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NIOS • Class X • Mathematics (211) • Ch 5

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Linear Equations

In NIOS Secondary (Class 10) Mathematics (211), Linear Equations represents one of the most foundational and highest-scoring modules in algebra. This comprehensive chapter systematically covers: Linear Equations in One Variable: Standard form $ax + b = 0$ ($a \neq 0$), fundamental axioms of equality, systematic transposition method, and single-unknown practical word problems. Linear Equations in Two Variables: Standard form $ax + by + c = 0$ (where $a^2 + b^2 \neq 0$), representation of linear equations as continuous straight lines on the Cartesian plane, and infinite pairs of solutions $(x, y)$. Simultaneous Linear Equations (Pair of Linear Equations in Two Variables): $$a_1 x + b_1 y + c_1 = 0 \quad \text{and} \quad a_2 x + b_2 y + c_2 = 0$$ Geometric Classification & Consistency Criteria: Intersecting Lines (Unique Solution / Consistent): $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ Coincident Lines (Infinitely Many Solutions / Dependent Consistent): $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Parallel Lines (No Solution / Inconsistent): $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Algebraic Solving Techniques: Method of Substitution: Expressing one variable in terms of the other and substituting into the remaining equation. Method of Elimination (by Equating Coefficients): Multiplying equations by non-zero constants to eliminate one variable through addition or subtraction. Cross-Multiplication Method (2312 Rule): Determinant-based algebraic solving: $$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{-y}{a_1 c_2 - a_2 c_1} = \frac{1}{a_1 b_2 - a_2 b_1}$$ Equations Reducible to Linear Form: Techniques to transform rational algebraic relations ($\frac{a}{x} + \frac{b}{y} = c$) into linear form using auxiliary variables $u = \frac{1}{x}, v = \frac{1}{y}$. Real-World Applied Word Problems: Age relationships, two-digit numbers ($10x + y$), fractional values, upstream and downstream relative boat velocities, and commercial cost calculations.

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How Do Air Traffic Controllers and GPS Satellites Prevent Mid-Air Collisions with Linear Equations?

At any given moment, over 10,000 commercial jetliners are cruising through international airspace at speeds exceeding 900 km/h. How do automated radar systems and flight computers ensure that two airplanes crossing paths do not collide?

Each aircraft\x27s straight-line flight corridor on a navigation grid is expressed as a linear equation in two variables: $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$. By solving this pair of simultaneous linear equations in real time, navigation supercomputers compute the exact coordinate point $(x, y)$ where the two flight corridors intersect!

If the ratio of their heading coefficients satisfies $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the paths intersect at a single point, and automated collision avoidance algorithms (TCAS) instantly assign different cruising altitudes. Without the mathematics of simultaneous linear equations, modern aerospace navigation, GPS triangulation, electronic circuit design, and algorithmic financial trading would be impossible!

Why This Chapter Matters

In the NIOS Class 10 Secondary Examination (Course 211), Linear Equations carries between 6 to 8 marks across objective questions, 2-mark algebraic solving questions, and a compulsory 4 or 5-mark application word problem (such as boat & stream, two-digit numbers, or age relations). Beyond school examinations, linear systems form the backbone of competitive exams (SSC, Banking, NDA, CUET) and university-level engineering, economics, and data analytics.

Before You Begin (Prerequisites)

Basic arithmetic operations with positive and negative directed integers.
Algebraic expressions, like and unlike terms, and distributive multiplication.
Cartesian coordinate system and plotting points $(x, y)$ on a graph.
Solving basic one-variable linear equations by transposition.

What You Will Learn (Core Objectives)

Express and identify linear equations in one variable ($ax + b = 0$) and two variables ($ax + by + c = 0$).
Determine whether a pair of linear equations is consistent, dependent, or inconsistent using coefficient ratios.
Solve simultaneous linear equations accurately using the Method of Substitution.
Solve pairs of linear equations using the Method of Elimination by equating coefficients.
Apply the Cross-Multiplication (2312) method to determine unique solutions efficiently.
Transform equations reducible to linear systems using auxiliary variable substitution.
Formulate and solve real-world word problems involving ages, two-digit numbers, fractions, geometry, and upstream/downstream motion.

Chapter Roadmap & Progression

1 1. Linear Equations in One and Two...

2 2. Pair of Linear Equations & Geome...

3 3. Algebraic Methods: Substitution...

