Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)
Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)
2nd Edition
ISBN: 9780076602919
Author: McGraw-Hill Education
Publisher: McGraw-Hill Education
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Unitary Method
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Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
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Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
Book Icon
Chapter 2.4, Problem 45HP

(a)

To determine

Find the correct solution of the equation:y1 2  (  g + 5 )  =  22

(a)

Expert Solution
Check Mark

Answer to Problem 45HP

3 g  =  6

Explanation of Solution

Given:

The Equation: 2  (  g + 5 )  =  22

Concept Used:

Addition or Subtraction Property of Equality:

If a  =  b , then a + c =  b + c or If a  =  b , then a − c =  b − c

The property that states that if you add or subtract the same number to both sides of an equation, the sides remain equal (i.e., the equation continues to be true.)

Multiplication and Division Properties of Equality:

If a  =  b , then a · c  =  b · c

If a  =  b , then a ÷ c  =  b ÷ c , where c is not equal to 0.

In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. 

Distributive method of multiplication: a  (  b  +  c  )  =  a  ·  b  +  a  ·  c

To solve the equation, we need to open the parenthesis by using the Distributive method of multiplication.

Calculation:

The steps and the solution given in the question is not correct. Correct steps are given below.

The equation: 2  (  g + 5 )  =  22

Step 1: Open the Parenthesis: 2  (  g + 5 )  =  2  ·  g  +  2  ·  5  = 2 g  +  10

Error: In our question it has given as 2  (  g + 5 )  = 2 g  +  5 , the constant term in the parenthesis should be multiplied with 2.

    Steps according to the questionActual correct steps Explanation of each step
    2  (  g + 5 )  =  222  (  g + 5 )  =  22Equation
    2  (  g + 5 )  = 2 g  +  5 Error: the constant term in the parenthesis should be multiplied with 2. Correct should be 2g+10 2 · g +  2 · 5 =  22 2 g  +  10  =  22

    Open the parenthesis by using the distributive method of multiplication.

    2 g  +  5 − 5=  22 − 5  Actually the constant term should be 10 in the left side. Error: 5 should be subtract from both side 2g  +10  −  10  =  22  −10Subtract 10 from both sides.
    2g  =  12Simplify the left and right side.

    In the left: +10  −  10  =  0

    In the right: 22  −10  =  12

    2g2  =  122Divide by 2, the coefficient of the variable g.
    g  =  6Solution of the equation.

Thus, the correct solution of the equation: 2  (  g + 5 )  =  22 is g=6

(b)

To determine

Find the correct solution of the equation. 5 d  =  2d  −  18

(b)

Expert Solution
Check Mark

Answer to Problem 45HP

  −6

Explanation of Solution

Given:

The equation: 5 d  =  2d  −  18

Concept Used:

Addition or Subtraction Property of Equality:

If a=bthen a+c=b+c or If a=b then a+c=b+c

The property that states that if you add or subtract the same number to both sides of an equation, the sides remain equal (i.e., the equation continues to be true.)

Multiplication and Division Properties of Equality:

If a=bthen a•c=b•c

If a=bthen a÷c=b÷c where c is not equal to 0.

In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. 

Calculation:

The steps and the solution of the equation 5 d  =  2d  −  18is given in the question are correct.

    Steps from the questionExplanation of each step
    5 d  =  2d  −  18The original equation.
    5d  −  2d  =  2d  −  2d  −  18 Subtract 2d from both sides .
    3d  =  −  18Simplify the left and right side.

    In the left: 5d  −  2d  =  3d

    In the right: 2d  −  2d  = 0 

    3d3  =  −183

    Divide each side by 3, the coefficient the variable ‘d’

    d  =  −6 The correct solution of the equation

Thus, the correct solution of the equation 5 d  =  2d  −  18 is d=−  6

(c)

To determine

Find the correct solution of the equation: −6z+13=7z

(c)

Expert Solution
Check Mark

Answer to Problem 45HP

  z  =  1

Explanation of Solution

Given:

The equation: −6z+13=7z

Concept Used:

Addition or Subtraction Property of Equality:

If a=bthen a+c=b+c or If a=b then a+c=b+c

The property that states that if you add or subtract the same number to both sides of an equation, the sides remain equal (i.e., the equation continues to be true.)

