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#15 Compulsory Algebra & Equations Math 15

Linear Equations in One & Two Variablesएक एवं दो चरों वाले रेखीय समीकरण

Master linear equations in one and two variables with rigorous algebraic manipulations, systematic word problem translation, and advanced speed-math shortcuts designed for top ranks in the Super TET Junior Mathematics exam.

Preparation Status

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Syllabus Focus & High-Yield Strategy: Master linear equations in one and two variables with rigorous algebraic manipulations, systematic word problem translation, and advanced speed-math shortcuts designed for top ranks in the Super TET Junior Mathematics exam.

1. Fundamental Axioms, Definitions & Foundational Properties

Definition of Linear Equation: An algebraic equation in which the highest power (degree) of the variable(s) involved is exactly $1$. The general form in one variable is $ax + b = 0$, where $a \neq 0$ and $a, b \in \mathbb{R}$.
Linear Equation in Two Variables: An equation expressible in the standard form $ax + by + c = 0$, where $a, b$, and $c$ are real numbers, and both $a^2 + b^2 \neq 0$. Its graph on a Cartesian plane is always a straight line.
Addition and Subtraction Axiom: If $A = B$, then $A + C = B + C$ and $A - C = B - C$. Adding or subtracting the same quantity from both sides maintains equality without altering the solution set.
Multiplication and Division Axiom: If $A = B$ and $C \neq 0$, then $AC = BC$ and $\frac{A}{C} = \frac{B}{C}$. Multiplying or dividing both sides by a non-zero real number preserves the equivalence of the equation.
Exam Pro-Tip: Always check that the coefficient of the leading variable does not evaluate to zero under any substituted parameter to avoid undefined denominators.
Memory Mnemonic: BAL — Balance All Lines by keeping operations identical on both sides.

2. Formulas, Standard Theorems & Algorithmic Procedures

Category / CaseFormula / Mathematical RuleTopper Calculation Shortcut
System of Two Variables$\begin{cases} a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0 \end{cases}$Use Cross-Multiplication: $\frac{x}{b_1c_2 - b_2c_1} = \frac{-y}{a_1c_2 - a_2c_1} = \frac{1}{a_1b_2 - a_2b_1}$
Consistency Criteria$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)Instantly check ratio slopes; if unequal, single intersection point.
Substitution Method Algorithm: Solve one equation for one variable in terms of the other, and substitute that expression into the second equation to reduce it to a single-variable linear equation.
Elimination Method Algorithm: Multiply one or both equations by suitable non-zero constants so that the coefficients of one variable are additive inverses, then add the equations to eliminate that variable.
Graphical Intersection Theorem: Every solution $(x, y)$ of a linear equation in two variables corresponds to a unique point on the line representing that equation, and vice versa.
15-Second Speed-Math Trick: When solving symmetric equations of the form $ax + by = c$ and $bx + ay = d$, directly add and subtract the two equations to instantly yield $(a+b)(x+y) = c+d$ and $(a-b)(x-y) = c-d$.

3. Word Problem Modeling & Practical Applications

Age Problems Modeling: Define present ages as variables $x$ and $y$. Translate statements like "Five years hence" to $(x + 5)$ and "Three years ago" to $(y - 3)$, establishing a direct system of linear equations.
Digit & Number Problems: A two-digit number is represented algebraically as $10x + y$, where $x$ is the tens digit and $y$ is the units digit. Reversing the digits yields $10y + x$.
Speed, Distance & Time Models: Apply $Time = \frac{Distance}{Speed}$. For upstream and downstream motion, let boat speed be $u$ and stream speed be $v$; downstream speed is $(u + v)$ and upstream speed is $(u - v)$.
Work & Time Equations: If a person completes a work in $n$ days, their 1-day work fraction is $\frac{1}{n}$. Combine fractional work rates into linear equations for simultaneous completion.
UP Exam High-Yield Tip: In number reversal problems, the sum of a two-digit number and the number formed by reversing its digits is always divisible by $11$, and their difference is always divisible by $9$.

4. Mathematical Edge Cases, Sign Conventions & Common Traps

Critical Distinction: Dependent systems ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$) have infinitely many solutions, whereas inconsistent systems ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$) yield zero solutions (parallel lines).
Edge Case of Zero Denominators: In cross-multiplication or algebraic fractions, ensure the denominator $a_1b_2 - a_2b_1 \neq 0$; if it equals zero, the lines are either parallel or coincident.
Common Calculation Trap: Forgetting to distribute a negative sign across a compound binomial numerator during transposition (e.g., $-(2x - 3)$ becoming $-2x - 3$ instead of $-2x + 3$).
Trap Prevention Trick: Always verify your derived $(x, y)$ values by substituting them back into both original equations before marking the final answer bubble.
Memory Mnemonic: SIGN — Subtract Inside Grouped Numerators with extreme caution regarding brackets.

