Day 12 — Linear equations

Updated

July 30, 2026

Day 12 — Linear equations

Stage II · concept day
Goal: Use equality axioms to solve multi-step linear equations; classify identities, contradictions, and conditionals; solve word problems; rearrange formulas.

Why this matters

Solving $ax+b=c$ is the prototype of state equations and constraint solving. Every unit conversion, break-even model, and “solve for $n$ in a complexity identity” uses the same reversible steps. Mistakes are usually illegal operations (dividing by zero, losing solutions) rather than arithmetic.

Theory

Equality axioms / properties

Equality on numbers is an equivalence relation and respects operations:

Reflexive: $a=a$
Symmetric: $a=b\Rightarrow b=a$
Transitive: $a=b$ and $b=c\Rightarrow a=c$
Addition property: $a=b\Rightarrow a+c=b+c$
Multiplication property: $a=b\Rightarrow ac=bc$
Substitution: if $a=b$, either may replace the other in expressions

Cancellation (careful): if $ac=bc$ and $c\neq 0$, then $a=b$. If $c=0$, no information.

Linear equations in one variable

A linear equation in $x$ can be written \[ax+b=0\qquad\text{or}\qquad ax+b=cx+d\] with $a,b,c,d$ constants, $x$ appearing only to the first power (after simplification).

Solution strategy:

Expand / clear parentheses.
Combine like terms on each side.
Collect variable terms on one side, constants on the other (add inverses).
Divide/multiply by the coefficient of $x$ if nonzero.
Check in the original equation.

Conditional, identity, contradiction

After simplifying $ax+b=cx+d$ to $px=q$ form:

Case	Meaning	Solution set
$p\neq 0$	Conditional equation	$\{q/p\}$ one solution
$p=0$ and $q=0$	Identity (always true)	all reals (or stated domain)
$p=0$ and $q\neq 0$	Contradiction (never true)	empty set $\emptyset$

Examples: $2x+1=2x+1$ identity; $2x+1=2x+3$ contradiction; $2x+1=5$ conditional $x=2$.

Clearing fractions and decimals

Multiply both sides by a common denominator (LCD) of all fractions to obtain an integer-coefficient equation.
For decimals, multiply by a power of $10$.

Word problems → equations

Define variable(s) with units.
Translate relationships (sum, difference, rate $\times$ time, percent of).
Solve.
Check reasonableness and original wording.

Classic patterns: total = parts; distance $=rt$; mixture; consecutive integers; “one number is $3$ more than twice another.”

Formula rearrangement

Treat the target variable as $x$ and apply the same properties.

Example: $v=u+at$ solve for $t$: $t=\dfrac{v-u}{a}$ if $a\neq 0$.
$F=\dfrac{9}{5}C+32$ solve for $C$: $C=\dfrac{5}{9}(F-32)$.

Equivalent equations

Two equations are equivalent if they have the same solution set.
Allowed steps (add same quantity, multiply by nonzero constant) produce equivalent equations.
Multiplying by an expression that can be zero may add extraneous solutions; dividing may lose solutions—more critical for nonlinear/rational cases (Day 17).

Worked examples

Example 1 — Two-step

$3x-7=11$. Add $7$: $3x=18$. Divide by $3$: $x=6$. Check: $18-7=11$.

Example 2 — Variables both sides

$5x+3=2x-9$. Subtract $2x$: $3x+3=-9$. Subtract $3$: $3x=-12$. $x=-4$.

Example 3 — Distribute

$2(3x-4)+5=3(x+1)$. $6x-8+5=3x+3$. $6x-3=3x+3$. $3x=6$. $x=2$.

Example 4 — Fractions

$\dfrac{x}{2}+\dfrac{x}{3}=5$. Multiply by $6$: $3x+2x=30$. $5x=30$. $x=6$.

Example 5 — Identity

$4(x-1)+2=2(2x+1)-4$. Left $4x-4+2=4x-2$. Right $4x+2-4=4x-2$. Equal for all $x$.

Example 6 — Contradiction

$3(x+2)=3x+5$. $3x+6=3x+5$. $6=5$. Never. $\emptyset$.

Example 7 — Word: consecutive

Three consecutive integers sum to $54$. Let $n+(n+1)+(n+2)=54$. $3n+3=54$. $3n=51$. $n=17$. Triple $17,18,19$.

Example 8 — Rate

Car $A$ at $60$ km/h, car $B$ at $80$ km/h, start together opposite directions. When $420$ km apart? $(60+80)t=420$. $t=3$ h.

Example 9 — Percent

A number $n$ increased by $15\%$ is $92$. $1.15n=92$. $n=80$.

Example 10 — Formula

$A=\frac{1}{2}(b_1+b_2)h$ solve for $b_2$: $2A/h=b_1+b_2$, so $b_2=\frac{2A}{h}-b_1$ ($h\neq 0$).

Example 11 — Decimals

$0.2x+0.05(100-x)=8$. Multiply by $100$: $20x+5(100-x)=800$. $20x+500-5x=800$. $15x=300$. $x=20$.

