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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)
$a = 0$
→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)
$$0 = k$ (where $k \neq 0$)$=Inconsistent — no solution(unitless)
$ax + b = ax + b$
→Identity — simplifies to $0 = 0$.
$ax + b = ax + c$ (where $b \neq c$)
→Inconsistent — simplifies to $b = c$, which is false.
Conditional: The normal case — one unique solution, e.g., $3x - 6 = 0$ gives $x = 2$.
Identity: Both sides simplify to the same expression. Often appears when you cross-multiply fractions that are actually equal.
Inconsistent: Both sides simplify to different constants. The variable vanishes, leaving a false statement.
Equations involving fractions can be cleared quickly using cross multiplication. Multiply diagonally across the equals sign to eliminate denominators.
$$\frac{a}{b} = \frac{c}{d} \quad \Rightarrow \quad ad = bc$$
Cross-multiplying removes both fractions in one step, converting the equation into a polynomial form.
$a, b$=Numerator and denominator of the left fraction(unitless)
$c, d$=Numerator and denominator of the right fraction(unitless)
$b = 0$ or $d = 0$
→The original fraction is undefined — the equation has no meaning.
Single Fraction: $\frac{2x + 1}{3} = 5$ — multiply both sides by 3 to get $2x + 1 = 15$.
Two Fractions: $\frac{x + 2}{3} = \frac{x - 1}{4}$ — cross multiply to get $4(x + 2) = 3(x - 1)$.
Watch the Signs: When clearing negative denominators, the signs flip: $\frac{x}{-2} = 3$ becomes $x = -6$.
Equation Solving Checklist
1
Remove brackets by distributing
2
Clear fractions (LCD method or cross multiply)
3
Collect like terms on each side
4
Transpose variables to one side, constants to the other
5
Solve for the variable
6
Verify by substitution
Linear Equations in Two Variables
A linear equation in two variables relates $x$ and $y$ through the form $ax + by + c = 0$. Each equation represents a straight line on the coordinate plane, and any point on that line is a solution — an $(x, y)$ pair that satisfies the equation.
$$ax + by + c = 0 \quad (a, b \text{ not both zero})$$
Two variables need two equations to find a unique solution. A single equation has infinitely many solutions (every point on the line).
$a, b$=Coefficients of $x$ and $y$(unitless)
$c$=Constant term(unitless)
$(x, y)$=A solution — coordinates of a point on the line(unitless)
$a = 0$
→Horizontal line: $by + c = 0$, i.e., $y = -c/b$.
$b = 0$
→Vertical line: $ax + c = 0$, i.e., $x = -c/a$.
Infinite Solutions: A single linear equation in two variables describes an entire line — every point on that line is a valid $(x, y)$ pair.
Unique Solution Requires Two Equations: Two non-parallel lines intersect at exactly one point, giving one $(x, y)$ solution.
[Parallel Lines]: If two lines have the same slope but different intercepts, they never intersect — no solution.
[Coincident Lines]: If two equations describe the same line (one is a multiple of the other), every point is a solution — infinitely many.
The slope-intercept form $y = mx + c$ is the most useful way to write a linear equation because it immediately reveals the slope and the y-intercept of the line.
$$y = mx + c$$
The slope $m$ controls how steeply the line rises, and $c$ tells you where it crosses the y-axis.
$m$=Slope (rate of change: rise/run)(unitless)
$c$=y-intercept (value of $y$ when $x = 0$)(unitless)
$(0, c)$=The point where the line crosses the y-axis(unitless)
$m > 0$
→Line rises from left to right.
$m < 0$
→Line falls from left to right.
$m = 0$
→Horizontal line: $y = c$.
Converting to Slope-Intercept: From $ax + by + c = 0$, solve for $y$: $y = -\frac{a}{b}x - \frac{c}{b}$, so $m = -\frac{a}{b}$.
Parallel Lines: Same slope $m$, different $c$. Example: $y = 2x + 1$ and $y = 2x - 3$.
Perpendicular Lines: Slopes are negative reciprocals: $m_1 \cdot m_2 = -1$. Example: $y = 3x$ and $y = -\frac{1}{3}x$.
