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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Chapter 2.4, Problem 46HP

(a)

To determine

Find the value of k for which the equation k ( 3x − 2 ) = 4 − 6x is an identity.

(a)

Expert Solution
Check Mark

Answer to Problem 46HP

  k =  −  2

Explanation of Solution

Given:

The equation: k ( 3x − 2 ) = 4 − 6x

Concept Used:

From the equation: k ( 3x − 2 ) = 4 − 6x , we need to find the value of k .

Using a rule: b  − a  =  − a + b =  −  ( a −  b )

Calculation:

According to the rule: b  − a  =  − a + b =  −  ( a −  b )

The right side we can write: 4  −  6x  =  −6x  + 4  =  −  (  6x  − 4 ) 

The equation becomes: k ( 3x − 2 ) = −  (  6x  − 4 ) 

    Steps Explanation of each step.
    k ( 3x − 2 ) = 4 − 6xOriginal Equation
    k ( 3x − 2 ) = −  (  6x  − 4 ) According to the rule: b  − a  =  − a + b =  −  ( a −  b )

    The right side we can write: 4  −  6x  =  −6x  + 4  =  −  (  6x  − 4 ) 

    k( 3x − 2 )( 3x − 2 )  =  −  (  6x  − 4 ) ( 3x − 2 )Divide each side by the coefficient ( 3x − 2 ) of the variable k
    k  =  −  (  6x  − 4 ) ( 3x − 2 )  =  −  2 (  3x  − 2 ) ( 3x − 2 )2 is the GCF of (  6x  − 4 ) in the right side.

    Write in factor form : −  (  6x  − 4 )  =  −  2 (  3x  − 2 )

    k =  −  2Value of k for which the equation k ( 3x − 2 ) = 4 − 6x is an identity

Thus, the value of k =  −  2 for which the equation k ( 3x − 2 ) = 4 − 6x is an identity.

(b)

To determine

Find the value of k for which the equation 15y−10+k=2(ky−1)−y is an identity.

(b)

Expert Solution
Check Mark

Answer to Problem 46HP

  k=8

Explanation of Solution

Given: The equation: 15y−10+k=2(ky−1)−y

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

Calculation:

    Steps Explanation of each step
    15y−10+k=2(ky−1)−yOriginal Equation
    15y−10+k=2ky−2−yOpen the parenthesis by using distributive method of multiplication.
    15y−15y−10+k=2ky−2−y−15ySubtract 15y from both the side.
    −10+k=2ky−2−16ySimplify. In the left 15y−15y = 0 and in the right −y−15y=−16y
    −10+10+k=2ky−2+10−16yAdd 10 to both the sides and simplify.
    k=2ky+8−16yIn the left − 10 + 10 = 0 and

    In the right − 2 + 10 = 8

    k−2ky=2ky−2ky+8−16ySubtract 2ky from both the sides.
    k−2ky=8−16yIn the right 2ky − 2ky = 0
    k(1−2y)=8(1−2y)In the left k is the GCF of the term k − 2ky and write in factor form.

    Similarly 8 is the GCF of the terms 8 − 16y and write in factor form.

    k(1−2y)(1−2y)=8(1−2y)(1−2y)Divide each side by the coefficient (1−2y) of k.
    k=8Value of k

Thus, the value of k=8 for which the equation 15y−10+k=2(ky−1)−y is an identity.

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Chapter 2.4, Problem 45HP
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Chapter 2.4, Problem 47HP

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. 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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