Multi-step equations are those that generally involve distributions and combining terms. Examples include equations such as: 3(𝑥+2)=3(𝑥−2)+3𝑥
At this juncture, students are also introduced to equations that result in identities in which solutions include all real numbers, and equations that have no solutions. These concepts although fascinating to some students, provide ready excuses to others who can’t solve an equation. Students assume if they can’t solve an equation, it falls in the category of “no solution”.
Students get confused with these problems because of the number of steps. I don’t assign a lot of problems for this topic; students need to learn to take their time and do them one step at a time.
Warm-Ups. Warm-Ups include previous types of complex equations and word problems.
1. Write an equation and solve. At a store, the owner raises the cost of a shirt by $8. The cost of 3 shirts now costs $126. What was the original cost of the shirt? Answer: Let x = original cost of a shirt. Then x+8 = increased cost of a shirt. 3(x+8) = 126; 3x+24=126; 3x = 102; x=$34 OR x+8 = 42; x = 34.
2. Solve. 5(x+5)= -17 Answer: 5x+25 = -27; 5x = -2; x = -2/5
3. Solve. -3/5(x-10)=15 Answer: x-10 = 15 ∙(-5/3); x-10 = -25; x = -15
4. Solve. 8a – 5 = 3a + 25 Answer: 5a=30; a = 6
5. Write an equation and solve. 7 times a numbers is 12 more than 3 times the number. Find the number. Answer: 7x = 12 + 3x; 4x=12; x=3
Hints to use for Problem 1 include “What are you trying to find?”, “How do you represent the original cost of a shirt?”, and “How do you represent the increased cost of a shirt?”
For problem 3, students may forget the negative sign when multiplying both sides by the reciprocal of -3/5. It is worthwhile to show students how to check if the answer is correct: -3/5(-15-10) =? 15; -3/5(-25) =?15; -3(-5) =? 15; 15=15
General Method. I start the lesson by writing an equation on the board and asking students how they would go about solving it. I also tell them that they have all the skills necessary to do this; they just have to take things in order:
There will be some hesitancy but usually there is a student who will ask “Should we distribute?” to which I usually reply “Yes. What are you going to distribute?”
The answer will usually come in the form of another question: “The left side?”
“That will work,” I reply. “Let’s do it.”
They will get 20x + 50. For the right side they will get 30x + 15. The entire thing now looks like:
“Solve it,” I’ll say. The answer is 10x = 35; x = 3.5. (Some will balk at an answer with a decimal; I assure them it’s fine. I also tell them to get used to it, and encourage them to check their answer by plugging in 3.5 in the original equation.)
I emphasize that the only thing different here than in previous lessons is that there are distribution steps on both sides of the equation. But once the parentheses are removed by distribution, the equation is in the form that they’ve seen and solved before.
“This next is slightly different,” I’ll announce:
“What’s different than the last one?”
I want them to see that in addition to distributions on both sides, there’s also a number on the left hand side added to the distribution. I’ll take it step by step with them, distribution the 3(y-5) first and then asking what do I do with the 25? They should get 3y+10 = 2y+11 which is easily solved: y = 1.
The next is the problem which in the previous lesson I said they would be doing:
I will first assure them they can do it, and then give them steps as necessary. But I want them to do this independently, writing it in their notebooks, as I walk around the classroom.
My basic hints are: Distribute on left side, and then combine like terms. Do the same on the right side. Can you solve it now?
Answer: 10y+15-4y = 6 + 3y + 39; 6y + 15 = 3y + 45; 3y = 30; y =10
Problems with Many Solutions and No Solutions. I will put a problem on the board and say “Let’s see what you come up with.”
This problem ends up with either 10 = 10, or 3y = 3y. I inform the students that this type of problem is called an “identity” and it is always true for all numbers; that is, no matter what y equals, the equation will always be true. I inevitably get the following question whose basic form is as follows: “So if we get a problem like this on a test, do we put ‘all numbers’?” To which I answer “Yes, but you still have to show your work.”
I then show them a problem that has no solution:
Solving this leads to 14 = 32, which we know to be false. “When you solve an equation and get an answer that makes no sense, the equation has no solutions. Again, someone will ask what the appropriate answer should be on a test. “You tell me,” I’ll say.
“No solution?”
“Correct.” I pause for dramatic effect. “But you still have to show your work.”
I then have them do some examples, writing three or four problems on the board and having students solve them at their seats. There will be a “no solution” and “all numbers” problem as well as one which has a solution. I walk around the room to see who gets them correct and ask them to come up to the board to work them out for the class.
Word Problems. Although textbooks provide some word problems, I find that this lesson presents sufficient challenge to students as is. Adding word problems to the mix is likely to overwhelm them and they will not be receptive to strategies. I provide more word problems at a later time when they have absorbed and mastered the techniques discussed in this lesson.
Homework. Problems should focus on the various types of multi-step problems covered in this lesson as well as those presented earlier lessons. Some problems should be of the “no solution” and “all numbers” type.