Forms of a Linear Equation
•
Standard: $ax + by + c = 0$ — best for classification and comparison
•
Slope-Intercept: $y = mx + c$ — best for graphing and identifying slope
•
Point-Slope: $y - y_1 = m(x - x_1)$ — best when given a point and slope
•
Two-Point: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$ — best when given two points
Solving Simultaneous Equations
The substitution method solves one equation for one variable, then plugs that expression into the other equation. This gives a single equation in one variable, which you solve normally.
$$\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}$$
By aligning coefficients and adding/subtracting, one variable is eliminated, leaving a single-variable equation.
$a_1, a_2$=Coefficients of $x$ in the two equations(unitless)
$b_1, b_2$=Coefficients of $y$ in the two equations(unitless)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
→Parallel lines — no solution (inconsistent system).
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
→Coincident lines — infinitely many solutions.
Match Coefficients: Multiply each equation so that the coefficients of the variable you want to eliminate are equal (or negatives).
Add or Subtract: If matched coefficients have the same sign, subtract. If opposite signs, add.
LCM Shortcut: To eliminate $x$ when coefficients are 3 and 5, multiply the first by 5 and the second by 3.
Best For: When both equations are in standard form ($ax + by = c$) and coefficients are clean integers or easy fractions.
Age Problems
Age problems describe relationships between people's ages at the same or different times. The key insight is that age difference stays constant — if Ali is 5 years older than Sara now, he will always be 5 years older.
$$\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}}}$$
Rates are additive — combine individual work rates to get the team rate, then take the reciprocal to find the time to complete the job together.
$t_1, t_2$=Time for each person to complete the job alone(hours)
$\frac{1}{t_1}, \frac{1}{t_2}$=Individual work rates (fraction of job per hour)(job/hour)
$t_{\text{together}}$=Time for both working together(hours)
One person leaves mid-job
→Split the work into two phases: together phase (both rates) and solo phase (one rate). Set up separate time equations for each phase.
Pipe Problems: An inlet pipe fills at rate $\frac{1}{t_\text{fill}}$, a drain empties at rate $\frac{1}{t_\text{drain}}$. Together: $\frac{1}{t_\text{fill}} - \frac{1}{t_\text{drain}} = \frac{1}{t_\text{net}}$.
Partial Completion: If workers finish $\frac{2}{3}$ of a job together and one leaves, find the time for the remaining $\frac{1}{3}$ done by the remaining worker.
Negative Work: A leak or drain works against filling — subtract its rate from the total.
Application Strategy
Worked Examples
Problem Statement
A store clerk arranges pens into packs. After giving 3 packs to each of 5 friends, she has 7 packs left. If she started with 25 packs, how many packs did each friend receive, and is the count consistent?
linear equationsolution
Strategic Toolkit
$$5x + 7 = 25$$
Step 1
Set up the equation
Let $x$ be the number of packs each friend received. Total given = $5x$, plus 7 remaining = 25 total.
$$5x + 7 = 25$$
Step 2
Isolate the variable term
Subtract 7 from both sides to move the constant away from the variable term.
$$5x = 25 - 7 = 18$$
Step 3
Solve for x
Divide both sides by 5 to find the number of packs per friend.
$$x = \frac{18}{5} = 3.6$$
Step 4
Interpret the result
Since $x = 3.6$ is not a whole number, each friend receiving 3.6 packs does not make physical sense with whole packs — the starting number or the remainder must be adjusted.
Problem Statement
A recipe requires mixing two solutions. If $\frac{x}{3} + \frac{x}{4} = 14$, find the value of $x$.
fractionscross multiplication
Strategic Toolkit
$$\frac{7x}{12} = 14$$
Step 1
Find the LCD
The denominators are 3 and 4. Their least common multiple is 12.
$$\text{LCD} = 12$$
Step 2
Clear the fractions
Multiply every term by 12 to eliminate both denominators at once.
$$12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 14$$
Step 3
Simplify each term
Distribute 12 into each fraction: $4x + 3x = 168$.
$$4x + 3x = 168$$
Step 4
Combine and solve
Combine like terms and divide to find $x$.
