Course Extras: Jacob’s Elementary Algebra

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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 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 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.

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.

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).

#37 – The equation comes from setting the sum of the two distances AB and BC equal to 215, which we were told was the total distance.
Midterm Review, Set III, Problems #38-39 – Explanation of Answer
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

The key to are the formulas given on page 405. Trinomial squares will be in one of those two forms.
On problem 7b, it’s in the form of a^2 – 2ab+ b^2 (the only difference in the two forms is if there’s a minus or plus sign in front of the second monomial). We’re trying to find the 3rd term in our trinomial square, which would be the b^2 in the formula. We know that the middle polinomial in the formula is 2ab; we also know that a is the square root of the first term (a^2). In this case, our first term is x^2, so the square root is x. So we can replace a with x, giving us 2xb. We also know that the 2nd term in this trinomial is 22x. So set them as equal: 2xb = 22x. Solving for b gets us 11. If b is 11, then b^2 is 11^2, which is 121. So the third term in this trinomial square is 121–it has to be in order for it to be in the form of a^2 – 2ab + b^2.
Problem 7d, is the same idea. Only this time, there’s a plus sign so it’s in the form of a^2 + 2ab + b^2. The square root of the first term this time is 6x, so we would replace a with 6x, giving us that 2(6x)b = 12x. Solving for b gives us 1. 1^2 is still 1.
We’re just using the knowledge that the first term is a^2 and the last b^2, and the middle is 2ab, to figure out what that third term.
On Problem 8C, it’s in the form of a^2 + 2ab + b^2. So to find a, take the square root of the first term. The square root of 4x^2 is 2x. To find b, take the square root of the 3rd term. The square root of 1 is 1. So now we just use what we’re told on page 405 that trinomials in that form equal (a + b)^2 and plug in a and b, giving us (2x + 1)^2 as the answer. If we check ourselves by squaring 2x +1, we’ll get the original trinomial.
8d looks like it’s in the form of a^2 – 2ab + b^2, but it’s not. The square root of the third term is 9, and of the first x. The coefficient of the 2nd term has to be in the form of 2ab for this to work. But 2x(9) would be 18x, not 9x. This isn’t a trinomial we can use that formula to factor, as it’s not a trinomial squared. We could have also seen that by factoring and then checking our work by squaring the factored answer–when we did, we’d see it wouldn’t give the same trinomial we started with.
8g, is in the form of a^2 + 2ab + b^2. a = square root of the first term = square root of 49x^2 = 7x; b = square root of the third term = square root of y^2 = y. Plugging it in to (a + b)^2 gives us (7x + y)^2. This would check out if we squared 7x + y.
10a -10c are the same idea–don’t let the fact that x isn’t square fool you. Here’s 10a worked out: a = square root of the first term = square root of x^4 = x^2; b = square root of the third term = square root of 121 = 11. Plugging it in to (a + b)^2 gives us (x^2 + 11)^2. This would check out if we squared 7x + y.

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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