This page is designed to help provide support for Kate’s Elementary Algebra eCourse, a video supplement that walk students through Elementary Algebra in an engaging, visual/auditory way! View samples on MasterBooksAcademy.com.
Corrections/Clarifications to Solutions
Chapter 4 Lesson 2, Problem 4g – Explanation of Answer
The text explains on the bottom of page 128 and top of 129 that “…we cannot write the absolute value of a variable, such as x, by dropping the + or – sign, because x itself has no sign. In this case, we must use the following algebraic definition of absolute value. If x is positive or zero, then |x| = x. If x is negative, then |x| = -x (or the opposite of x).”
To make sense of this, let’s say x = -4. If you’d just said the absolute value of x was x, then you’re saying the absolute value is -4…which it isn’t. By saying the absolute value of x is -x, then when you plug in -4, you’ll get – – 4, which is 4.
The algebraic rule just makes sure we actually get the absolute value when we plug in the value for the variable.
Chapter 2, Lesson 4, Set I, Problem 1c – There should be two more rows of both the two circles and the four circles on Figure 4. No figure matches the expression right now.
Chapter 2, Lesson 5, Set III, Problem 15 (optional problem) – In the chart in 15a, the x numbers should start at 0. Then in 15d, 49 should be plugged in instead of 50 into the equation (because the first week is when x = 0, the second would be when x = 1, etc.), yielding $324.50 instead of $325. 15b and 15c are correct as they are.
Chapter 2, General Note
In this chapter (such as on page 84, question 4), there are problems where students are asked to figure out formulas. In these cases, the goal is to look at them and think “what do I have to do to x to get y?” While there are more complicated relationships, all of these problems where students have to find the formula should be that you add, subtract, multiply, or divide x by something to get y. Start by looking at if the x values are greater than or less than the y values. If greater than, you have to subtract or divide x by some value to get y. Then look at what the difference is between one of the x and y values and see if it’s the same for all the others; if not, try division. If x is less than y, you have to add or multiply by some value to get y. Look at what you’d have to add to x to get to y in one column and see if it holds true for the others; if not, look at what you’d have to multiply x to get to y.
Chapter 2, Summary & Review, Problem 8, Clarification
The equation 25x/6 means the same thing as 25/6 all times x. (To see how, just picture simplifying 25/6 times x. You’d view x as x/1 and multiply the numerators and denominators, which would give you 25x/6.) So 25/6 is the constant of variation.
Chapter 3, Lesson 1, Set I, Problem 3 – This problem is using information from previous math courses.
On 3a, if she catches 3 a day, then the time it would take her to catch say 12 would be 12 divided by 3, or 12/3 (fractions are another way of showing division). We’re using x to stand for any amount she could need to catch, so we’d have x divided by 3, or x/3.
On 3b, if she does 3 a day and there are 7 days in a week, then she’d catch 3 * 7, or 21, each week. So in y weeks, she’d catch 21 times that number of weeks, or 21y (which means 21 times y).
On 3c, the question refers back to Lesson 4 of Chapter 2. If we use y for the mice caught and x for the number of days, we’d have y = 3x. So that would be a direct variation.
Test 2B, Problem 12
Answer should be y = 4(x-1) + 1. However, this is a hard function to guess. The question table should really be changed to match the current answer of y= 4x + 1:
x 1 2 3 4 5
y 6 9 13 21
Test 3A, Extra credit Problems – Explanation of Answer
Since a car uses 4 tires at a time, we take the mileage (18,000) and multiply it by 4 to find the total cumulative mileage all the tires would have gone collectively. We then divide that by 5, since it was shared evenly between 5 tires. Part b is showing that using x to stand for an unknown mileage.
Chapter 6, Lesson 3, Set II, Problem 4e
In the answer, b should equal -5.
Chapter 6, Lesson 4, Problem 5g – Explanation of Answer
The y intercept is when x is 0, as that is when the line will be touching the y-axis.
Plugin in 0 for x gets 0 + 0y = 14. Since 0 times any value is 0, we have 0 = 14, which isn’t true. Thus, there’s no y-intercept.