$$7x = 168 \quad \Rightarrow \quad x = 24$$
Problem Statement
A cinema sells adult tickets for Rs. 200 and child tickets for Rs. 100. If 50 tickets were sold in total for Rs. 8000, how many of each type were sold?
substitution methodsystem of equations
Strategic Toolkit
$$\begin{cases} x + y = 50 \\ 200x + 100y = 8000 \end{cases}$$
Step 1
Define variables and write equations
Let $x$ = adult tickets, $y$ = child tickets. Equation 1 gives the count, Equation 2 gives the revenue.
$$x + y = 50, \quad 200x + 100y = 8000$$
Step 2
Isolate one variable
From the first equation, express $y$ in terms of $x$.
$$y = 50 - x$$
Step 3
Substitute into the second equation
Replace $y$ with $(50 - x)$ in the revenue equation.
$$200x + 100(50 - x) = 8000$$
Step 4
Expand and simplify
Distribute 100 and combine the $x$ terms.
$$200x + 5000 - 100x = 8000 \quad \Rightarrow \quad 100x = 3000$$
Step 5
Solve for both variables
Divide by 100 to find $x$, then substitute back to find $y$.
$$x = 30, \quad y = 50 - 30 = 20$$
Step 6
Verify
Check: 30 adult tickets at Rs. 200 = Rs. 6000, plus 20 child tickets at Rs. 100 = Rs. 2000. Total = Rs. 8000. Confirmed.
$$30 \times 200 + 20 \times 100 = 6000 + 2000 = 8000 \checkmark$$
Problem Statement
The sum of two numbers is 45 and their difference is 15. Find the two numbers.
elimination methodsystem of equations
Strategic Toolkit
$$\begin{cases} x + y = 45 \\ x - y = 15 \end{cases}$$
Step 1
Set up the system
Let the numbers be $x$ (larger) and $y$ (smaller). Write the sum and difference as equations.
$$x + y = 45 \quad (1)$$ $$x - y = 15 \quad (2)$$
Step 2
Add the equations
The $y$ terms have opposite coefficients ($+1$ and $-1$), so adding eliminates $y$.
$$(x + x) + (y - y) = 45 + 15$$
Step 3
Solve for x
The $y$ terms cancel, leaving a simple equation in $x$.
$$2x = 60 \quad \Rightarrow \quad x = 30$$
Step 4
Find y by back-substitution
Substitute $x = 30$ into either original equation to find $y$.
$$30 + y = 45 \quad \Rightarrow \quad y = 15$$
Problem Statement
A father is 3 times as old as his son. In 12 years, he will be twice as old as his son. Find their current ages.
age differencelinear equation
Strategic Toolkit
$$\begin{cases} F = 3S \\ F + 12 = 2(S + 12) \end{cases}$$
Step 1
Define variables
Let $S$ = son's current age, $F$ = father's current age. Translate both statements into equations.
$$F = 3S, \quad F + 12 = 2(S + 12)$$
Step 2
Substitute from the first equation
Replace $F$ with $3S$ in the second equation.
$$3S + 12 = 2(S + 12)$$
Step 3
Expand and solve for S
Distribute the 2, collect $S$ terms, and solve.
$$3S + 12 = 2S + 24 \quad \Rightarrow \quad S = 12$$
Step 4
Find F
Use $F = 3S$ to get the father's age.
$$F = 3 \times 12 = 36$$
Step 5
Verify with the future condition
In 12 years: father = 48, son = 24. Is 48 twice 24? Yes.
$$48 = 2 \times 24 \checkmark$$
Problem Statement
The sum of the digits of a two-digit number is 9. If 27 is added to the number, the digits reverse. Find the original number.
tens digitunits digit
Strategic Toolkit
$$\begin{cases} t + u = 9 \\ (10t + u) + 27 = 10u + t \end{cases}$$
Step 1
Define variables
Let $t$ = tens digit, $u$ = units digit. The number is $10t + u$ and its reverse is $10u + t$.
$$\text{Number} = 10t + u$$
Step 2
Write the equations
The digit sum gives one equation, and the reversal condition gives another.
$$t + u = 9 \quad (1)$$ $$10t + u + 27 = 10u + t \quad (2)$$
Step 3
Simplify the reversal equation
Move all terms to one side and combine like terms.