Multiplication and Division Properties of Equality:

If a=b then a•c=b•c

If a=b then a÷c=b÷cwhere c is not equal to 0.

In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. 

Calculation:

The steps and the solution given in the question is not correct. Correct steps are given below.

    Steps according to the questionActual correct steps Explanation of each step
    −6z+13=7z−6z+13=7zEquation
    −6z+13−6z=7z−6z Error: the variable term in the left side −6z should be added to both sides. In fact we do always the opposite operation. If it is subtracted, then always add the term to both sides. In actual: In the left: −6z  +  6z  = 0  In the right: 7z+6z  =  13z−  6z+6z  +  13=7z+6z  +  13

    Add −6z to both sides.

    In the left: −6z  +  6z  = 0 

    In the right: 7z+6z  =  13z

    Actually it should be:13=  13z13=  13zSimplify

    In the left: −6z  +  6z  = 0 

    In the right: 7z+6z=13z

    13z13  =   1313Divide each side by 13, the coefficient of ‘z’
    z  =  1Solution of the equation..

Thus, the correct solution of the equation −6z+13=7z is z=1

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Chapter 2.4, Problem 44HP
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Chapter 2.4, Problem 46HP

Chapter 2 Solutions

Algebra 1, Homework Practice Workbook (MERRILL ALGEBRA 1)

