Holt Mcdougal Larson Algebra 2: Student Edition 2012
Holt Mcdougal Larson Algebra 2: Student Edition 2012
1st Edition
ISBN: 9780547647159
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Expert Solution & Answer
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Chapter 3.6, Problem 21E
Solution

To calculate: The value of x for the equation 4x+103+3=15 and also check the solution.

The resultant answer is x=37 .

Given information:

The given equation is 4x+103+3=15 .

Formula used: Raise each side of the equation to the same power to eliminate the radical.

Calculation:

Write the original equation:

  4x+103+3=15

Subtract 2 from each side of the equation,

  4x+103+33=1534x+103=12

Now divide both sides by -4,

  4x+1034=124x+103=3

Cube each side to eliminate the radical an simplify,

  x+1033=33x+10=27x=2710x=37

Now check x=37 in the original equation.

  4x+103+3=15

Now substitute -37 for x and simplify,

  437+103+3=?154273+3=?1543+3=?1512+3=?1515=15

Chapter 3 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 3.1 - Prob. 11GPCh. 3.1 - Prob. 12GPCh. 3.1 - Prob. 13GPCh. 3.1 - Prob. 14GPCh. 3.1 - Prob. 15GPCh. 3.1 - Prob. 16GPCh. 3.1 - Prob. 17GPCh. 3.1 - Prob. 18GPCh. 3.1 - Prob. 19GPCh. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10ECh. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - 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