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Homework Help Complex Problems – Moving Deeper into Concepts

As you move deeper into each textbook (Algebra, Geometry, Calculus, etc), the problems will contain several concepts you’ve learned earlier – all in one problem.  This complexity can make it difficult to find the “how-to” or examples in the book to help us understand how to work it.

Recently, we received this question from our student support page: There doesn’t seem to be any examples or teaching that I can find that helps me solve this problem.

This is a common question – not this particular problem, but in general, as the problems develop into including several simple concepts stacked into compound calculations. However, the explanations are there, they just may be back a few, or several, chapters.

For example, Jacobs Algebra Chapter 12 Summary and Review Problem 14h.

Concepts include but are not limited to:

  • Chapter 12: Square Roots. Simplify radical as much as possible. Example of this step page 480-481
  • Chapter 5: Equations in One Variable. Specifically for this review problem, Equivalent Equations (Lesson 3) page 162-163.
  • Chapter 12: Square Roots: Radical Equations. Page 505 has examples of squaring both sides to eliminate the radical.

Math builds on itself. As you learn more and more concepts, the problems reach back and build on calculating and analyzing skills learned earlier in the book or even in an earlier course. These complex problems can be hard to find “how-to” when we just can’t see it! The solutions manual is a good resource to help with steps, but sometimes even with those steps, we need to see where it was explained or taught.

We are here to help! Send us your homework questions and let us help. Filling out this form makes it easy to be sure you’ve told us what we need to know to help you, but you can also send an email to support@askdrcallahan.com.

Be sure to tell us:

  1. the course,
  2. the chapter,
  3. the lesson,
  4. the problem,
  5. YOUR issue as best as you can explain it.

We love to help.

 

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Algebra Chapter 16 Lesson 2 Set 1 Problem #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.Algebra Chapter 16 Lesson 2 Set 1 Problem 3

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

Here is how to solve. Simultaneous equations are covered 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

Substitute this equation into the second equation to get

86 = 2h + 4r

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

 

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!

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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.