2.8 is the x-intercept, but that is not what you would use to find the y-intercept. The x-intercept tells you where the line intersects the x-axis, but that’s not necessarily when it will intercept the y-axis.
0 is the y-value for when the line intersects the x-axis, which will always be 0 since it’s on the x-axis. But that’s not what the problem is asking for.
The picture shows the different intercepts to lend clarity.
Chapter 6, Lesson 4, Problem 7d – Explanation of Answer
The problem is asking if (2,8) is a solution to the equation. Since it yields 12 = 6, the answer is no. The line that equation represents doesn’t go through point (2,8). If it had worked out like it did in 8a, then the answer would have been yes, as that would have been a valid solution/point on the line.
Checking helps you see if you got a valid answer; if the check yields something that doesn’t equal like 12 = 6, then you know what you plugged in isn’t a solution/point on the line. In 7d, the value given wasn’t a valid one.
Chapter 6, Summary & Review, Problems 14a.-f. – Explanation of Answer
The equations that were graphed are given below for ease of grading. Keep in mind that when checking graphs, look at if it intersects the vertical axis at the same place as that shown in the solution and if the line is angled the same direction/would go through the same points if extended to the same part of the graph. It is okay if students graphed different parts of the line, so long as it would go through the same points if extended to what is shown.
- 14a. y = -3/2x + 3
- 14b. y = 10 + 2x
- 14c. y = -1/5x
- 14d. y = -5x + 8
- 14e. y = 2/5x -1
- 14f. x = 6
Test 6a, Question 13, Problem 13 – Explanation of Answer
You have to rearrange the equation so that x and y are on the left. Then you think through the values for a, b, and c. See page 212 in the textbook for a review of standard form.
Test 6a, Question 13, Problem 23 – Explanation of Answer
You graph by inserting values for x or y and seeing what y or x values you get, then graphing those points. 0 is an easy value to insert.
Chapter 7, Lesson 7, Set I, Problem 2 – Explanation of Answer
To solve this, start by graphing the equation. The Solution’s Manual says one possible answer as there are a number of answers for each question. The way to figure it out is to graph the equation and then just draw straight lines that either
- 2a) don’t intersect (setting x + 2y equal to any number but 5 does this; for example, x + 2y = 4),
- 2b) intersect everywhere (i.e., are the same line) or
- 2c) intersect at point (1,2).
Then record the equation of each line, as that equation is an answer to the problem.
Chapter 7, Lesson 7, Set I, Problem 2b – Correction
2x + 2y = 10 is an error. Correct answers include 2x + 4y = 10 or any other equation that gets the same line.
Chapter 8, Review, Set I, Problem 12b
In the denominator of the solution shown, there should be a negative sign in the exponent, making it 2 x 10^-1.
Midterm Review, Set II, Problems 36 and 37 – Explanation of Answer
The equation is found using the distance = rate times time (d=rt) formula. We know to use it because it asks for an expression for the distance. The rate for that section we were told is 40 mph (rate is like speed). Notice that the distance AB is represented as 50x–so they’re using x to stand for the time it takes to go that distance, and plugging x for t and 50 (the speed for that section) for r into the d = rt formula, getting d = 50x (just the 50x part is listed, as that’s how to find the distance). The total time we’re told is 5 hr total, so the time for BC would be 5 – x–that is, 5 hr minus the time it took for AB. If we plug that and 40 mph into the formula and leave off units, we get d = 40(5 – x). The answer key just listed the right-hand side, as we were asked for an expression for distance, not the full equation (although I’d count the full equation as correct).
See the explanation of Set II, problems 36 and 37. The thinking is similar, except this time the x is standing for the rate (r), as we know the times.
Midterm Test, Problem 4
Read the question very carefully. Note that it is asking for positive integers. See page 91 for a definition of what those are. Note that there is an error in the answer. The correct answer is a.
Chapter 10, Lesson 6 – Explanation of Some Problems with Trinomial Squares
Separate out the second part:
2x^2/(x + 3) + (-x-3)/(x+3)We can view negative signs as multiplications of -1. Think of the -x – 3 in the second fraction as -1x + -1(3). We can factor out the negative 1:
2x^2/(x + 3) + -1(x + 3)/(x + 3)Now we see the x + 3 in the numerator and the denominator cancel out, leaving us with just this:
2x^2/(x+3) – 1
When you have x(2-x)/2(x-2), you can factor out a -1 in the numerator to get -x(-2+x) in the numerator which can be rewritten as -x(x – 2), and the (x-2) would then cancel out with the denominator, leaving -x/2.
