Problem

Evaluate: $\frac{-4(4+2)-2^{3}}{|-12-4(5)|}$

02:21

Problem 48 Medium Difficulty

Solve $5(x+1) \leq 2(x-3) .$ Write the solution set in interval notation and graph it.

Answer

$\left(-\infty,-\frac{11}{3}\right]$

Related Courses

Algebra

Elementary and Intermediate Algebra

Chapter 3

Graphing Linear Equations and Inequalities in Two Variables; Functions

Section 1

Graphing Using the Rectangular Coordinate System

Discussion

You must be signed in to discuss.

Top Algebra Educators

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Video Transcript

we want to both solve and Graff. The solution, set for five times X plus one, is less than or equal to two times X minus three, and we're gonna graph it on a number line, but also write our solution in set notation. So let's go ahead and start by distributing on both sides. So five times X will be five x five times. One will be five, and that's less than or equal to two times X, which is to X and two times negative three, which is negative. Six. As usual, we want to get our integers on one side and are variables on the other. So I'm gonna go ahead and subtract this to X. That will give me three. X Plus five is less than or equal to negative six. And since my goal is to get the ex alone, I'm gonna start by subtracting five here and I get three acts is less than or equal to negative 11. And then I again want the ex alone. So three times X will be done undone by me dividing by three. So I have X is less than or equal to negative 11 3rd Okay, since I want it in. Ah, interval notation. That means I'm going to use parentheses or brackets and I want to think about how far down this could possibly go cause it's less than so. The ultimate low, which isn't actually a boundary but is a boundary in a sense, is negative. Infinity. But since I can't actually hit negative infinity, I use a parentheses. It's a soft enclosure, and it's gonna go all the way up to negative 11 3rd which it will include, So that will be a bracket. So this is my interval notation answer. And then I want to graph it. So what I'm gonna do is I'm gonna plot zero, and I'm gonna plot negative 11 3rd right here. Since it is including negative 11 3rd I use a solid circle, and then I'm gonna go in the less than direction, which is that way. So that is my graph of the solution set negative infinity up to negative 11 3rd inclusive

We have video lessons for 94.35% of the questions in this textbook