Answer | 8 x + 3 | = 0 = one solution − 2 x + 4 = 2 x + 4 = one solution 12 x − 2 x = 10 x − 8 = no solution − 6 = 5 x − 9 = one solution 0 = | x + 13 | + 2 = no solution 9 = 3 | 2 x − 11 | =two solution 3 x − 12 = 3 ( x − 4 ) + 1 no solution − 4 ( x + 4 ) = − 4 x − 16 = infinite solution 7 − 2 x = 3 − 2 ( x − 2 ) = infinite solution Explanation Given: | 8 x + 3 | = 0 − 2 x + 4 = 2 x + 4 12 x − 2 x = 10 x − 8 − 6 = 5 x − 9 0 = | x + 13 | + 2 9 = 3 | 2 x − 11 | 3 x − 12 = 3 ( x − 4 ) + 1 − 4 ( x + 4 ) = − 4 x − 16 7 − 2 x = 3 − 2 ( x − 2 ) Calculation: Determine the number of solutions for each equation: | 8 x + 3 | = 0 = one solution − 2 x + 4 = 2 x + 4 = one solution 12 x − 2 x = 10 x − 8 = no solution − 6 = 5 x − 9 = one solution 0 = | x + 13 | + 2 = no solution 9 = 3 | 2 x − 11 | =two solution 3 x − 12 = 3 ( x − 4 ) + 1 no solution − 4 ( x + 4 ) = − 4 x − 16 = infinite solution 7 − 2 x = 3 − 2 ( x − 2 ) = infinite solution Conclusion: Thus, the equations are categorized.

BIG IDEAS MATH Integrated Math 1: Student Edition 2016

16th Edition

ISBN: 9781680331127

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 1, Problem 8CA

To determine

To place: each equation into one of the four categories.

Expert Solution & Answer

Answer to Problem 8CA

|8x+3|=0 = one solution

−2x+4=2x+4 = one solution

12x−2x=10x−8 = no solution

−6=5x−9 = one solution

0=|x+13|+2 = no solution

9=3|2x−11| =two solution

3x−12=3(x−4)+1 no solution

−4(x+4)=−4x−16 = infinite solution

7−2x=3−2(x−2) = infinite solution

Explanation of Solution

Given:

|8x+3|=0

−2x+4=2x+4

12x−2x=10x−8

−6=5x−9

0=|x+13|+2

9=3|2x−11|

3x−12=3(x−4)+1

−4(x+4)=−4x−16

7−2x=3−2(x−2)

BIG IDEAS MATH Integrated Math 1: Student Edition 2016, Chapter 1, Problem 8CA

Calculation:

Determine the number of solutions for each equation:

|8x+3|=0 = one solution

−2x+4=2x+4 = one solution

12x−2x=10x−8 = no solution

−6=5x−9 = one solution

0=|x+13|+2 = no solution

9=3|2x−11| =two solution

3x−12=3(x−4)+1 no solution

−4(x+4)=−4x−16 = infinite solution

7−2x=3−2(x−2) = infinite solution

Conclusion:

Thus, the equations are categorized.

Chapter 1, Problem 7CA

Chapter 1, Problem 9CA

Chapter 1 Solutions

BIG IDEAS MATH Integrated Math 1: Student Edition 2016

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - Prob. 57E Ch. 1.1 - Prob. 58E Ch. 1.1 - Prob. 59E Ch. 1.1 - Prob. 60E Ch. 1.1 - 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