First multiply by a fraction worth 1 to get a common denominator, and then multiply by a fraction worth 1 to get rid of the fraction in the denominator.
Test 14, Question 40, Clarification
To square that, you would take (SQRTx + 1)((SQRTx + 1), which results in x + 2SQRTx +1–in other words, it doesn’t help us solve the problem. I’m guessing he thought it equaled x + 1, but you have to square the entire side, not just each term separately.
You generally want to get x on a side by itself before squaring. If we subtract 1 first, we get this:
SQRTx = SQRT5 – 1.
Then we square both sides. SQRTx squared is x, and (SQRT5 – 1) squared is (SQRT5 – 1)(SQRT5 – 1). When you distribute that multiplication and simplify, you get the answer on the answer sheet.
Chapter 15, Lesson 2, Problem 3, Clarification
The key to this type of problem is to think about how everything relates to the variables you’re working with. Here, we’re told to use x and y:
x = Mr. Dither’s present age
y = Dagwood’s present age
So go back and read the paragraph to think about what you know in terms of that. “Suppose that Mr. Dithers is now 42 years older than Dogwood.”
This sentence is referring to their present ages. We’re told Mr. Dither’s (x) is 42 more than Dagwood (y).
x = y + 42
The next part is “nine years ago he was three times as old as Dagwood was at the time.”
How can we write this in terms of their current age?
9 years ago, Mr. Dither’s age would have been his current age minus 9, or x – 9.
Dagwood’s would be his current age minus 9, or y – 9.
We’re told Mr. Dither was 3 times as old as Dagwood’s age then, so
x – 9 = 3(y – 9)
One thing you can do if you’re stuck on word problems is to try to solve them with an actual value. For example, make Mr. Dither an actual age. See what you have to do to solve, and then do the same thing with the unknown instead. Or try writing out the relationships in words.
Chapter 16, Lesson 1, Problem 3 – You can also set up the equation like this: (6 mi/hr)(x) = (54 mi/hr)(40 min – x) Note that we’re setting these as equal because the distance up and down the mountain is the same, and we find distance by multiplying speed times time (x here is standing for the time up the mountain and 40 min – x for the time down).
You can then rewrite 6 mi/hr as 6 mi/60 min, since 60 min equals 1 hour, and 54 mi/hr as 54 mi/60 min. Then when you do the math, the units work out. You could at that point just rewrite without units (6/60)x = (54/60)(40 – x) to make it easier, but you’ve already checked to see that the units will work out. The Solutions Manual wrote 6(x/60) = 54((40 – x)/60, which means the same thing but makes it harder to see where the 60 came from.
Chapter 16, Lesson 3, Problem 5f (page 639) – (Note that this is not a problem I recommend solving in the eCourse schedule; however, here’s a clarification if you’ve attempted it.) The answer you should get when working out the problem there is x < 3 (rather than x > 3 as in the Solutions Manual), but note that we already were told to solve for when x < 0, so we really know just that x < 0.
Questions and Answers
I struggle with word problems; what do I do?
See the new “Notes for Students” section in the eCourse for some hints that may help.
My graph doesn’t match the one in the Solutions Manual, and I can’t figure out why.
It may just be that the Solutions Manual graphed a different range of input values. See the newly updated Course File/Download for a whole section explaining how to grade graphs.
I struggle figuring out when I can simplify fractions and when I can’t. What do I do?
See the new “Notes for Students” section in the eCourse for some hints that may help.
What do I do if I struggle with problems including negative numbers?
See the new “Notes for Students” section in the eCourse for some hints that may help.
When should I list both the positive and negative square root?
See the new “Notes for Students” section in the eCourse for some hints that may help.
I’m a sinner saved by grace through Jesus Christ. I’m also a writer—I’ve been writing ever since I could hold a pen. I… 