Comprehensive Mathematical Matrix & Method Comparison

Parameter / DimensionMethod / Concept AMethod / Concept BExam Speed & Accuracy Takeaway
Substitution MethodBest when coefficient of one variable is $\pm 1$Time-consuming if fractions are introduced earlyUse only when isolation requires $\le 1$ algebraic step.
Elimination MethodBest for integer coefficients requiring quick scalingRequires careful sign management during subtractionFastest method for standard competitive exam formats.
Cross-MultiplicationDirect formula application for standard two-variable formProne to sign errors in $a_1b_2 - a_2b_1$ denominatorMemorize cyclic determinant pattern: $b_1, c_1, a_1, b_1$.
Graphical MethodVisualizes intersection, parallelism, and coincidenceImpractical for exact fractional roots in timed testsUse conceptually for consistency tests rather than plotting.
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Question 1 of 10 Super TET Junior Mathematics

1. What is the solution for $x$ in the linear equation $\frac{3x + 5}{2} - \frac{x - 2}{3} = \frac{47}{6}$?

Correct Answer: Option A

Multiply both sides by the least common multiple, 6: $3(3x+5)-2(x-2)=47$. Expanding gives $9x+15-2x+4=47$, so $7x+19=47$, $7x=28$, and $x=4$. Therefore, Option A is correct.

Question 2 of 10 Super TET Junior Mathematics

2. If the system of linear equations $kx + 3y = k - 3$ and $12x + ky = k$ has infinitely many solutions, what is the value of $k$?

Correct Answer: Option A

For infinitely many solutions, the coefficient and constant ratios must be equal. From $\frac{k}{12}=\frac{3}{k}$, we get $k^2=36$, so $k=\pm6$. When $k=6$, all three ratios equal $\frac{1}{2}$. When $k=-6$, the constant ratio is $\frac{3}{2}$ while the coefficient ratios are $-\frac{1}{2}$. Thus $k=6$, Option A.

Question 3 of 10 Super TET Junior Mathematics

3. The sum of the digits of a two-digit number is $9$. If $27$ is subtracted from the number, the digits are reversed. What is the original number?

Correct Answer: Option A

Let the number be $10x + y$. Given $x + y = 9$. When $27$ is subtracted, digits reverse: $(10x + y) - 27 = 10y + x \implies 9x - 9y = 27 \implies x - y = 3$. Adding the two equations: $(x+y) + (x-y) = 9 + 3 \implies 2x = 12 \implies x = 6$. Then $y = 3$. The original number is $10(6) + 3 = 63$. Distractors like $81$ give $8-1=7 \neq 3$.

Question 4 of 10 Super TET Junior Mathematics

4. Father's age is three times the sum of the ages of his two children. After 5 years, his age will be twice the sum of their ages. What is the father's present age?

Correct Answer: Option A

Let the sum of children's present ages be $S$, and father's age be $F$. Given $F = 3S$. After 5 years, each child's age increases by $5$, so sum of their ages becomes $S + 10$. Father's age becomes $F + 5$. Given $F + 5 = 2(S + 10)$. Substitute $F = 3S$: $3S + 5 = 2S + 20 \implies S = 15$. Father's present age $F = 3(15) = 45$ years.

Question 5 of 10 Super TET Junior Mathematics

5. Solve for $x$ and $y$: $2x + 3y = 11$ and $2x - 4y = -24$. What is the value of $y - x$?

Correct Answer: Option C

Subtract the second equation from the first: $(2x + 3y) - (2x - 4y) = 11 - (-24) \implies 7y = 35 \implies y = 5$. Substitute $y = 5$ into $2x + 3(5) = 11 \implies 2x + 15 = 11 \implies 2x = -4 \implies x = -2$. We need $y - x = 5 - (-2) = 5 + 2 = 7$.

Question 6 of 10 Super TET Junior Mathematics

6. A boat goes $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10$ hours. In $13$ hours, it can go $40\text{ km}$ upstream and $55\text{ km}$ downstream. What is the speed of the stream?

Correct Answer: Option A

Let upstream speed be $u$ and downstream speed be $d$. Equations are $\frac{30}{u} + \frac{44}{d} = 10$ and $\frac{40}{u} + \frac{55}{d} = 13$. Let $\frac{1}{u} = A$ and $\frac{1}{d} = B$. Then $30A + 44B = 10$ and $40A + 55B = 13$. Solving this linear system yields $A = \frac{1}{5}$ and $B = \frac{1}{11}$, meaning $u = 5\text{ km/h}$ and $d = 11\text{ km/h}$. Speed of stream = $\frac{d - u}{2} = \frac{11 - 5}{2} = 3\text{ km/h}$.

