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Secondary Math I · Module 4 · Equations and Inequalities

4.3 — Solving Equations Literally

Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 4, Section 4.3
Independently created to support students using MVP Math. Not official publisher material and not affiliated with or endorsed by the Mathematics Vision Project.

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Practice solving equations with numbers and letters, explain each step, and see how operations affect inequalities.

1. In class: A Practice Understanding Task

In your MVP packetSection 4.3 · Main task · p. 10 · Problems 1–10

The same steps work with numbers or letters

The packet pairs a numerical equation with a similar equation containing another variable. Use the methods from Section 4.1 and Section 4.2: isolate the requested variable by doing the same operation to both sides. Multiplying by the denominator first often keeps the arithmetic simpler.

For example, Main 1 and 2 both begin by multiplying by 5:

Main 1Main 2
\(\frac{3x+2}{5}=7\)\(\frac{3x+2y}{5}=7\)
\(3x+2=35\)\(3x+2y=35\)
\(3x=33\)\(3x=35-2y\)
\(x=11\)\(x=\frac{35-2y}{3}\)

The steps are the same: multiply by 5, subtract the added term, then divide by 3. In the second equation, the answer depends on \(y\).

Worked example 1 — Collect the x-terms on one side

Original problem — Main 8: Solve \(2(3x+4y)=4x+12y\) for \(x\).

When \(x\) occurs on both sides, first distribute, then subtract an x-term from both sides so that the remaining x-terms are together:

\[\begin{aligned}2(3x+4y)&=4x+12y\\6x+8y&=4x+12y\\2x+8y&=12y\\2x&=4y\\x&=2y\end{aligned}\]

After distributing, subtract \(4x\) from both sides, subtract \(8y\) from both sides, and divide both sides by 2.

Explain what each step undoes

Main 9–10 ask for explanations as well as algebra. Name the operation and explain why it helps isolate \(x\).

Worked example 2 — Clear a denominator in a literal equation

Original problem — Main 9: Solve \(\frac{ax+b}{c}-d=e\) for \(x\), explaining each step.

EquationReason for the step
\(\frac{ax+b}{c}-d=e\)Original equation
\(\frac{ax+b}{c}=e+d\)Add \(d\) to both sides to undo subtracting \(d\).
\(ax+b=c(e+d)\)Multiply both sides by \(c\) to remove the denominator.
\(ax=c(e+d)-b\)Subtract \(b\) from both sides.
\(x=\frac{c(e+d)-b}{a}\)Divide both sides by \(a\) to leave \(x\) alone.

Here \(a\) and \(c\) must be nonzero.

Worked example 3 — Undo a square root

Original problem — Main 10: Solve \(r\sqrt{\frac{mx}{n}}+s=t\) for \(x\), explaining each step.

First isolate the square root. Then square both sides: squaring undoes a square root, as in \((\sqrt{9})^2=9\).

EquationReason for the step
\(r\sqrt{\frac{mx}{n}}+s=t\)Original equation
\(r\sqrt{\frac{mx}{n}}=t-s\)Subtract \(s\) from both sides.
\(\sqrt{\frac{mx}{n}}=\frac{t-s}{r}\)Divide both sides by \(r\).
\(\frac{mx}{n}=\left(\frac{t-s}{r}\right)^2\)Square both sides to remove the square root.
\(mx=n\left(\frac{t-s}{r}\right)^2\)Multiply both sides by \(n\).
\(x=\frac{n}{m}\left(\frac{t-s}{r}\right)^2\)Divide both sides by \(m\).

Note: A square root is nonnegative, so \(\frac{t-s}{r}\) must be greater than or equal to 0 for the original equation to have a real solution. The steps above assume \(r\), \(m\), and \(n\) are nonzero.

Try the in-class problems before using these results to check your work:

ProblemSolution for xProblemSolution for x
1\(x=11\)2\(x=\frac{35-2y}{3}\)
3\(x=12\)4\(x=\frac{3(11+5y)}4\)
5\(x=12\)6\(x=15-y\)
7\(x=2\)8\(x=2y\)

2. Ready — How operations affect inequalities

In your MVP packetSection 4.3 · Ready · p. 11 · Problems 1–8

An inequality compares quantities using symbols such as \(<\), \(>\), \(\le\), or \(\ge\). Adding or subtracting the same number on both sides preserves their order, as does multiplying or dividing both sides by a positive number.

For Ready 1–7, begin each row with the original \(-9<2\). Keep the printed \(<\) symbol while testing whether the resulting statement is true or false. Rows 6–7 show why that symbol must be reversed when you want to preserve a true inequality.

Check your work — Ready 1–8 answer key

  1. \(-16<-5\) — True

    Subtracting 7 preserves the order.

  2. \(6<17\) — True

    Adding 15 preserves the order.

  3. \(-19<-8\) — True

    Adding −10 preserves the order.

  4. \(-90<20\) — True

    Multiplying by positive 10 preserves the order.

  5. \(-\frac95<\frac25\) — True

    Dividing by positive 5 preserves the order.

  6. \(54<-12\) — False

    Multiplying by −6 reverses the order: the true inequality is \(54>-12\).

  7. \(3<-\frac23\) — False

    Dividing by −3 reverses the order: the true inequality is \(3>-\frac23\).

  8. Multiplying or dividing both sides by a negative number reverses the inequality. Adding or subtracting a negative number does not.

3. Set — Rearrange formulas

In your MVP packetSection 4.3 · Set · p. 11 · Problems 9–20

These problems practice the methods from Section 4.2 and the in-class examples above. Try them independently and show your steps; the key includes algebra so you can check your work. Any quantity used as a divisor must be nonzero.

