Linear Equations

Linear Equations in One Variable

A linear equation in one variable has the form $ax + b = 0$, where $a \neq 0$. The highest power of $x$ is 1, meaning the equation describes a single straight line when plotted. Solving it means finding the one value of $x$ that makes the statement true.

$$ax + b = 0 \quad \Rightarrow \quad x = -\frac{b}{a}$$

Rearrange the equation to isolate $x$ by moving the constant to the other side and dividing by the coefficient.

$a$=Coefficient of $x$ (must be non-zero)(unitless)

$b$=Constant term(unitless)

$x$=The unknown variable(unitless)

$a = 0$

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If $b \neq 0$, there is no solution (contradiction). If $b = 0$, every $x$ works (identity).

Degree: The highest power of the variable is 1 — that is what makes it "linear".

Unique Solution: A linear equation in one variable has exactly one solution (when $a \neq 0$).

Transpose Rule: Moving a term across the equals sign flips its sign: $x + 3 = 7$ becomes $x = 7 - 3$.

Not every equation can be solved. A conditional equation is true for exactly one value, while an identity is true for every value of the variable. An inconsistent equation has no solution at all.

$$\text{Identity: } 2(x + 3) = 2x + 6 \qquad \text{Inconsistent: } x + 1 = x + 2$$

After simplification, if the variable cancels out, check what remains to classify the equation.

$0 = 0$=Identity — true for all $x$(unitless)

$$0 = k$ (where $k \neq 0$)$=Inconsistent — no solution(unitless)

$ax + b = ax + b$

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Identity — simplifies to $0 = 0$.

$ax + b = ax + c$ (where $b \neq c$)

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Inconsistent — simplifies to $b = c$, which is false.

Conditional: The normal case — one unique solution, e.g., $3x - 6 = 0$ gives $x = 2$.

Identity: Both sides simplify to the same expression. Often appears when you cross-multiply fractions that are actually equal.

Inconsistent: Both sides simplify to different constants. The variable vanishes, leaving a false statement.

Equations involving fractions can be cleared quickly using cross multiplication. Multiply diagonally across the equals sign to eliminate denominators.

$$\frac{a}{b} = \frac{c}{d} \quad \Rightarrow \quad ad = bc$$

Cross-multiplying removes both fractions in one step, converting the equation into a polynomial form.

$a, b$=Numerator and denominator of the left fraction(unitless)

$c, d$=Numerator and denominator of the right fraction(unitless)

$b = 0$ or $d = 0$

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The original fraction is undefined — the equation has no meaning.

Single Fraction: $\frac{2x + 1}{3} = 5$ — multiply both sides by 3 to get $2x + 1 = 15$.

Two Fractions: $\frac{x + 2}{3} = \frac{x - 1}{4}$ — cross multiply to get $4(x + 2) = 3(x - 1)$.

Watch the Signs: When clearing negative denominators, the signs flip: $\frac{x}{-2} = 3$ becomes $x = -6$.

Equation Solving Checklist

Remove brackets by distributing

Clear fractions (LCD method or cross multiply)

Collect like terms on each side

Transpose variables to one side, constants to the other

Solve for the variable

Verify by substitution

Linear Equations in Two Variables

A linear equation in two variables relates $x$ and $y$ through the form $ax + by + c = 0$. Each equation represents a straight line on the coordinate plane, and any point on that line is a solution — an $(x, y)$ pair that satisfies the equation.

$$ax + by + c = 0 \quad (a, b \text{ not both zero})$$

Two variables need two equations to find a unique solution. A single equation has infinitely many solutions (every point on the line).

$a, b$=Coefficients of $x$ and $y$(unitless)

$c$=Constant term(unitless)

$(x, y)$=A solution — coordinates of a point on the line(unitless)

$a = 0$

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Horizontal line: $by + c = 0$, i.e., $y = -c/b$.

$b = 0$

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Vertical line: $ax + c = 0$, i.e., $x = -c/a$.

Infinite Solutions: A single linear equation in two variables describes an entire line — every point on that line is a valid $(x, y)$ pair.

Unique Solution Requires Two Equations: Two non-parallel lines intersect at exactly one point, giving one $(x, y)$ solution.