$$9t + 27 = 9u \quad \Rightarrow \quad t + 3 = u$$
Step 4
Substitute into the sum equation
Replace $u$ with $t + 3$ in $t + u = 9$.
$$t + (t + 3) = 9 \quad \Rightarrow \quad 2t = 6 \quad \Rightarrow \quad t = 3$$
Step 5
Find u and the number
Calculate $u$ from the sum, then form the original number.
$$u = 9 - 3 = 6, \quad \text{Number} = 10(3) + 6 = 36$$
Step 6
Verify
Reverse of 36 is 63. Is $36 + 27 = 63$? Yes.
$$36 + 27 = 63 \checkmark$$
Problem Statement
Two cyclists start from the same point and ride in opposite directions. One rides at 12 km/h and the other at 8 km/h. How long until they are 60 km apart?
distancespeed
Strategic Toolkit
$$T = \frac{D}{S_1 + S_2}$$
Step 1
Identify what is known
Speeds are $S_1 = 12$ km/h and $S_2 = 8$ km/h. The total separation distance needed is $D = 60$ km.
$$S_1 = 12, \quad S_2 = 8, \quad D = 60$$
Step 2
Use the combined rate
Since they travel in opposite directions, their separation rate is the sum of their speeds.
$$\text{Combined rate} = 12 + 8 = 20 \text{ km/h}$$
Step 3
Calculate time
Time = total distance divided by combined rate.
$$T = \frac{60}{20} = 3 \text{ hours}$$
Step 4
Verify individual distances
Cyclist 1 covers $12 \times 3 = 36$ km. Cyclist 2 covers $8 \times 3 = 24$ km. Total = 60 km. Confirmed.
$$36 + 24 = 60 \text{ km} \checkmark$$
Problem Statement
Pipe A can fill a tank in 6 hours and Pipe B can fill it in 4 hours. If both are opened together, how long will it take to fill the tank?
work rate
Strategic Toolkit
$$\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_{\text{together}}}$$
Step 1
Express individual rates
Pipe A fills at $\frac{1}{6}$ of the tank per hour. Pipe B fills at $\frac{1}{4}$ of the tank per hour.
$$r_A = \frac{1}{6}, \quad r_B = \frac{1}{4}$$
Step 2
Add the rates
When both pipes are open, their rates add together.
$$r_{\text{combined}} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$$
Step 3
Take the reciprocal for time
The time to fill the tank together is the reciprocal of the combined rate.
$$t_{\text{together}} = \frac{12}{5} = 2.4 \text{ hours} = 2 \text{ hours } 24 \text{ minutes}$$
Step 4
Alternative — using the shortcut formula
For two workers with times $a$ and $b$: $t_{\text{together}} = \frac{ab}{a+b}$.
$$\frac{6 \times 4}{6 + 4} = \frac{24}{10} = 2.4 \text{ hours} \checkmark$$
Problem Statement
A bag contains 50 coins consisting of Rs. 5 and Rs. 10 coins. If the total value is Rs. 400, how many of each type of coin are in the bag?
system_of_equationselimination method
Strategic Toolkit
$$\begin{cases} x + y = 50 \\ 5x + 10y = 400 \end{cases}$$
Step 1
Set up variables
Let $x$ = number of Rs. 5 coins, $y$ = number of Rs. 10 coins.
$$x + y = 50 \quad (\text{count})$$ $$5x + 10y = 400 \quad (\text{value})$$
Step 2
Simplify the value equation
Divide the entire value equation by 5 to get smaller numbers.
$$x + 2y = 80$$
Step 3
Eliminate x by subtraction
Subtract the simplified count equation from the value equation.
$$(x + 2y) - (x + y) = 80 - 50$$
Step 4
Solve for y and then x
The $x$ terms cancel, leaving $y$. Back-substitute for $x$.
$$y = 30, \quad x = 50 - 30 = 20$$
Step 5
Verify
20 coins at Rs. 5 = Rs. 100. 30 coins at Rs. 10 = Rs. 300. Total = Rs. 400. Confirmed.
$$20 \times 5 + 30 \times 10 = 100 + 300 = 400 \checkmark$$
Problem Statement
Determine whether the system $2x + 3y = 7$ and $4x + 6y = 10$ has a solution.
