This unit looks more closely at equations and introduces inequalities and uniform motion problems. The equations are more complex, involving more negative numbers and fractions, both within the equation itself, as well as the solutions.
In this particular lesson the focus is on literal equations—equations like I = prt, in which all components of the equation are letters. (The term “literal equations” is a vestige of algebra books written prior to the 1950’s in which the term “literal” was used instead of “variable” which is used now.) Understanding how such equations may be solved for other letters in the equation plays a role in seeing how a variable can be expressed in terms of the others in the equation. For example, the formula “distance = rate x time” or d = rt, can be solved to express rate in terms of distance and time (i.e., r = d/t), as well as time in terms of distance and rate (i.e., t = d/r).
This unit contains a variety of word problems, and introduces the first of the word problems involving uniform motion: motion in opposite directions, same direction, and round-trip. Later, in the unit on algebraic fractions (rational expressions) students are introduced to problems with wind and current in which the equation must be solved for either parameter. In the unit on systems of equations there are wind and current problems in which students must solve for both parameters. As students progress through the course, the earlier types of problems should be constantly re-introduced in homework assignments, warm-ups and tests. Otherwise the latest type of uniform motion problem becomes the one most familiar and likely the only type the student can solve.
This particular lesson is key in developing proficiency in solving more complex equations. Throughout, I emphasize leaving improper fractions in fractional form and not converting to mixed numbers or decimals.
Warm-Ups.
1. Write an equation and solve. Five times a certain number is 12 more than twice the number. What is the number? Answer: 5x = 12 + 2x; 3x = 12, x = 4
2. Solve. 22 – x = 7x – 2 Answer: 8x = 24; x = 3
3. An apple has twice as many calories as a peach. The two fruits together have 105 calories. How many calories are in each fruit? Answer:Let the number of calories in a peach. Then x + 2x = 105; 3x = 105; x = 35 (peach) and 2x = 70 (apple).
4. Solve. 2x + 5 = – 101 Answer: 2x = –106; x = –53
5. Solve. 5y + 7 – 1 – y = 17 Answer: 4y + 7 = 17; 4y = 10; y = 10/4 = 5/2
Problem 4 requires students to subtract 5 from both sides. It may be important to remind them that subtraction of 5 is the same as adding – 5. Also, some students may forget that –101 – 5 is the same as –101+ (– 5). Some tend to confuse it with -101 – (-5), and will add +5 to obtain -96. A helpful prompt to remind them is “I lost $101 and then lost $5; how much did I lose in total?”
Negative Numbers in Equations. Problem 4 of the Warm-Ups provides an entrée to the subject of negative numbers in equations. “Anyone remember what the opposite of a number is? Like the opposite of 5—what is that?” They should know that it is negative five, but I also prepare myself for no one volunteering an answer. “And the opposite of negative 3?”
They will be more receptive now that the ice has been broken, and volunteer that it is positive 3. “And what is the sum of a number and its opposite?” Zero. So far so good. Now I up the ante.
“How about the opposite of x?” I’ll usually hear “negative x”, and asking for the opposite of –x, I’ll hear “positive x”. I’m always on the look-out for a student or students who may ask “But what if x is already negative?”
“Good question. Let’s look at that. If x is -5, then –x is the opposite of -5, which is 5. We could write that as – (– x). We can read that as the opposite of negative x.”
“What if I have an equation like –x = 5? How would I solve this equation so that x is positive x?”
I should hear “Divide both sides by –1” but if I don’t I remind them that –x is the same as –1x. We just don’t write the -1.” Dividing by –1 results in x = –5.
More Examples. I work with them on the first two.
1. t/3 = -11; Answer: t = -33. I may remind them that t/3 = 1/3(t), and we multiply both sides by the reciprocal of 1/3 which is 3.
2. -6m + 2 = -7m Answer: m +2 = 0, m = –2. I advise adding 7m to each side to obtain a positive variable. This leaves 0 on the right hand side. They then need to subtract 2 (i.e., add –2) from (to) each side.
3. -7(r-1) = 0 Answer: I advise that (r – 1) can be written as (r + (–1) ) so it is easier to see that distributing the –7 results in –7r + 7. –7r +7=0; -7r = -7; r = 1
4. 7 = n/2 -1; Answer: n/2 = 8; n = 16
5. Challenge problem: x + 36 = 1 –4(x–5) Again, this involves distributing a negative number. A hint might be that the right hand side can be written as 1 + (–4)(x +(–5)). Answer: x + 36 = 1 -4x + 20; 5x = -15, and x = -3.
Literal Equations. “Suppose we had this problem to solve:”
“I know you know how to do this but before we do it, I’m going to write another on the board:”
“Now I want someone to tell me what to do to solve the first one.”
Someone will tell me to divide each side by 2. “Good. Now I’m going to do the equivalent on the second equation. What do I do to isolate x? I divide both sides by what?”
Students are hesitant but I will hear some say “Divide by a.”
“I’ll do that, and I get x = c/a. Remember we don’t use the divide sign here; when dividing we use the fraction bar. And the first equation we have x = 10/2 which is = 5. Can I simplify c/a in the second equation?”
I will hear “No” to which I agree. “So we just leave it x = c/a. This type of equation where there are letters, and we solve for another letter in the equation, is called a ‘literal equation’. We solve these equations following the same rules as for equations with numbers. How would you solve this?”
“How do we isolate x?” I have them write in their notebooks. Generally I will see the correct answer, though some will write x = a ÷ 3.
“Remember, we use the fraction bar, not the divide sign. The answer is x = a/3. Or, you can write it as x=1/3(a). ”
Examples. I work through a number of these with the class. I have them work in their notebooks and I walk around offering guidance as necessary.
1. Solve for b.
I ask them how they can rewrite x/2. They should now recall it is ½(x)
“How do we isolate x?” I will hear either “Multiply by the reciprocal of ½” or “Divide by ½ ” or “Multiply both sides by 2.” I’m happy with any of these. The answer will then be x = 2b.
2. I = prt I give them a hint: “Think of pr as one number, and it is the coefficient of t. What do we do to isolate t?” Answer: t = I/pr
Solve for We want to isolate the variable t. Think of pr as one number, which it is. Then we can divide both sides by pr
3. Solve for b: p = a + b + c Answer: b = p – a – c.
4. Solve for g:
Prompt: “First, what do we do to remove the denominator?” Multiply by 2.
“How do we now isolate g?” Divide by . Dividing by a variable squared may seem like something not allowed; I assure them they can. “It represents a number just like t represents a number, and as long as that number isn’t zero, we can divide both sides by it. And yes, t is not 0, so go ahead!
4. Solve for r.
“What’s the coefficient of r?” It is 2π. “We can think of 2π as one number.
Homework. The homework problems should be a mix of regular equations (which contain negatives and fractions like 2y/3 + 7 = 5) and literal equations. The latter should be kept fairly simple but have one or two challenging problems that can be worked through the next day, like: