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Algebra 16.2 #3

This is an error in the book by Master Books (ISBN: 978-0890519851) – the book with the blue cover showing the world.

The problem is discussing hens and rabbits with a total of 30 heads and 86 feet.

The solution manual is referring to the wrong problem which is related to money.

Here is how to solve. This does not match up exactly with this chapter but should have been in chapter 7. So if you forget how to solve these, refer back to chapter 7. 

heads = # hens + # rabbits

feet = 2 x # hens + 4x number of rabbits

So let h be the number of hens and r be the number of rabbits.

We can re-write the above as

heads = 30 = h + r

feet = 86 = 2h + 4r

30=h+r

So r = 30-h

Now substitute this into the second to get

86 = 2h + 4r

86 = 2h + 4(30-h) = 2h + 120 – 4h = -2h +120

 

So now rearrange to get

2h = 120-86 = 34

h = 17 (number of hens)

 

Now we can find the rabbits are

R = 30-h = 30-17 = 13

 

Checking our work

13 + 17 = 30 (total number of heads)

2 x 17 + 4 x 13 = 86 which are the number of feet.

Correct!

 

 

 

 

 

 

 

From: June <PrincessJune@protonmail.com>
Reply-To: June <PrincessJune@protonmail.com>
Date: Saturday, July 27, 2019 at 3:34 PM
To: Dale Callahan <dcallahan@askdrcallahan.com>
Subject: Re: A question on Algebra

 

Greetings!

 

I have two questions from Elementary Algebra.

 

In Chapter 16, Lesson 2, Set I, Question #3 and in Chapter 17, Lesson 2, Set II, Question #12, their respective answers don’t seem to match up with the questions. Is this a mistake? And if not, could you please explain how to obtain the correct answers?

 

Thank you very much.

 

Clara

Sent with ProtonMail Secure Email.

 

 

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Algebra Ch13 Test A Problem #13

Question from Grace:

We are using this test A as a practice test. We can’t figure out how to solve this.

Answer from Dr. Callahan:

Grace,

To solve this one, you take the square root of each side

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

The left side needs a little work to do this. Best to see if we can factor
16x^2 + 8x +1

As a hint, you know your factors would need to be the same since you are taking a square root.

So start with the first term 16x^2 – square root of it is 4x. So try that

(4x + ___) (4x + ___)

Then the last term is 1. So the obvious thing that can be multiplied by itself to get 1 is 1. So try

(4x + 1) (4x + 1)

Now when I multiply is out I get

16x^2 + 4x + 4x + 1 = 16x^2 + 8x +1

So that works!!

Now back to the problem

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

+/- SQRT ((4x + 1) (4x + 1)) = +/- SQRT (9)

Using the + we have
(4x + 1) = 3
And 4x = 2 so x = 1/2 is one answer

Using the – we have
(4x + 1) = -3
4x+1 = -3
4x = -4
x = -1

 

 

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Algebra Ch10.2 #3

Question from Makenna

I don’t understand how the perimeter can be 678 ft.
255 is the width x 2 = 510
505 is the length x 2 = 1010 + 510 = 1520

 

Answer from Dr. Callahan

Makenna,

You are correct. The 678 was from the older version of the book (which used a flag of 104 x 235. Flags have grown since then. 😉)

Just FYI – the older flag was flown in Detroit on the side of the J.L. Hudson Store – which no longer exists.
https://en.wikipedia.org/wiki/J._L._Hudson_Department_Store_and_Addition

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Algebra Ch5.6 #15

Ch5.6 #15

As before – all are similar so I will pick the harder one.

Givens

AB = CD

AB = 9(x-2)

CD = x + 3(x+5)

Therefore,

9(x-2) = x + 3(x+5)

Now multiply out and solve

9x – 18 = x + 3x + 15

9x – x – 3x  = 15 + 18

5x = 33

x = 6.6

Now go back and find the lengths

AB = 9(x-2) = 9(6.6 – 2) = 41.4

CD = AB = 41.4

CE = x = 6.6

ED = 3(x+5) = 3(6.6+5) = 34.8

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Algebra Ch5.6 #14

Now let’s look at Ch 5.6 #14.

Parts a-c are similar – so will do part c since the harder.

Area = x + 20 = (x-1)4    > I am just stating all things that are equal based on the given.

x+20 = (x-1)4

x+20 = 4x -4  > multiplied out

20 + 4 = 4x – x  > moved similar items to same side

24 =  3x

x = 8

Now we know x, we have to find the sides (ALWAYS GO BACK AND MAKE SURE YOU KNOW WHAT WAS ASKED)

x -1  = 8 – 1 = 7

4 = 4

 

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Algebra Ch5.6 #9

For problem 5.6, parts a-d are all pretty much the same. Here I will do part d)  since a hair more complex. The steps the same for the others.

Since AB = CD, we can substitute the following

AB is x + 6(x+2)

CD is 2(x+10)

To get

x + 6(x+2)  = 2(x+10)

So to find the length of AB and CD, I need to figure out x. So solve the equation.

x + 6(x+2)  = 2(x+10)

x + 6x + 12 = 2x + 20   – I just multiplied everything out

x + 6x – 2x = 20 – 12  > I moved all the x terms to the left and all the constants (numbers) to the right

5x = 8  > did the math

x = 8/5  = 1.6

Now I have x, but the problem asked for the lengths.

So now we can find the parts

AB = x + 6(x+2) = 1.6 + 6(1.6 + 2) = 1.6 + 6(3.6) = 23.2

AE = x = 1.6

EB = 6(x+2) = 21.6

CD = AB = 23.2

 

 

To do the others – follow the same process

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Algebra Ch5.1 #10x

This is an error in the Algebra Solutions Manual

Mary asked,

In Algebra 1, chapter 5 lesson 1 problem 10x, the answer key says the answer is true for all positive integers.  But isn’t it true for all integers? 1 to the power of x =1.  If it is a negative power, it is still one, right?

Mary
Dr. Callahan Answer:
You are correct. Should be for all real numbers – not just integers. I tend to test these in a calculator to make sure though 😉
So that is incorrect in the Solutions manual.
Just for completeness sake – my goto math tool is Wolfram – so here is there answer.
 
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Algebra Textbook Evaluations

Look for the following material or heading in the contents.

  • Variables
  • Exponents
  • Order of Operations
  • Equations and inequalities
  • Word problems (converting words into symbols)
  • Real Numbers
  • Adding
  • Subtracting
  • Multiplication
  • Division
  • Distributive property
  • Linear Equations (might be equations in one variable)
  • Graphing
  • Slope
  • Intercepts
  • Point-slope and or slope-intercept formulas
  • Systems of linear equations (or simultaneous equations)
  • Linear inequalities
  • Solving
  • Graphing
  • Absolute Values
  • Exponents
  • Products
  • Divisions
  • Scientific notation
  • Polynomials
  • Quadratic equations
  • Factoring
  • Radicals

Notes for use: All the material on the above list needs to be covered in algebra. Instead of moving fast, the students should understand these concepts pretty well.

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Algebra Ch3.4 Set I #9 Solutions Manual Error

Angela’s son brought this error in the Earth cover Algebra I Solutions Manual to our attention. Thank you!

The problem (Ch3.4 #9) wording changed completely from the salamander cover to the Earth cover. However, the solution in the old and new solutions manuals is the same.  In other words, both manuals have the solution to the salamander wording of the problem. The Earth solutions manual has the wrong solution to the Earth Ch3.4 #9 problem.

Angela’s son worked it perfectly and here is his correct solution to the Earth cover Algebra I Chapter 3.4 Set 1 #9 problem.