parallel_linesinconsistent_eq
Strategic Toolkit
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$
Step 1
Compare the coefficient ratios
Check if $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ and whether the constant ratio matches.
$$\frac{2}{4} = \frac{1}{2}, \quad \frac{3}{6} = \frac{1}{2}$$
Step 2
Check the constant ratio
If $\frac{c_1}{c_2}$ also equals $\frac{1}{2}$, the lines coincide. Otherwise, they are parallel.
$$\frac{7}{10} \neq \frac{1}{2}$$
Step 3
Conclude
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and never intersect. The system has no solution.
Memorize formulas
Linear Equation in One Variable
Solve $ax + b = 0$ by isolating $x$
Formula
$$x = -\frac{b}{a}$$
Slope-Intercept Form
Standard form for graphing a line — read slope and y-intercept directly
Formula
$$y = mx + c$$
Work Rate Formula
Combined time for two workers
Formula
$$t_{\text{together}} = \frac{t_1 \cdot t_2}{t_1 + t_2}$$
Distance-Rate-Time
Fundamental relation between distance, speed, and time
Formula
$$D = S \times T$$
Two-Digit Number
Value from tens and units digits
Formula
$$N = 10t + u$$
Combined Rate (Opposite Directions)
Separation speed when moving away from each other
Formula
$$S_{\text{combined}} = S_1 + S_2$$
Harmonic Mean (Average Speed)
Average speed over two equal distances at different speeds
Formula
$$S_{\text{avg}} = \frac{2S_1 S_2}{S_1 + S_2}$$
Review
1
Solve for $x$: $3(x - 2) + 5 = 2(x + 7)$
Ans
15
Review
Tap to reveal
Expand and simplify both sides
Distribute the coefficients on each side of the equation.
$$3x - 1 = 2x + 14$$
Isolate $x$
Subtract $2x$ from both sides, then add 1.
$$x = 15$$
2
A line passes through $(0, -3)$ and has slope $m = 2$. Which equation represents this line in slope-intercept form?
Ans
$y = 2x - 3$
Review
Tap to reveal
Identify slope and y-intercept from the given point
Since the point has $x = 0$, its $y$-value is directly the y-intercept $c$.
$$m = 2, \quad c = -3$$
Write in slope-intercept form
Substitute $m$ and $c$ into $y = mx + c$.
$$y = 2x - 3$$
3
A delivery truck covers 240 km in 4 hours. What is its average speed?
Ans
60 km/h
Review
Tap to reveal
Apply the distance-rate-time formula
Rearrange $D = S \times T$ to solve for speed.
$$S = \frac{D}{T} = \frac{240}{4} = 60 \text{ km/h}$$
4
Pipe A can fill a tank in 12 hours and Pipe B can fill it in 24 hours. If both pipes are opened together, how long will it take to fill the tank?
Ans
8 hours
Review
Tap to reveal
Apply the combined work rate formula
Use $t_{\text{together}} = \frac{t_1 \cdot t_2}{t_1 + t_2}$ with $t_1 = 12$ and $t_2 = 24$.
$$t = \frac{12 \times 24}{12 + 24} = \frac{288}{36} = 8 \text{ hours}$$
5
The tens digit of a two-digit number is 4 more than the units digit, and the sum of its digits is 10. What is the number?
Ans
73
Review
Tap to reveal
Set up equations for the digits
Let $t$ = tens digit, $u$ = units digit. Translate the conditions into equations.
$$t = u + 4, \qquad t + u = 10$$
Solve the system and construct the number
Substitute $t = u + 4$ into $t + u = 10$ to find $u$, then $t$, then $N = 10t + u$.
$$u = 3,\ t = 7,\ N = 10(7) + 3 = 73$$
6
A stationery shop sells pens and notebooks. Two pens and three notebooks cost Rs. 90, while four pens and one notebook cost Rs. 70. What is the cost of one notebook?