[Parallel Lines]: If two lines have the same slope but different intercepts, they never intersect — no solution.

[Coincident Lines]: If two equations describe the same line (one is a multiple of the other), every point is a solution — infinitely many.

The slope-intercept form $y = mx + c$ is the most useful way to write a linear equation because it immediately reveals the slope and the y-intercept of the line.

$$y = mx + c$$

The slope $m$ controls how steeply the line rises, and $c$ tells you where it crosses the y-axis.

$m$=Slope (rate of change: rise/run)(unitless)

$c$=y-intercept (value of $y$ when $x = 0$)(unitless)

$(0, c)$=The point where the line crosses the y-axis(unitless)

$m > 0$

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Line rises from left to right.

$m < 0$

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Line falls from left to right.

$m = 0$

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Horizontal line: $y = c$.

Converting to Slope-Intercept: From $ax + by + c = 0$, solve for $y$: $y = -\frac{a}{b}x - \frac{c}{b}$, so $m = -\frac{a}{b}$.

Parallel Lines: Same slope $m$, different $c$. Example: $y = 2x + 1$ and $y = 2x - 3$.

Perpendicular Lines: Slopes are negative reciprocals: $m_1 \cdot m_2 = -1$. Example: $y = 3x$ and $y = -\frac{1}{3}x$.

Forms of a Linear Equation

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Standard: $ax + by + c = 0$ — best for classification and comparison

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Slope-Intercept: $y = mx + c$ — best for graphing and identifying slope

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Point-Slope: $y - y_1 = m(x - x_1)$ — best when given a point and slope

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Two-Point: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$ — best when given two points

Solving Simultaneous Equations

The substitution method solves one equation for one variable, then plugs that expression into the other equation. This gives a single equation in one variable, which you solve normally.

$$\text{From Eq. 1: } x = \frac{c - by}{a} \quad \Rightarrow \quad \text{Substitute into Eq. 2}$$

Substitution turns a system of two equations into one equation in one variable, which can be solved directly.

$x$=Variable isolated from one equation(unitless)

$y$=The other variable (still present in Eq. 2)(unitless)

Step 1 — Isolate: Pick the variable with coefficient 1 (or simplest coefficient) and solve for it in one equation.

Step 2 — Substitute: Replace that variable in the second equation with the expression you found.

Step 3 — Solve: The second equation now has one variable — solve it, then back-substitute to find the other variable.

Best For: When one equation already has a variable with coefficient 1, or when coefficients are awkward (e.g., 7 and 13).

The elimination method adds or subtracts the two equations so that one variable cancels out entirely. You may need to multiply one or both equations by constants to make the coefficients match.

$$\begin{aligned} a_1 x + b_1 y &= c_1 \\ a_2 x + b_2 y &= c_2 \end{aligned}$$

By aligning coefficients and adding/subtracting, one variable is eliminated, leaving a single-variable equation.

$a_1, a_2$=Coefficients of $x$ in the two equations(unitless)

$b_1, b_2$=Coefficients of $y$ in the two equations(unitless)

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

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Parallel lines — no solution (inconsistent system).

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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Coincident lines — infinitely many solutions.

Match Coefficients: Multiply each equation so that the coefficients of the variable you want to eliminate are equal (or negatives).

Add or Subtract: If matched coefficients have the same sign, subtract. If opposite signs, add.

LCM Shortcut: To eliminate $x$ when coefficients are 3 and 5, multiply the first by 5 and the second by 3.

Best For: When both equations are in standard form ($ax + by = c$) and coefficients are clean integers or easy fractions.

Age Problems

Age problems describe relationships between people's ages at the same or different times. The key insight is that age difference stays constant — if Ali is 5 years older than Sara now, he will always be 5 years older.

$$\text{Current age of A} = \text{Current age of B} + d$$

Age difference $d$ is constant across all time periods. Use this invariant to link present and past/future ages.

$d$=Fixed age difference between two people(years)

Define Variables Clearly: Let $x$ = Ali's current age, $y$ = Sara's current age. Write the constant difference as one equation.

Future/Past: "In 5 years" means add 5 to each age. "5 years ago" means subtract 5.