Example 12 — Literal

Solve $ax+b=cx+d$ for $x$ (assuming $a\neq c$): $(a-c)x=d-b$, $x=\dfrac{d-b}{a-c}$.

Example 13 — Check necessity

$\frac{x}{x}=1$ is not true for $x=0$; after multiplying by $x$ carelessly you might think all $x$ work. Domain matters.

Example 14 — Break-even

Cost $C=500+12n$, revenue $R=20n$. Break-even $20n=500+12n$. $8n=500$. $n=62.5$ — if $n$ must be integer, interpret context (ceil units).

Exercises

Easy

Solve $5x+2=17$.
Solve $1-2x=9$.
Solve $3(x+4)=24$.
Solve $\frac{x}{5}= -3$.
Solve for $r$: $C=2\pi r$.

Medium

Solve $4(2x-1)-3=2(x+5)+x$.
Solve $\dfrac{2x-1}{3}=\dfrac{x+4}{2}$.
Classify: $5x-1=5(x+2)-11$ as identity / conditional / contradiction; solve if applicable.
Classify $2(x+3)=2x+5$.
Word: one number is $4$ less than $3$ times another; sum is $28$. Find the numbers.
Mixture: $20$ L of $10\%$ solution mixed with $x$ L of $30\%$ to get $25\%$. Equation and $x$.
Rearrange $v^2=u^2+2as$ for $s$.
Solve $0.15x+0.10(20)=0.12(x+20)$.

Hard

Prove: if $a=b$, then $a+c=b+c$ using “same quantity added” reasoning from field axioms (informal OK).
Solve for $x$ in terms of parameters: $a(x-b)=c(x+d)$ with stated conditions when unique.
For which $k$ does $kx+1=2x+k$ have (i) one solution (ii) all $x$ (iii) no solution?
Word: digit problem—two-digit number $10t+u$, digits sum to $12$, number $= 4$ times sum of digits? Set equations carefully.
Explain why multiplying both sides of an equation by $0$ is useless; give example.
Distance: runners $3$ km apart run toward each other at $8$ and $7$ km/h. Time to meet?
Show that $x=2$ is not a solution of $x+1=x+2$ without “simplifying to $1=2$” first—by direct substitution.

Challenge / CS-flavored

Memory: $n$ structures of $24$ bytes plus $16$ byte header uses $4096$ bytes total. Solve for max $n$.
Loop: for i in 0..n-1 runs how many times? Solve $n$ if runtime proportional to $n$ equals $1000$ units when $c n=1000$ and $c=4$.
Temperature conversion chain: solve $F$ given $C$, then invert.
Fixed point: $x=0.8x+20$ (e.g. damped update). Solve $x$.
Explain “solve for $n$” in $2^n \ge 10^6$ as not linear—why Day 8/20 methods differ.

Solution sets formally

The solution set of an equation in variable $x$ over $\mathbb{R}$ is $\{x\in\mathbb{R}: \text{equation holds}\}$.
Linear case yields $\emptyset$, a singleton, or $\mathbb{R}$ (after simplification to $px=q$).

Equivalent transformations (safe list)

Safe (preserve solution set over $\mathbb{R}$ for linear forms):

Add/subtract the same expression in $x$ to both sides
Multiply/divide both sides by a nonzero constant

Unsafe without casework:

Multiply/divide by an expression that might be $0$
Square both sides (can add solutions; nonlinear)

Literal equations catalog

Formula	Solve for
$I=Prt$	$t=\frac{I}{Pr}$
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$	$f=\frac{uv}{u+v}$ (with $u+v\neq 0$)
$y-y_1=m(x-x_1)$	$m=\frac{y-y_1}{x-x_1}$ ($x\neq x_1$)
$an+b=c$	$n=\frac{c-b}{a}$ ($a\neq 0$)

Word problem schema

Uniform motion: $d=rt$, chase/together/apart cases.
Mixture: pure amounts of solute: concentration $\times$ volume.
Work: rates add when parallel.
Digits: $10t+u$ structure.
Age: “in $k$ years” shifts both ages by $k$.

Extra exercises

Solve for $x$: $a(x+1)=b(x-1)$ with $a\neq b$.
For which $m$ is $mx+1=2x+m$ an identity?
Two trains $150$ km apart head toward each other at $70$ and $80$ km/h; time to meet.
Clear decimals: $0.05x+0.1(20-x)=1.2$.
Show that if two linear equations are equivalent, their graphs (as lines in $x$ only—degraded) match solution sets on the number line.

CS connection

Constraint solving, unit tests as equations, and configuration “solve for thread count given budget” are linear equation patterns. Formula rearrangement is API design: which quantity is input. Identity vs contradiction mirrors tautology vs unsatisfiable boolean constraints (Stage III).

Common pitfalls

Pitfall	What to do instead
Moving terms without sign change	Adding the additive inverse
Dividing by variable coefficient that might be $0$	Case-split on parameters
Forgetting to check	Substitute back
Losing domain restrictions	Track steps that multiply by expressions
Word problem variable undefined	Sentence: “let $x$ be …” with units
Treating identity as “no solution”	All values work

Fully worked extra examples

E15 — Fractions both sides.
$\frac{x+1}{3}=\frac{2x-1}{5}$. Cross: $5(x+1)=3(2x-1)$, $5x+5=6x-3$, $8=x$.