Question 7 of 10 Super TET Junior Mathematics

7. For what value of $k$ does the pair of linear equations $x + 2y = 3$ and $5x + ky = 15$ have infinitely many solutions?

Correct Answer: Option A

Condition for infinitely many solutions is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Here $\frac{1}{5} = \frac{2}{k} = \frac{3}{15}$. Since $\frac{3}{15} = \frac{1}{5}$, equating $\frac{1}{5} = \frac{2}{k}$ gives $k = 10$.

Question 8 of 10 Super TET Junior Mathematics

8. The denominator of a fraction is $4$ more than twice its numerator. If $1$ is subtracted from both numerator and denominator, the fraction becomes $\frac{4}{13}$. What is the original fraction?

Correct Answer: Option A

Let the numerator be $x$ and the denominator be $y$. The condition gives $y=2x+4$. After subtracting $1$ from each, $\frac{x-1}{y-1}=\frac{4}{13}$. Substitute $y=2x+4$: $\frac{x-1}{2x+3}=\frac{4}{13}$, so $13(x-1)=4(2x+3)$. Thus $13x-13=8x+12$, $5x=25$, and $x=5$. Then $y=14$, so the original fraction is $\frac{5}{14}$, Option A.

Question 9 of 10 Super TET Junior Mathematics

9. A railway half ticket costs half the full fare, but the reservation charge on a half ticket is the same as on a full ticket. One reserved first class full ticket from Station A to B costs $₹ 2530$, and one full and one half ticket cost $₹ 3810$. What is the base full fare?

Correct Answer: Option B

Let $F$ be the base full fare and $R$ the reservation charge. The full ticket gives $F+R=2530$. A full and a half ticket together cost $(F+R)+(\frac{F}{2}+R)=3810$, so $\frac{F}{2}+R=1280$. Subtracting the first equation gives $\frac{F}{2}=1250$, hence $F=2500$. Option B is correct; the reservation charge is $30$, which verifies both totals.

Question 10 of 10 Super TET Junior Mathematics

10. Places A and B are $100\text{ km}$ apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in $5$ hours. If they travel towards each other, they meet in $1$ hour. What is the speed of the faster car?

Correct Answer: Option A

Let speeds be $x$ and $y$ ($x > y$). Relative speed in same direction = $x - y$. Distance = $100\text{ km}$, time = $5\text{ hrs} \implies x - y = \frac{100}{5} = 20$. Relative speed in opposite direction = $x + y$. Time = $1\text{ hr} \implies x + y = \frac{100}{1} = 100$. Adding the two equations: $2x = 120 \implies x = 60\text{ km/h}$ and $y = 40\text{ km/h}$. The faster car's speed is $60\text{ km/h}$.

5-Minute Exam Simulation Mode

Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.

Time Limit:

5 Minutes (300s)

Marking Scheme:

+3 Marks | -1 Negative

Passing Benchmark:

21+ Marks (70%)

High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)

#1
A linear equation in one variable $ax + b = 0$ has a unique solution $x = -a^{-1}b$ provided $a \neq 0$.
#2
The graph of every linear equation in two variables $ax + by + c = 0$ is a straight line in the Cartesian plane.
#3
A system of two linear equations has a unique solution when $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
#4
Parallel lines represent an inconsistent system of linear equations with zero solutions ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$).
#5
Coincident lines represent a dependent system with infinitely many solutions ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$).
#6
Two-digit numbers are algebraically modeled as $10x + y$, where $x$ is the tens digit and $y$ is the units digit.
#7
Upstream speed is $(u - v)$ and downstream speed is $(u + v)$, where $u$ is boat speed and $v$ is stream speed.
#8
Substitution, elimination, and cross-multiplication are the three core analytical tools for solving simultaneous linear equations.

Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)

Consistency Check Acronym

Memory Key: UNI

U - Unique ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$), N - No solution (Parallel), I - Infinite (Coincident).

Word Problem Translation Anchor

Memory Key: READ

R - Recognize variables, E - Establish equations, A - Apply method, D - Double-check solution.

Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)

Negative Marking Alert #1:

Negative Marking Trap 1: Failing to invert signs correctly when distributing a negative multiplier across parentheses during algebraic expansion.

Negative Marking Alert #2:

Negative Marking Trap 2: Confusing parallel line conditions (zero solutions) with coincident line conditions (infinite solutions).

Negative Marking Alert #3:

Negative Marking Trap 3: Mixing up upstream speed $(u - v)$ with downstream speed $(u + v)$ in word problems.

Negative Marking Alert #4:

Negative Marking Trap 4: Assuming a system with equal coefficient ratios $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ always has zero solutions without checking the constant ratio $\frac{c_1}{c_2}$.

Self-Assessment Checklist

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