Ch. 2.1 - Prob. 1AGPCh. 2.1 - Prob. 1BGPCh. 2.1 - Prob. 2GPCh. 2.1 - Prob. 3GPCh. 2.1 - Prob. 4AGPCh. 2.1 - Prob. 4BGPCh. 2.1 - Prob. 5GPCh. 2.1 - Prob. 1CYUCh. 2.1 - Prob. 2CYUCh. 2.1 - Prob. 3CYU
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Prob. 27PPSCh. 2.6 - Prob. 28PPSCh. 2.6 - Prob. 29PPSCh. 2.6 - Prob. 30PPSCh. 2.6 - Prob. 31PPSCh. 2.6 - Prob. 32PPSCh. 2.6 - Prob. 33PPSCh. 2.6 - Prob. 34PPSCh. 2.6 - Prob. 35PPSCh. 2.6 - Prob. 36PPSCh. 2.6 - Prob. 37PPSCh. 2.6 - Prob. 38PPSCh. 2.6 - Prob. 39PPSCh. 2.6 - Prob. 40PPSCh. 2.6 - Prob. 41PPSCh. 2.6 - Prob. 42PPSCh. 2.6 - Prob. 43PPSCh. 2.6 - Prob. 44PPSCh. 2.6 - Prob. 45PPSCh. 2.6 - Prob. 46HPCh. 2.6 - Prob. 47HPCh. 2.6 - Prob. 48HPCh. 2.6 - Prob. 49HPCh. 2.6 - Prob. 50HPCh. 2.6 - Prob. 51PFACh. 2.6 - Prob. 52PFACh. 2.6 - Prob. 53PFACh. 2.6 - Prob. 54PFACh. 2.6 - Prob. 55PFACh. 2.6 - Prob. 56PFACh. 2.6 - Prob. 57PFACh. 2.6 - Prob. 58PFACh. 2.6 - Prob. 59PFACh. 2.6 - Prob. 60PFACh. 2.7 - Prob. 1AGPCh. 2.7 - Prob. 1BGPCh. 2.7 - Prob. 1CGPCh. 2.7 - Prob. 1DGPCh. 2.7 - Prob. 2AGPCh. 2.7 - Prob. 2BGPCh. 2.7 - Prob. 3GPCh. 2.7 - Prob. 4GPCh. 2.7 - Prob. 1CYUCh. 2.7 - Prob. 2CYUCh. 2.7 - Prob. 3CYUCh. 2.7 - Prob. 4CYUCh. 2.7 - Prob. 5CYUCh. 2.7 - Prob. 6CYUCh. 2.7 - Prob. 7CYUCh. 2.7 - Prob. 8PPSCh. 2.7 - Prob. 9PPSCh. 2.7 - Prob. 10PPSCh. 2.7 - Prob. 11PPSCh. 2.7 - Prob. 12PPSCh. 2.7 - Prob. 13PPSCh. 2.7 - Prob. 14PPSCh. 2.7 - Prob. 15PPSCh. 2.7 - Prob. 16PPSCh. 2.7 - Prob. 17PPSCh. 2.7 - Prob. 18PPSCh. 2.7 - Prob. 19PPSCh. 2.7 - Prob. 20PPSCh. 2.7 - Prob. 21PPSCh. 2.7 - Prob. 22PPSCh. 2.7 - Prob. 23PPSCh. 2.7 - Prob. 24PPSCh. 2.7 - Prob. 25PPSCh. 2.7 - Prob. 26PPSCh. 2.7 - Prob. 27PPSCh. 2.7 - Prob. 28PPSCh. 2.7 - Prob. 29PPSCh. 2.7 - Prob. 30PPSCh. 2.7 - Prob. 31PPSCh. 2.7 - Prob. 32PPSCh. 2.7 - Prob. 33PPSCh. 2.7 - Prob. 34PPSCh. 2.7 - Prob. 35HPCh. 2.7 - Prob. 36HPCh. 2.7 - Prob. 37HPCh. 2.7 - Prob. 38HPCh. 2.7 - Prob. 39HPCh. 2.7 - Prob. 40HPCh. 2.7 - Prob. 41PFACh. 2.7 - Prob. 42PFACh. 2.7 - Prob. 43PFACh. 2.7 - Prob. 44PFACh. 2.7 - Prob. 45PFACh. 2.7 - Prob. 46PFACh. 2.7 - Prob. 47PFACh. 2.7 - Prob. 48PFACh. 2.7 - Prob. 49PFACh. 2 - Prob. 1MCQCh. 2 - Prob. 2MCQCh. 2 - Prob. 3MCQCh. 2 - Prob. 4MCQCh. 2 - Prob. 5MCQCh. 2 - Prob. 6MCQCh. 2 - Prob. 7MCQCh. 2 - Prob. 8MCQCh. 2 - Prob. 9MCQCh. 2 - Prob. 10MCQCh. 2 - Prob. 11MCQCh. 2 - Prob. 12MCQCh. 2 - Prob. 13MCQCh. 2 - Prob. 14MCQCh. 2 - Prob. 15MCQCh. 2 - Prob. 16MCQCh. 