4 4. The Cross-Multiplication Method...

5 5. Equations Reducible to Linear Fo...

Complete Concept Guide (100% Curriculum Coverage)

1. Linear Equations in One and Two Variables

2. Pair of Linear Equations & Geometric Consistency Criteria

Geometric & Algebraic Classification

A. System of Simultaneous Linear Equations:

When two linear equations in the same two variables are considered together, they form a Pair of Linear Equations in Two Variables:

$$a_1 x + b_1 y + c_1 = 0$$ $$a_2 x + b_2 y + c_2 = 0$$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers such that $a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$.

B. Consistency, Geometric Behavior & Number of Solutions:

The nature of the lines and the number of solutions can be determined without drawing the graph by simply comparing the ratios of their coefficients:

Ratio Comparison	Graphical Representation	Algebraic Interpretation	Consistency Status
$$\mathbf{\frac{a_1}{a_2} \neq \frac{b_1}{b_2}}$$	Intersecting Lines (Intersect at exactly one point)	Unique Solution (Exactly one ordered pair $(x, y)$)	Consistent
$$\mathbf{\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}$$	Coincident Lines (Lines overlap completely)	Infinitely Many Solutions	Consistent & Dependent
$$\mathbf{\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}}$$	Parallel Lines (Lines never intersect)	No Solution (Solution set is empty $\emptyset$)	Inconsistent

Crucial Examination Rule: Before writing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$, make sure that both equations are written in the same standard format—either with constant terms on the LHS ($ax + by + c = 0$) or with constant terms on the RHS ($ax + by = c$). Mixing formats causes fatal sign errors!

3. Algebraic Methods: Substitution and Elimination

4. The Cross-Multiplication Method (The 2312 Rule)

5. Equations Reducible to Linear Form & Applied Word Problems

Word Problems & Applications

A. Equations Reducible to Linear Form:

Equations with variables in denominators, such as:

$$\frac{a_1}{x} + \frac{b_1}{y} = c_1 \quad \text{and} \quad \frac{a_2}{x} + \frac{b_2}{y} = c_2$$

are non-linear. They are transformed into standard linear equations by introducing auxiliary variables: let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. After solving for $u$ and $v$, invert to find $x = \frac{1}{u}$ and $y = \frac{1}{v}$.

B. Standard Applied Word Problem Mathematical Models:

Problem Type	Variable Setup	Governing Mathematical Model
1. Age Problems	Present age of father $= x$ yrs, son $= y$ yrs.	$n$ years ago: $(x - n), (y - n)$ $m$ years hence: $(x + m), (y + m)$
2. Two-Digit Numbers	Tens digit $= x$, Units digit $= y$.	Original Number $= 10x + y$ Reversed Number $= 10y + x$
3. Fractions	Numerator $= x$, Denominator $= y$.	Original Fraction $= \frac{x}{y}$ Adjusted: $\frac{x + k_1}{y + k_2} = \text{value}$
4. Boat & Stream	Boat speed in still water $= u$ km/h, Stream speed $= v$ km/h.	Downstream Speed $= (u + v)$ km/h Upstream Speed $= (u - v)$ km/h Time $= \frac{\text{Distance}}{\text{Speed}}$
5. Work and Time	1 person finishes in $x$ days, 2nd finishes in $y$ days.	1 day\x27s work: $\frac{1}{x} + \frac{1}{y} = \frac{1}{\text{Total Days}}$

Key Formulas, Identities & Theorems

General Form of Linear Pair in Two Variables

$$a_1 x + b_1 y + c_1 = 0 \quad \text{and} \quad a_2 x + b_2 y + c_2 = 0$$

Standard canonical representation where coefficients are real numbers.

Condition for Unique Solution (Intersecting Lines)

$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \quad [\text{Consistent System}]$$

Lines intersect at exactly one point $(x, y)$.

Condition for Infinitely Many Solutions (Coincident Lines)

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \quad [\text{Consistent & Dependent System}]$$

Both equations represent the exact same line.

Condition for No Solution (Parallel Lines)

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \quad [\text{Inconsistent System}]$$

Lines are strictly parallel with identical slope but distinct intercepts.

Cross-Multiplication Formula (2312 Rule)

$$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{-y}{a_1 c_2 - a_2 c_1} = \frac{1}{a_1 b_2 - a_2 b_1}$$

Direct determinant solution for unique intersection.

