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Secondary Math I · Module 4 · Equations and Inequalities

4.1 — Cafeteria Actions and Reactions

Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 4, Section 4.1
Independently created to support students using MVP Math. Not official publisher material and not affiliated with or endorsed by the Mathematics Vision Project.

Human review completed .

Solve multi-step equations by undoing operations, and explain why each step keeps the same solution.

1. In class: A Develop Understanding Task

In your MVP packetSection 4.1 · Main task · pp. 1–2 · Problems 1–3

Use the cafeteria actions to understand and justify the steps in solving an equation.

What does it mean to solve an equation?

An equation states that two expressions are equal. A solution is a value that makes the equation true when substituted for the variable. For example, 5 is a solution of \(3x+2=17\), because \(3(5)+2=17\).

Solving means finding the value or values that make the statement true. We do this by writing simpler equivalent equations—equations with the same solutions—until the unknown value is clear.

For instance, from \(3x+2=17\), subtract 2 from both sides to get \(3x=15\). Then divide both sides by 3 to get \(x=5\). Subtracting 2 from only one side would change the equation.

Worked example 1 — How many trays were in each carton?

In your MVP packetSection 4.1 · Main 1 · p. 1

In the packet, Elvira unloads 4 cartons, students take 24 trays, and the remaining trays are divided equally among 3 serving lines. Each line ends up with 12 trays. Let \(x\) be the number of trays in each carton.

ActionNumber of trays
Unload 4 cartons\(4x\)
24 trays are taken\(4x-24\)
Divide the remaining trays among 3 lines\(\frac{4x-24}{3}\) per line

Since each line has 12 trays, the equation is

\[\frac{4x-24}{3}=12\]

Inverse operations undo one another: adding undoes subtracting, and multiplying undoes dividing. Here the last action was dividing by 3. Undo that first, then undo subtracting 24, then multiplying by 4.

StepEquivalent equationMeaning in the story
Multiply both sides by 3\(4x-24=36\)There were 36 trays before they were split among the lines.
Add 24 to both sides\(4x=60\)There were 60 trays before students took 24.
Divide both sides by 4\(x=15\)Each carton contained 15 trays.

Check: \(\frac{4(15)-24}{3}=\frac{36}{3}=12\). The answer fits both the equation and the story.

Worked example 2 — Why the order of actions matters

In your MVP packetSection 4.1 · Main 2 · pp. 1–2

The Post-It notes describe doubling the number of students, adding 4 students, and leaving one-third of the group at the table. In the order printed, the final number is 12:

\[\frac{2x+4}{3}=12\]

Undo the actions in reverse order: multiply both sides by 3, subtract 4 from both sides, and divide both sides by 2.

\[\begin{aligned}\frac{2x+4}{3}&=12\\2x+4&=36\\2x&=32\\x&=16\end{aligned}\]

There were 16 students initially. Checking the actions gives \(16\to32\to36\to12\).

Now add 4 before doubling. This gives \(\frac{2(x+4)}{3}=12\). Undo division by 3, then multiplication by 2, then addition of 4:

\[\begin{aligned}\frac{2(x+4)}{3}&=12\\2(x+4)&=36\\x+4&=18\\x&=14\end{aligned}\]

This time, the starting number is 14 students. The actions give \(14\to18\to36\to12\). Both stories end at 12, but they begin with different numbers. The operation order changes the equation and can change its solution.

Read the structure before choosing a step

In your MVP packetSection 4.1 · Main 3 · p. 2

Parentheses and fraction bars show which operations apply to a whole group. In \(\frac{2(x+4)}{3}\), addition of 4 comes before doubling. In \(\frac{2x+4}{3}\), doubling comes before addition of 4. In both, division by 3 is last.

Worked example 3 — A different final operation

Original equation from Main 3: \(2\left(\frac{x}{3}+4\right)=12\)

The actions are: divide the starting number by 3, add 4, then double. Undo the final doubling first:

\[\begin{aligned}2\left(\frac{x}{3}+4\right)&=12\\\frac{x}{3}+4&=6\\\frac{x}{3}&=2\\x&=6\end{aligned}\]

The steps divide both sides by 2, subtract 4 from both sides, and multiply both sides by 3. Checking gives \(6\to2\to6\to12\).

