1](-2, 1](−2,1].Remember, integers are whole numbers.In this interval, the integers are −1-1−1, 000, and 111.

STEP 13

So, (A∩B)∩Z={−1,0,1}(A \cap B) \cap \mathbb{Z} = \{-1, 0, 1\}(A∩B)∩Z={−1,0,1}.We found them!

STEP 14

The set of integers that belong to both A and B is {−1,0,1}\{-1, 0, 1\}{−1,0,1}.

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Studdy solves anything!

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1. Solve for set A
2. Solve for set B
3. Find the intersection of A and B
4. Find the integers in the intersection

We have $|x+2| \leq 3$ .
This means the stuff inside the absolute value, which is $x+2$ , can be between $-3$ and $3$ , inclusive.
So, we can write it as $-3 \leq x+2 \leq 3$ .

To get $x$ by itself, let's add $-2$ to all parts of the inequality: $-3 + (-2) \leq x+2 + (-2) \leq 3 + (-2)$ .
This simplifies to $-5 \leq x \leq 1$ .

This means $x$ can be any number from $-5$ to $1$ , including $-5$ and $1$ .
We can write this as $A = [-5, 1]$ .
Boom!

We have $1 < \frac{3x+8}{2} \leq 13$ .
Let's tackle this one piece at a time!

Combining $-2 < x$ and $x \leq 6$ , we get $-2 < x \leq 6$ , which means $B = (-2, 6]$ .

We have $A = [-5, 1]$ and $B = (-2, 6]$ .
The intersection, $A \cap B$ , is where both inequalities are true.
Think of it like overlapping the two intervals on a number line!

The overlap starts at $-2$ (not including $-2$ because it's not in B) and goes up to $1$ (including $1$ because it's in both A and B).
So, $A \cap B = (-2, 1]$ .

We're looking for the integers in $(-2, 1]$ .
Remember, integers are whole numbers.
In this interval, the integers are $-1$ , $0$ , and $1$ .

So, $(A \cap B) \cap \mathbb{Z} = \{-1, 0, 1\}$ .
We found them!

The set of integers that belong to both A and B is $\{-1, 0, 1\}$ .