Two-Digit Number Value Formula

$$\text{Original Number} = 10x + y \quad ; \quad \text{Reversed Number} = 10y + x$$

Where $x$ is the tens digit and $y$ is the units digit.

Relative Speeds in Water (Boat & Stream)

$$v_{\text{downstream}} = u + v \quad ; \quad v_{\text{upstream}} = u - v$$

Where $u$ is speed in still water and $v$ is stream velocity ($u > v$).

Conceptual Solved Examples & Case Studies

Example 1

Solve the following pair of linear equations by the Method of Elimination:
$$3x + 4y = 10 \quad \text{--- (1)}$$
$$2x - 2y = 2 \quad \text{--- (2)}$$

Step-by-Step Solution:

Step-by-Step Solution:
Step 1: Equalize coefficients of $y$.
Equation (1) has $+4y$ and equation (2) has $-2y$. Multiply equation (2) by $2$:
$$2 \times (2x - 2y) = 2 \times 2 \implies 4x - 4y = 4 \quad \text{--- (3)}$$
Step 2: Add equation (1) and equation (3) to eliminate $y$:
$$(3x + 4y) + (4x - 4y) = 10 + 4$$
$$7x = 14 \implies x = \frac{14}{7} = \mathbf{2}$$
Step 3: Substitute $x = 2$ into equation (1):
$$3(2) + 4y = 10 \implies 6 + 4y = 10 \implies 4y = 4 \implies y = \mathbf{1}$$
Verification: In (2): $2(2) - 2(1) = 4 - 2 = 2$ (Confirmed).
$$\mathbf{\text{Answer: } x = 2, \quad y = 1}$$

Example 2

Solve the following system by Substitution, and hence find the value of $m$ for which $y = mx + 3$:
$$2x + 3y = 11 \quad \text{--- (1)}$$
$$2x - 4y = -24 \quad \text{--- (2)}$$

Step-by-Step Solution:

Step-by-Step Solution:
Step 1: Express $x$ in terms of $y$ from equation (1):
$$2x = 11 - 3y \implies x = \frac{11 - 3y}{2} \quad \text{--- (3)}$$
Step 2: Substitute expression (3) into equation (2):
$$2\left(\frac{11 - 3y}{2}\right) - 4y = -24$$
$$(11 - 3y) - 4y = -24 \implies 11 - 7y = -24$$
$$-7y = -24 - 11 = -35 \implies y = \frac{-35}{-7} = \mathbf{5}$$
Step 3: Substitute $y = 5$ into expression (3):
$$x = \frac{11 - 3(5)}{2} = \frac{11 - 15}{2} = \frac{-4}{2} = \mathbf{-2}$$
Step 4: Find $m$ using $y = mx + 3$:
$$5 = m(-2) + 3 \implies 5 - 3 = -2m \implies 2 = -2m \implies m = \mathbf{-1}$$
$$\mathbf{\text{Answer: } x = -2, \quad y = 5, \quad m = -1}$$

Example 3

Solve the following pair of equations using the Cross-Multiplication Method:
$$8x + 5y = 9 \quad \text{and} \quad 3x + 2y = 4$$

Step-by-Step Solution:

Step-by-Step Solution:
Step 1: Write both equations in standard form $ax + by + c = 0$:
$$8x + 5y - 9 = 0 \quad (a_1 = 8, b_1 = 5, c_1 = -9)$$
$$3x + 2y - 4 = 0 \quad (a_2 = 3, b_2 = 2, c_2 = -4)$$
Step 2: Set up the cross-multiplication ratios:
$$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{-y}{a_1 c_2 - a_2 c_1} = \frac{1}{a_1 b_2 - a_2 b_1}$$
Compute denominators:
$$b_1 c_2 - b_2 c_1 = 5(-4) - 2(-9) = -20 + 18 = -2$$
$$a_1 c_2 - a_2 c_1 = 8(-4) - 3(-9) = -32 + 27 = -5$$
$$a_1 b_2 - a_2 b_1 = 8(2) - 3(5) = 16 - 15 = 1$$
Step 3: Solve for $x$ and $y$:
$$\frac{x}{-2} = \frac{-y}{-5} = \frac{1}{1}$$
$$x = -2 \times 1 = \mathbf{-2} \quad ; \quad y = 5 \times 1 = \mathbf{5}$$
$$\mathbf{\text{Answer: } x = -2, \quad y = 5}$$