The first note always describes the unknown starting group, and the last note says 12 remain. The order of the three middle notes is:

Equation, in packet orderMiddle three actionsStarting students
\(\frac{2(x+4)}{3}=12\)Add 4 → double → divide by 3\(x=14\)
\(2\left(\frac{x}{3}+4\right)=12\)Divide by 3 → add 4 → double\(x=6\)
\(\frac{2x+4}{3}=12\)Double → add 4 → divide by 3\(x=16\)

Choose a step by what it undoes. You can also rewrite an expression using distribution, as in Set below. Whatever method you use, each new equation must have the same solutions as the previous one.

2. Ready — Graph lines and check points

In your MVP packetSection 4.1 · Ready · p. 3 · Problems 1–6

For a refresher on drawing a line through two well-separated points, see Section 3.2. An ordered pair is a solution when substituting its x- and y-coordinates makes the equation true. Try the problems before checking the answers.

Check your work — Ready 1–6 answer key

Plots not shown. The two points for each line are sample plotting points; other correct choices are valid.

  1. Given point: yes; \(5(1)-2=3\)

    Two plotting points: \((-1,-7)\) and \((2,8)\)

  2. Given point: no; at \(x=0\), the equation gives \(y=8\), not 7.

    Two plotting points: \((-4,10)\) and \((8,4)\)

  3. Given point: yes; \(-2+4=2\)

    Two plotting points: \((-4,8)\) and \((8,-4)\)

  4. Given point: yes; \(1+2=3\)

    Two plotting points: \((-8,-6)\) and \((6,8)\)

  5. Given point: yes; \(\frac52(2)-7=-2\)

    Two plotting points: \((0,-7)\) and \((6,8)\)

  6. Given point: no; at \(x=2\), the equation gives \(y=-\frac83\), not −5.

    Two plotting points: \((-6,8)\) and \((6,-8)\)

3. Set — Equivalent expressions and equations

In your MVP packetSection 4.1 · Set · p. 4 · Problems 7–15

Equivalent expressions have the same value for every value of the variable. Rewriting an expression does not change its value, so it can also be used within an equation without changing its solutions.

Use parentheses and fraction bars carefully

The distributive property multiplies each term in a group: \(a(b+c)=ab+ac\). A minus sign before parentheses subtracts every term in the group. Set 7 compares \(14-(3a+2)\) with \(14-3a-2\); they are equivalent because both the \(3a\) and the 2 are subtracted.

Likewise, dividing a sum or difference by a number divides every term. Set 9 uses

\[\frac{x-7}{4}=\frac{x}{4}-\frac74\]

In Set 10, that same reasoning gives

\[\frac{3(w-9)}5=\frac{3w-27}5=\frac{3w}5-\frac{27}5\]

The final term is \(-\frac{27}5\), not \(-27\). The fraction bar applies to both terms in the numerator.

Start by clearing the denominator

The earlier problems ask you to rewrite expressions, so splitting a fraction into separate terms can be useful there. When solving an equation, it is often easier to multiply both sides by the denominator first. This removes the fraction before you do the rest of the arithmetic.

Worked example 4 — Clear the denominator before solving

Original problem — Set 13: Solve \(\frac{4(x-2)}5=20\).

Worked solution: Multiply both sides by 5, then divide by 4 and add 2:

\[\begin{aligned}\frac{4(x-2)}5&=20\\4(x-2)&=100\\x-2&=25\\x&=27\end{aligned}\]

There is no need to distribute the 4 or divide each term in the numerator by 5.

Worked example 5 — Keep the numerator together

Original problem — Set 15: Solve \(\frac{4x-2}5=20\).

Worked solution: Multiply both sides by 5, then add 2 and divide by 4:

\[\begin{aligned}\frac{4x-2}5&=20\\4x-2&=100\\4x&=102\\x&=\frac{102}4=\frac{51}2=25.5\end{aligned}\]

Compare the two examples: \(4(x-2)\) and \(4x-2\) are different expressions, so the equations have different solutions.

