Course Extras: Jacob’s Elementary Algebra

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

Corrections/Clarifications to Solutions

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 3, Lesson 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.

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.

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.


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