In a previous lesson, I taught how to combine like terms which included expressions that had a distribution within it. In this lesson it is taken to the next step which is to combine like terms in equations, which is really just more of the same. It’s fairly straightforward and students usually take to it, although the distribution step can be confusing when it is being subtracted. For example, the expression 5 –(x -2) simplifies to 5 – x + 2. Students need to think of a minus sign outside parentheses as the same thing as -1 outside the parentheses.
This lesson also includes solving equations that contain fractions, such as:
This was previously discussed in Unit 14 for seventh grade. All terms are multiplied by a common denominator. For the equation above, all terms would be multiplied by 8, resulting in 5m + 6 = 56, which is an easier equation to solve. I go into slightly more detail in this lesson, but it is the same technique.
Warm-Ups.
1. Combine like terms. 2x + 3x Answer: 5x
2. Combine like terms. k/2 + k/6 Answer: ½ (k) + 1/6(k) = 3/6(k) + 1/6(k) = 2/3(k)
3. Solve. n/19 = 7 Answer: n = 19(7) = 133
4. Solve. -3/5(x) = 7/5 Answer: x = 7/5 ∙ (-5/3) = -21
5. Write as an equation and solve. The sum of twice a number and 16 is 86. Answer: 2x + 16 = 86; 2x = 70
Problem 2 again requires students to see k/2 as ½(k). The terms can be combined as shown, or by writing k/2 as 3k/6. This can be explained by asking students what do we multiply the denominator by to get 6. Then the numerator is multiplied by 3 as well. Problem 3 can be looked at as 1/19(n) = 7, but by now students should be shown that each side can be multiplied by 19 without rewriting it with a fractional coefficient. Problem 5 is a multi-step equation which students may have questions on. One prompt may be “What should we do with 16? Can we move it?” Doing so results in the equation 2x = 70, which is then easily solved.
Combining like terms and solving. I start out by writing an equation on the board and ask how we would solve it.
“This is similar to Problem 5 of the Warm-Ups. What did we need to do to solve this?”
They will (hopefully) recall that the numbers were moved to the other side, which is what is done here.
“Tell me what I need to do then.” I have the students dictate the steps which should be “Add 4 to each side, and then divide by 10.”
Subtracting 4 from each side results in 10x =55; x = 5.5.
“This is a two-step equation which we did in seventh grade, and which I hope you remember how to do.” (These were discussed in Unit 13 for seventh grade).
I now write another equation on the board:
“What should I do to solve this one?”
If someone says “Combine like terms” that would be a good day, indeed. Assuming this doesn’t happen, I will prompt: “Are there like terms that we can combine?”
They will undoubtedly see that 7x and 3x can be combined. In case they don’t see that the 12 and 39 can be added, I’ll prompt: “What about 12 + 39. Can anything be done there, or shall we just leave those guys hanging for dear life?”
Once terms are combined the equation looks familiar! 10x -4 = 51
“Yes, it’s the same equation we solved before, just made a bit more complicated. So it went from a two-step equation to a multi-step equation. If we have an equation we need to combine all like terms first before we begin other steps.
Examples.
1. p + 3p = 19.6 Answer: 4p = 19.6; p = 4.9
2. 2x + 3x + 45 = 90 Answer: 5x = 45; x = 9
3. 0.07t – 0.5 = 2.3
This one I work through with the class. “How can we eliminate the decimals from this equation? What would we multiply the first term by to eliminate the decimal. Most students see that multiplying 0.07t by 100 will do it.
“If we multiply one term by 100, we have to multiply all terms by 100.” This may not be immediately obvious so I rewrite it. “What we’re doing is we’re multiplying both sides by 100. If we think of it like this, you can see it:
“Now what are we left with when we do that?”
“It’s easy now.” I left them solve it; 7t = 280; t = 40
More complex problems. I write another problem on the board.
“What should we do?”
They should see that they need to distribute first. Doing so yields:
“Solve it.” Combining like terms and simplifying: x = 11
On the board and in their notebooks:
STEP 1. Distribute as necessary and combine similar terms on both sides of equation.
STEP 2. Undo addition/subtraction operations first, to isolate the variable term
STEP 3. Undo multiplication/division operations to find value of the variable
“How would you solve this one?” ¾(p) + 1/8 = 2.
I’ll usually hear someone say to find a common denominator.
“Do we really need to here?” I’ll ask. I’ll have them work in their notebooks and walk around. What I advise is to combine terms—in this case move the 1/8 to the other side. The result is ¾(p) = 2 – 1/8, which simplifies to ¾(p) = 15/8. (Some may still need help converting the 2 into eighths, and writing 2 as 2/1 helps in this regard, though I secretly wish they begin to see you just multiply by the chosen denominator. Have patience I tell myself, but some students stubbornly cling to writing the 1 in the denominator.
“The equation ¾(p) = 15/8 is solved how?” They should remember from the previous lesson to multiply both sides by the reciprocal of the coefficient, but chances are good that I still need to remind them of that. I tell myself again: Have patience, you were a novice once too. The answer is p = 5/2 in lowest terms.
“Is there a way we can do this by eliminating fractions, like we eliminated decimals? When we eliminated decimals in the last problem 0.07t – 0.5 = 2.3 we multiplied by 100. Let’s take a closer look at this. We’ll write this problem with fractions instead of decimals. What’s 0.07 as a fraction?”
I go through each decimal in the equation and end up with 7/100(t) – 5/10 = 2 3/10
“When we multiplied each term by 100, we were really doing this:”
“Cross cancel and we end up with 7t – 5 = 23. So now with ¾(p) +1/8 = 2, what should be the common denominator here?”
Hearing “eight”, I tell them “Multiply each term by 8/1 and see what you get.” Again, I check notebooks. They should get 6p + 1 = 16. Solving it, they get p = 5/2.
Examples. I work through the first one with them.
1. n – 3/5n + 6 = 27 – 9; Answer: Simplifying first: 2/5(n) + 6 = 18; 2/5(n) =12; n = 30
2. 5/8(m) + ¾ = 7 (I have them multiply by a common denominator, in this case eight) 5m + 6 = 56; 5m = 50, m = 10
3. k/2 – k/6 = 3 (I remind them that k/2 is ½(k), k/6 = 1/6(k). I advise them to do it in whatever way they want and walk around to provide guidance) 3k – k = 18, 2k = 18; k = 9
4. 57 = 8y + 25 (In this case, the equation is in a different order than they’re used to, but it’s solved in the same way as they have done) Answer: 8y = 32, y = 4
Homework. Homework should be a mix of equations with fractions, distributions, and easier one-step and two-step that don’t require combining terms, so they get all types of problems.