2 - Prob. 17MCQCh. 2 - Prob. 18MCQCh. 2 - Prob. 19MCQCh. 2 - Prob. 20MCQCh. 2 - Prob. 21MCQCh. 2 - Prob. 22MCQCh. 2 - Prob. 23MCQCh. 2 - Prob. 24MCQCh. 2 - Prob. 25MCQCh. 2 - Prob. 1SGRCh. 2 - Prob. 2SGRCh. 2 - Prob. 3SGRCh. 2 - Prob. 4SGRCh. 2 - Prob. 5SGRCh. 2 - Prob. 6SGRCh. 2 - Prob. 7SGRCh. 2 - Prob. 8SGRCh. 2 - Prob. 9SGRCh. 2 - Prob. 10SGRCh. 2 - Prob. 11SGRCh. 2 - Prob. 12SGRCh. 2 - Prob. 13SGRCh. 2 - Prob. 14SGRCh. 2 - Prob. 15SGRCh. 2 - Prob. 16SGRCh. 2 - Prob. 17SGRCh. 2 - Prob. 18SGRCh. 2 - Prob. 19SGRCh. 2 - Prob. 20SGRCh. 2 - Prob. 21SGRCh. 2 - Prob. 22SGRCh. 2 - Prob. 23SGRCh. 2 - Prob. 24SGRCh. 2 - Prob. 25SGRCh. 2 - Prob. 26SGRCh. 2 - Prob. 27SGRCh. 2 - Prob. 28SGRCh. 2 - Prob. 29SGRCh. 2 - Prob. 30SGRCh. 2 - Prob. 31SGRCh. 2 - Prob. 32SGRCh. 2 - Prob. 33SGRCh. 2 - Prob. 34SGRCh. 2 - Prob. 35SGRCh. 2 - Prob. 36SGRCh. 2 - Prob. 37SGRCh. 2 - Prob. 38SGRCh. 2 - Prob. 39SGRCh. 2 - Prob. 40SGRCh. 2 - Prob. 41SGRCh. 2 - Prob. 42SGRCh. 2 - Prob. 43SGRCh. 2 - Prob. 44SGRCh. 2 - Prob. 45SGRCh. 2 - Prob. 46SGRCh. 2 - Prob. 47SGRCh. 2 - Prob. 48SGRCh. 2 - Prob. 49SGRCh. 2 - Prob. 50SGRCh. 2 - Prob. 51SGRCh. 2 - Prob. 52SGRCh. 2 - Prob. 53SGRCh. 2 - Prob. 54SGRCh. 2 - Prob. 55SGRCh. 2 - Prob. 56SGRCh. 2 - Prob. 57SGRCh. 2 - Prob. 58SGRCh. 2 - Prob. 59SGRCh. 2 - Prob. 60SGRCh. 2 - Prob. 61SGRCh. 2 - Prob. 62SGRCh. 2 - Prob. 63SGRCh. 2 - Prob. 64SGRCh. 2 - Prob. 65SGRCh. 2 - Prob. 66SGRCh. 2 - Prob. 67SGRCh. 2 - Prob. 68SGRCh. 2 - Prob. 69SGRCh. 2 - Prob. 70SGRCh. 2 - Prob. 71SGRCh. 2 - Prob. 72SGRCh. 2 - Prob. 73SGRCh. 2 - Prob. 74SGRCh. 2 - Prob. 75SGRCh. 2 - Prob. 76SGRCh. 2 - Prob. 77SGRCh. 2 - Prob. 78SGRCh. 2 - Prob. 79SGRCh. 2 - Prob. 1PTCh. 2 - Prob. 2PTCh. 2 - Prob. 3PTCh. 2 - Prob. 4PTCh. 2 - Prob. 5PTCh. 2 - Prob. 6PTCh. 2 - Prob. 7PTCh. 2 - Prob. 8PTCh. 2 - Prob. 9PTCh. 2 - Prob. 10PTCh. 2 - Prob. 11PTCh. 2 - Prob. 12PTCh. 2 - Prob. 13PTCh. 2 - Prob. 14PTCh. 2 - Prob. 15PTCh. 2 - Prob. 16PTCh. 2 - Prob. 17PTCh. 2 - Prob. 18PTCh. 2 - Prob. 19PTCh. 2 - Prob. 20PTCh. 2 - Prob. 21PTCh. 2 - Prob. 22PTCh. 2 - Prob. 23PTCh. 2 - Prob. 24PTCh. 2 - Prob. 25PTCh. 2 - Prob. 1CRCh. 2 - Prob. 2CRCh. 2 - Prob. 3CRCh. 2 - Prob. 4CRCh. 2 - Prob. 5CRCh. 2 - Prob. 6CRCh. 2 - Prob. 7CRCh. 2 - Prob. 8CRCh. 2 - Prob. 9CRCh. 2 - Prob. 10CRCh. 2 - Prob. 11CRCh. 2 - Prob. 12CRCh. 2 - Prob. 13CRCh. 2 - Prob. 14CRCh. 2 - Prob. 15CR

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