Check your work — Set 9–20 answer key

  1. \(Q=25\pi h\)

    Divide by \(25\pi\): \(h=\frac{Q}{25\pi}\)

  2. \(Q=\pi r^2h\)

    Divide by \(\pi r^2\): \(h=\frac{Q}{\pi r^2}\)

  3. \(y=7m+6\)

    \(y-6=7m\)

    \(m=\frac{y-6}{7}\)

  4. \(y=mx+b\)

    \(y-b=mx\)

    \(m=\frac{y-b}{x}\)

  5. \(A=3(z+7)\)

    \(\frac A3=z+7\)

    \(z=\frac A3-7\)

  6. \(A=(z+7)w\)

    \(\frac Aw=z+7\)

    \(z=\frac Aw-7\)

  7. \(x+2=28\)

    \(x=26\)

  8. \(x+2y=28\)

    \(x=28-2y\)

  9. \(2x-45=30\) (Multiply both sides by 5.)

    \(2x=75\)

    \(x=\frac{75}2=37.5\)

  10. \(2x-45y=30\) (Multiply both sides by 5.)

    \(2x=30+45y\)

    \(x=\frac{30+45y}{2}\)

  11. \(3(x-2)=48\) (Multiply both sides by 4.)

    \(x-2=16\)

    \(x=18\)

  12. \(3(x-2y)=48\) (Multiply both sides by 4.)

    \(x-2y=16\)

    \(x=16+2y\)

4. Go — Intercepts and justified equation solutions

In your MVP packetSection 4.3 · Go · pp. 12–13 · Problems 21–32

Read intercepts from tables and graphs

An x-intercept is a point where the graph meets the x-axis, so its y-coordinate is 0. A y-intercept is a point where it meets the y-axis, so its x-coordinate is 0. For a refresher, see Section 3.3. Give each intercept as an ordered pair.

Name the operation used on both sides

Go 26–32 asks you to justify your steps. For example, “subtract 3 from both sides” uses the subtraction property of equality. Addition, multiplication, and division have corresponding properties. Distribution rewrites an expression without changing its value. The in-class examples show how to explain why a step helps.

Try Go before opening the key. For 24–25, the answers use one unit per grid spacing; the packet does not label the numerical scales.

Check your work — Go 21–32 answer key

Intercepts — 21–25

Plots not shown.

  1. x-intercept: \((2,0)\)

    y-intercept: \((0,4)\)

  2. x-intercept: \((18,0)\)

    y-intercept: \((0,-6)\)

  3. x-intercept: \((2,0)\)

    y-intercept: \((0,4)\)

  4. x-intercept: \((3,0)\)

    y-intercept: \((0,9)\)

  5. x-intercept: \((-3,0)\)

    y-intercept: \((0,4)\)

Solutions and justifications — 26–32

Equivalent steps and simplifications are also valid.

  1. \(4x=12\) — add 10 to both sides (addition property)

    \(x=3\) — divide both sides by 4 (division property)

  2. \(-27=3x\) — subtract 11 from both sides (subtraction property)

    \(x=-9\) — divide both sides by 3 (division property)

  3. \(-2x=4\) — subtract 6 from both sides (subtraction property)

    \(x=-2\) — divide both sides by −2 (division property)

  4. \(5x+3=18\) — subtract \(x\) from both sides (subtraction property)

    \(5x=15\) — subtract 3 from both sides (subtraction property)

    \(x=3\) — divide both sides by 5 (division property)

  5. \(x-10=12\) — subtract \(2x\) from both sides (subtraction property)

    \(x=22\) — add 10 to both sides (addition property)

  6. \(12x=15-3y\) — subtract \(3y\) from both sides (subtraction property)

    \(x=\frac{15-3y}{12}=\frac{5-y}4\) — divide both sides by 12 (division property), then simplify

  7. \(X=\frac9{B+7}\) — divide both sides by \(B+7\) (division property)

    Here \(B\ne-7\). The packet uses uppercase \(X\) and \(B\) in this problem.

Additional learning resources

ResourceWhat it supports
Khan Academy — Intro to equations with variables on both sidesCollecting variable terms on one side using the same operations on both sides. Supports Main 7–8 and Go 29–30.
Khan Academy — One-step inequalities examplesMultiplying and dividing inequalities, including why a negative multiplier or divisor reverses the inequality. Supports Ready.

Teacher’s review of the source packet

No substantial mathematical error was found in this section. The graph scales in Go 24–25 are unlabeled; the answer key uses the usual convention of one unit per grid spacing.

Coverage — Section 4.3

Packet locationWhere supported
Main 1–8 · p. 10Numerical/literal comparison, links to earlier methods, worked variable-on-both-sides solution, and visible checking answers
Main 9–10 · p. 10Worked solutions with reasons for each step, including undoing a square root
Ready 1–8 · p. 11Inequality rules and an answer key explaining each true/false result
Set 9–20 · p. 11Linked formula-solving instruction and an answer key with algebraic steps
Go 21–25 · p. 12Intercept reminder, earlier lesson link, and numerical answers; plots omitted
Go 26–32 · p. 13Equality-property reminder and solutions with justifications

Source: Mathematics Vision Project, Secondary Mathematics One, Module 4: Equations and Inequalities (May 2016), Section 4.3, printed pp. 10–13; teacher notes for 4.3. Packet equations and short excerpts are attributed under CC BY 4.0. Equations use MathJax.

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