Avoid Two Variables: If possible, express one person's age in terms of the other: let Ali's age = $x$, then Sara's = $x - d$.

Number and Digit Problems

A two-digit number has a tens digit $t$ and a units digit $u$. The number itself equals $10t + u$. Swapping the digits gives $10u + t$.

$$\text{Number} = 10t + u, \qquad \text{Swapped} = 10u + t$$

The place-value system means the tens digit contributes ten times as much as the units digit to the number's value.

$t$=Tens digit (0–9)(digit)

$u$=Units digit (0–9)(digit)

Digit Sum: $t + u$ = sum of digits. The problem often states this directly.

Swap Property: The difference between a number and its reverse is always a multiple of 9: $(10t + u) - (10u + t) = 9(t - u)$.

Three-Digit: For number $htu$, value = $100h + 10t + u$.

Money and Currency Problems

Money problems involve two unknowns — the count of each denomination — and two conditions — the total number of items and the total monetary value. This naturally gives a system of equations.

$$\begin{cases} x + y = N \;\; (\text{total count}) \\ a x + b y = T \;\; (\text{total value}) \end{cases}$$

One equation tracks how many items there are; the other tracks how much they are worth. Solve simultaneously.

$x, y$=Number of items of each denomination(count)

$a, b$=Value of each denomination (e.g., Rs. 10, Rs. 20)(currency)

$N$=Total number of items(count)

$T$=Total monetary value(currency)

Denomination Table: Organize information as a table — Type | Count | Value per item | Total value — to avoid confusion.

Exchange Problems: When currency is exchanged (e.g., dollars to rupees), the rate is the slope of the linear relationship.

Profit/Loss Problems: Profit on cost price uses CP as the base: $SP = CP \times (1 + \frac{r}{100})$.

Distance, Rate, and Time

The fundamental relationship between distance, speed, and time is $D = S \times T$. Most travel problems are built by expressing distance in two different ways and setting them equal.

$$D = S \times T, \qquad S = \frac{D}{T}, \qquad T = \frac{D}{S}$$

Given any two of the three quantities, you can find the third. Units must be consistent throughout.

$D$=Distance traveled(km, m, miles)

$S$=Speed (rate of travel)(km/h, m/s)

$T$=Time taken(hours, seconds)

Same Distance, Different Speeds: If two people travel the same distance, $S_1 T_1 = S_2 T_2$. The slower person takes more time.

Round Trip: The distance out equals the distance back, but speed may differ. Set $S_{\text{out}} \times T_{\text{out}} = S_{\text{back}} \times T_{\text{back}}$.

Head Start: If one person starts earlier, their extra time is $T_{\text{start}}$. They travel $S \times T_{\text{start}}$ before the other begins.

Meeting Problem: When two people travel toward each other, their combined rate closes the gap: $T = \frac{D}{S_1 + S_2}$.

Work-Rate Problems

If a person can complete a job in $t$ hours, their work rate is $\frac{1}{t}$ of the job per hour. When multiple people work together, their rates add up.

$$\text{Work done per hour} = \frac{1}{t}, \qquad \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_{\text{together}}}$$

Rates are additive — combine individual work rates to get the team rate, then take the reciprocal to find the time to complete the job together.

$t_1, t_2$=Time for each person to complete the job alone(hours)

$\frac{1}{t_1}, \frac{1}{t_2}$=Individual work rates (fraction of job per hour)(job/hour)

$t_{\text{together}}$=Time for both working together(hours)

One person leaves mid-job

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Split the work into two phases: together phase (both rates) and solo phase (one rate). Set up separate time equations for each phase.

Pipe Problems: An inlet pipe fills at rate $\frac{1}{t_\text{fill}}$, a drain empties at rate $\frac{1}{t_\text{drain}}$. Together: $\frac{1}{t_\text{fill}} - \frac{1}{t_\text{drain}} = \frac{1}{t_\text{net}}$.

Partial Completion: If workers finish $\frac{2}{3}$ of a job together and one leaves, find the time for the remaining $\frac{1}{3}$ done by the remaining worker.

Negative Work: A leak or drain works against filling — subtract its rate from the total.