Ans
Rs. 22
Review
Tap to reveal
Set up the system of equations
Define variables for pen and notebook costs, then translate the price information.
$$2p + 3n = 90, \qquad 4p + n = 70$$
Solve by substitution
Express $n$ from the second equation and substitute into the first.
$$n = 70 - 4p \implies 2p + 3(70 - 4p) = 90 \implies p = 12,\ n = 22$$
7
Solve the system $2x + y = 7$ and $3x - y = 8$. What is $x + y$?
Ans
$(3, 1)$
Review
Tap to reveal
Add equations to eliminate $y$
The $+y$ and $-y$ terms cancel when the equations are added.
$$5x = 15 \implies x = 3$$
Substitute to find $y$
Plug $x = 3$ into either original equation.
$$2(3) + y = 7 \implies y = 1, \quad \text{so } (x, y) = (3, 1)$$
8
A father is 4 times as old as his son. In 10 years, the father will be 2.5 times as old as his son. How old is the son now?
Ans
10 years
Review
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Set up equations for current and future ages
Define variables and write the age relationship for both time periods.
$$f = 4s, \qquad f + 10 = 2.5(s + 10)$$
Solve for the son's age
Substitute $f = 4s$ into the second equation and solve.
$$4s + 10 = 2.5s + 25 \implies 1.5s = 15 \implies s = 10$$
9
A two-digit number is 4 times the sum of its digits. If 36 is added to the number, the digits reverse their positions. Find the original number.
Ans
48
Review
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Translate both conditions into equations
Use $N = 10t + u$ to express each condition algebraically.
$$10t + u = 4(t + u), \qquad 10t + u + 36 = 10u + t$$
Solve the system for $t$ and $u$
Simplify each equation and solve simultaneously.
$$2t = u, \quad u - t = 4 \implies t = 4,\ u = 8$$
Construct the number
Combine the digits using $N = 10t + u$.
$$N = 10(4) + 8 = 48$$
10
A cashier has Rs. 1,200 in Rs. 50 and Rs. 100 notes. If there are 18 notes in total, how many Rs. 100 notes are there?
Ans
6
Review
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Set up the system
Define variables for each type of note and write one equation for count and one for total value.
$$x + y = 18, \qquad 50x + 100y = 1200$$
Solve by substitution
Express $x$ in terms of $y$ and substitute into the value equation.
$$x = 18 - y \implies 50(18 - y) + 100y = 1200 \implies y = 6$$
11
Two trains depart from the same station simultaneously. One travels north at 80 km/h and the other travels south at 120 km/h. After how many hours will they be 500 km apart?
Ans
2.5 hours
Review
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Calculate the combined speed
For opposite directions, add the individual speeds.
$$S_{\text{combined}} = 80 + 120 = 200 \text{ km/h}$$
Find the time
Use $D = S \times T$ rearranged to $T = D / S$.
$$T = \frac{500}{200} = 2.5 \text{ hours}$$
12
Two workers can complete a task together in 6 hours. If one worker alone takes 10 hours, how long would the second worker take alone?
Ans
15 hours
Review
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Set up the work rate equation
The combined rate equals the sum of individual rates.
$$\frac{1}{10} + \frac{1}{t_2} = \frac{1}{6}$$
Solve for $t_2$
Isolate $\frac{1}{t_2}$ and take the reciprocal.
$$\frac{1}{t_2} = \frac{1}{6} - \frac{1}{10} = \frac{1}{15} \implies t_2 = 15 \text{ hours}$$
13
Two cyclists start from points A and B, 70 km apart, and ride toward each other. One cycles at 15 km/h and the other at 20 km/h. How far from point A will they meet?
Ans
30 km
Review
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Find when they meet
The combined speed determines how quickly the gap closes.
$$t = \frac{70}{15 + 20} = \frac{70}{35} = 2 \text{ hours}$$
Calculate distance from A
Multiply the slower cyclist's speed by the meeting time.
$$d = 15 \times 2 = 30 \text{ km from A}$$
14
A driver travels from city X to city Y at 60 km/h and returns from Y to X at 40 km/h. If the distance between the cities is the same each way, what is the average speed for the entire round trip?