In Set 14, \(4(\frac{x}5-2)=20\), dividing both sides by 4 first is also convenient. Choose steps that simplify the equation; you do not have to split a fraction into separate terms.

Study the examples, then try Set independently before checking the key.

Check your work — Set 7–15 answer key
  1. Yes. Subtracting the group subtracts both terms: \(14-(3a+2)=14-3a-2\).
  2. Yes. \(2(2b-5)=4b-10\) by distribution.
  3. Yes. Dividing the whole numerator by 4 gives \(\frac{x}{4}-\frac74\).
  4. No. \(\frac{3(w-9)}5=\frac{3w}5-\frac{27}5\), which differs from \(\frac{3w}5-27\).
  5. No. \(3(x-5)=3x-15\). The left sides \(3x-15\) and \(3x-5\) differ by 10 for every \(x\), so they cannot both equal 35 at the same \(x\).
  6. Circle the second and third expressions: \(\frac{4t}2-10\) and \(2t-10\). Both simplify to \(2t-10\).
  7. \(x=27\)
  8. \(x=35\)
  9. \(x=\frac{51}2=25.5\)

4. Go — Check proposed solutions

In your MVP packetSection 4.1 · Go · p. 4 · Problems 16–21

Substitute the proposed value into both sides, evaluate, and check whether they are equal. See the solution definition above for a reminder. Try the problems on your own first; the key includes substitution work because the packet asks you to show your work.

Packet clarification — Go 21

The packet gives \(b=5\), but the equation uses \(x\). To check the likely intended question, use \(x=5\).

Check your work — Go 16–21 answer key
  1. Yes. \(4(-3)+3=-9\), matching the right side.
  2. Yes. \(\frac34(\frac43)+\frac12=1+\frac12=\frac32\), matching the right side.
  3. No. \(2.5(2)-10=-5\), which is not \(-0.5\).
  4. No.

    Left: \(2(5-2(-5))=2(15)=30\)

    Right: \(20-2((-5)-1)=20-2(-6)=32\)

  5. Yes.

    Left: \(4(\frac14)=1\)

    Right: \(\frac14+\frac34=1\)

  6. With the intended value \(x=5\): yes.

    Left: \(6(5)-2=28\)

    Right: \(4(5+2)=28\)

    As printed, \(b=5\) does not specify a value for \(x\).

Additional learning resources

ResourceWhat it supportsScope
Khan Academy — Multi-step equations reviewWorked equation-solving steps and the operations used to keep both sides equal.Supports the in-class lesson and Set. Some examples have the variable on both sides.
Khan Academy — Equivalent expressions: negative numbers & distributionRecognizing equivalent expressions using distribution and negative signs.Close match to Set 7–12.

Teacher’s review of the source packet

Go 21 · p. 4: The given value is \(b=5\), while the equation is \(6x-2=4(x+2)\). The supplement identifies the mismatch and checks the likely intended value \(x=5\). No other substantial mathematical correction is needed.

Coverage — Section 4.1

Packet locationSkills supportedWhere supported
Main 1 · p. 1Translate actions to an equation; reverse operations and justify equivalent equationsDefinitions and tray example
Main 2–3 · pp. 1–2Operation order, grouping, and different equation structuresSeating examples and note-order table
Ready 1–6 · p. 3Graph linear equations and test ordered pairsEarlier graphing link, checking reminder, and answer key; plots omitted
Set 7–12 · p. 4Equivalent expressions, distribution, fraction bars, and equivalent equationsSet explanations and reasoned key
Set 13–15 · p. 4Solve grouped multi-step equationsIn-class examples, two Set examples clearing denominators, and key
Go 16–21 · p. 4Check proposed solutions by substitutionReminder and key showing work; Go 21 interpretation labeled

Source: Mathematics Vision Project, Secondary Mathematics One, Module 4: Equations and Inequalities (May 2016), Section 4.1, printed pp. 1–4; teacher notes for 4.1. Packet equations and short excerpts are attributed under CC BY 4.0. Equations use MathJax and require an internet connection.

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