E16 — Identity recognition.
$2(x-4)+8=2x$: left $2x-8+8=2x$. Identity; solution set $\mathbb{R}$.

E17 — Literal with factoring.
$ax-c=bx+d$ → $ax-bx=d+c$ → $x(a-b)=d+c$. If $a\neq b$, $x=\frac{d+c}{a-b}$. If $a=b$ and $d+c=0$, all $x$; if $a=b$ and $d+c\neq 0$, empty.

E18 — Mixture.
$10$ L of $20\%$ acid + $x$ L of $50\%$ acid → $30\%$ mixture:
$0.2\cdot 10+0.5x=0.3(10+x)$, $2+0.5x=3+0.3x$, $0.2x=1$, $x=5$.

End-of-day synthesis problems

S1. Solve $\frac{2x-1}{4}-\frac{x+3}{6}=1$ by LCD.

S2. Classify and solve $k x+3=2x+k$ in cases on $k$.

S3. Word: ticket prices adults $\$a$ children $\$c$; $3$ adults $+2$ children $=\$46$; $1$ adult $+4$ children $=\$36$. Wait—single variable day: instead, one number is thrice another; sum $64$. Find them.

S4. Rearrange $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$ for $R_1$.

S5. Prove: adding $c$ to both sides preserves solutions (equality axiom narrative).

S6. Solve $0.25(x-3)=0.1x+1.1$.

S7. Identity check: expand both sides of $2(x+3)-x=x+6$.

S8. Distance: walk $4$ km/h for $t$ hours and bike $12$ km/h for $2-t$ hours cover $20$ km total. Equation for $t$.

Checkpoint

Use $+c$ and $\times c$ ($c\neq 0$) properties deliberately
Solve multi-step linear equations with parentheses and fractions
Classify identity / contradiction / conditional
Translate a word problem to an equation and solve
Rearrange a multi-variable formula for a chosen variable
Case-split a parameter equation (S2 style)

Write two takeaways in your own words.

Deep dive — linear equations, cases, word models

D1 — Clear fractions with LCD carefully.
$\dfrac{2x-1}{4}-\dfrac{x+3}{6}=1$. LCD $12$:
$3(2x-1)-2(x+3)=12$, $6x-3-2x-6=12$, $4x-9=12$, $4x=21$, $x=\frac{21}{4}$.
Check by substitution into the original.

D2 — Parameter cases (full write-up).
$kx+3=2x+k$. Collect: $kx-2x=k-3$, $x(k-2)=k-3$.
- If $k\neq 2$: unique $x=\dfrac{k-3}{k-2}$.
- If $k=2$ and $k-3=0$ i.e. $2=3$ false: wait, when $k=2$, right side $k-3=-1\neq 0$ → $0\cdot x=-1$ contradiction → $\emptyset$.
- Identity would need $k=2$ and $k-3=0$ simultaneously—impossible here. (Compare $kx+k=2x+2$ for a true identity case when $k=2$.)

D3 — Literal formula rearrangement.
$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$. Solve for $R_1$:
$\dfrac{1}{R_1}=\dfrac{1}{R}-\dfrac{1}{R_2}=\dfrac{R_2-R}{R R_2}$, so $R_1=\dfrac{R R_2}{R_2-R}$ when $R_2\neq R$.

D4 — Mixture model.
$10$ L of $20\%$ acid + $x$ L of $50\%$ → $30\%$ blend:
$0.2\cdot 10+0.5x=0.3(10+x)$, $2+0.5x=3+0.3x$, $0.2x=1$, $x=5$.

D5 — Motion toward each other.
Distance $150$ km, speeds $70$ and $80$ km/h: $(70+80)t=150$, $t=1$ h.
Opposite directions with same numbers: still add speeds for separation rate.

D6 — Equivalent equations discipline.
Adding the same expression to both sides preserves the solution set.
Multiplying by a nonzero constant preserves it.
Multiplying by an expression that can vanish (e.g. $x$) can add or drop solutions—defer full rational-equation care to Day 17, but never multiply a linear training equation by $0$.

Extra practice set

Solve $\dfrac{x+1}{3}=\dfrac{2x-1}{5}$ and check.
For which $m$ is $mx+1=2x+m$ an identity? Conditional? Empty?
Two numbers sum to $64$; one is thrice the other. Find them.
Rearrange $v^{2}=u^{2}+2as$ for $a$.
Clear decimals: $0.05x+0.1(20-x)=1.2$.

Tomorrow

Day 13 — Inequalities. Flip rule when multiplying by negatives; compound inequalities; interval notation; absolute value inequalities; 1D systems of inequalities.

Formula	Solve for
\(I=Prt\)	\(t=\frac{I}{Pr}\)
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)	\(f=\frac{uv}{u+v}\) (with \(u+v\neq 0\))
\(y-y_1=m(x-x_1)\)	\(m=\frac{y-y_1}{x-x_1}\) (\(x\neq x_1\))
\(an+b=c\)	\(n=\frac{c-b}{a}\) (\(a\neq 0\))