Ans
48 km/h
Review
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Apply the harmonic mean formula
For equal distances at two different speeds, the average speed is the harmonic mean.
$$S_{\text{avg}} = \frac{2 S_1 S_2}{S_1 + S_2} = \frac{2 \times 60 \times 40}{60 + 40}$$
Calculate the result
Evaluate the expression.
$$S_{\text{avg}} = \frac{4800}{100} = 48 \text{ km/h}$$
15
For what value of $k$ is the system $2x + 3y = 7$ and $4x + ky = 14$ inconsistent (has no solution)?
Ans
6
Review
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Write the inconsistency condition
For no solution, the coefficient ratios must match but the constant ratio must differ (or all match, meaning identical parallel lines).
$$\frac{2}{4} = \frac{3}{k}$$
Solve for $k$ and verify
Cross-multiply to find $k$, then check all three ratios.
$$\frac{1}{2} = \frac{3}{k} \implies k = 6 \quad \text{(verified: all ratios } = \tfrac{1}{2})$$
16
A person walks from town P to town Q at 5 km/h starting at 6:00 AM. Another person cycles from Q to P at 15 km/h starting at 8:00 AM. They meet at 10:00 AM. What is the distance between P and Q?
Ans
50 km
Review
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Calculate distance covered before the cyclist starts
The walker travels alone from 6:00 AM to 8:00 AM.
$$d_1 = 5 \times 2 = 10 \text{ km}$$
Calculate distance covered while both are moving
From 8:00 AM to 10:00 AM, they approach each other at a combined speed.
$$d_2 = (5 + 15) \times 2 = 40 \text{ km}$$
Add the two segments
The total distance is the sum of the solo and combined segments.
$$D = 10 + 40 = 50 \text{ km}$$
17
The perimeter of a rectangle is 34 cm and its diagonal is 13 cm. What are the length and width of the rectangle?
Ans
12 cm and 5 cm
Review
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Write the two equations
Perimeter gives $l + w$, and the diagonal gives $l^2 + w^2$ via Pythagoras.
$$l + w = 17, \qquad l^2 + w^2 = 169$$
Find $lw$ using the squared sum identity
Expand $(l + w)^2$ to find the product $lw$.
$$(l+w)^2 = l^2 + 2lw + w^2 \implies 289 = 169 + 2lw \implies lw = 60$$
Solve for $l$ and $w$
Use $(l - w)^2 = l^2 - 2lw + w^2$ to get $l - w$, then solve the system.
$$(l-w)^2 = 169 - 120 = 49 \implies l - w = 7 \implies l = 12,\ w = 5$$
18
Which of the following equations is an identity (true for all values of $x$)?
Ans
$2(x + 3) = 2x + 6$
Review
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Simplify each equation
Expand and combine like terms on both sides to check for validity.
$$2(x + 3) = 2x + 6 \implies 2x + 6 = 2x + 6 \quad \text{(always true)}$$
19
A and B can complete a job in 12 days, B and C in 15 days, and A and C in 20 days. How many days would A alone take to complete the job?
Ans
30 days
Review
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Write the three rate equations
Express each pair's combined work rate.
$$\frac{1}{A}+\frac{1}{B}=\frac{1}{12},\quad \frac{1}{B}+\frac{1}{C}=\frac{1}{15},\quad \frac{1}{A}+\frac{1}{C}=\frac{1}{20}$$
Add all three equations
The sum of the left sides is twice the total combined rate.
$$2\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)=\frac{1}{5}\implies \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{10}$$
Isolate A's rate
Subtract the B-C equation from the total to find A's individual rate.
$$\frac{1}{A}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}\implies A=30\text{ days}$$
20
A boat travels 48 km downstream in 3 hours and 48 km upstream in 6 hours. What is the speed of the boat in still water?
Ans
12 km/h
Review
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Calculate downstream and upstream speeds
Use $S = D/T$ for each leg of the journey.
$$S_{\text{down}} = \frac{48}{3} = 16,\qquad S_{\text{up}} = \frac{48}{6} = 8$$
Set up and solve the still-water equations
Downstream adds current speed ($b + c = 16$), upstream subtracts it ($b - c = 8$).
$$(b+c)+(b-c) = 16+8 \implies 2b = 24 \implies b = 12 \text{ km/h}$$
Question 1 of 20
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1
Solve for $x$: $3(x - 2) + 5 = 2(x + 7)$