Example 4

A boat covers 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can cover 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Step-by-Step Solution:

Step-by-Step Solution:
Step 1: Define variables:
Let speed of boat in still water $= u$ km/h, and speed of stream $= v$ km/h ($u > v$).
Speed upstream $= (u - v)$ km/h, Speed downstream $= (u + v)$ km/h.
Let $\frac{1}{u - v} = x$ and $\frac{1}{u + v} = y$.
Step 2: Form equations from given conditions:
$$\text{Case 1: } \frac{30}{u-v} + \frac{44}{u+v} = 10 \implies 30x + 44y = 10 \quad \text{--- (1)}$$
$$\text{Case 2: } \frac{40}{u-v} + \frac{55}{u+v} = 13 \implies 40x + 55y = 13 \quad \text{--- (2)}$$
Step 3: Eliminate $x$ by multiplying (1) by 4 and (2) by 3:
$$120x + 176y = 40 \quad \text{--- (3)}$$
$$120x + 165y = 39 \quad \text{--- (4)}$$
Subtract (4) from (3):
$$11y = 1 \implies y = \frac{1}{11}$$
Substitute $y = \frac{1}{11}$ into (1):
$$30x + 44\left(\frac{1}{11}\right) = 10 \implies 30x + 4 = 10 \implies 30x = 6 \implies x = \frac{6}{30} = \frac{1}{5}$$
Step 4: Solve for $u$ and $v$:
$$u - v = \frac{1}{x} = 5 \quad \text{--- (5)}$$
$$u + v = \frac{1}{y} = 11 \quad \text{--- (6)}$$
Add (5) and (6): $2u = 16 \implies u = \mathbf{8\text{ km/h}}$.
Substitute into (6): $8 + v = 11 \implies v = \mathbf{3\text{ km/h}}$.
$$\mathbf{\text{Answer: Speed of boat in still water} = 8\text{ km/h}, \quad \text{Speed of stream} = 3\text{ km/h}}$$

Example 5

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the original number.

Step-by-Step Solution:

Step-by-Step Solution:
Step 1: Set up variables:
Let the tens digit be $x$ and the units digit be $y$.
Original Number $= 10x + y$
Reversed Number $= 10y + x$
Step 2: Translate conditions into equations:
Condition 1: Sum of digits is 9:
$$x + y = 9 \quad \text{--- (1)}$$
Condition 2: 9 times original number equals 2 times reversed number:
$$9(10x + y) = 2(10y + x)$$
$$90x + 9y = 20y + 2x \implies 90x - 2x + 9y - 20y = 0$$
$$88x - 11y = 0$$
Divide by 11:
$$8x - y = 0 \implies y = 8x \quad \text{--- (2)}$$
Step 3: Substitute (2) into (1):
$$x + 8x = 9 \implies 9x = 9 \implies x = \mathbf{1}$$
Substitute $x = 1$ into (2):
$$y = 8(1) = \mathbf{8}$$
Step 4: Compute original number:
$$\text{Original Number} = 10(1) + 8 = \mathbf{18}$$
Verification: $1 + 8 = 9$. Reversed is 81. $9 \times 18 = 162$ and $2 \times 81 = 162$. Confirmed!
$$\mathbf{\text{Answer: The required number is } 18}$$

Common Misconceptions & Examiner Traps

Common Misconception

Comparing ratio coefficients when equations have mismatched standard forms.

Scientific Reality & Correction

Always transpose both equations so all constants are on the LHS ($ax + by + c = 0$) or both on RHS ($ax + by = c$) before computing ratios.

Common Misconception

Forgetting the negative sign in the middle numerator of the cross-multiplication formula.

Scientific Reality & Correction

Use the standard 2312 mnemonic or memorise that the $y$-column carries an alternating negative sign: $\frac{-y}{a_1c_2 - a_2c_1}$.

Common Misconception

Representing a two-digit number simply as the algebraic product $xy$.

Scientific Reality & Correction

A two-digit number with tens digit $x$ and units digit $y$ has place value $10x + y$. E.g., 53 is $10(5) + 3$, not $5 \times 3$.

Common Misconception

Reversing upstream and downstream speeds in boat problems.

Scientific Reality & Correction

Stream helps the boat downstream ($u + v$), but opposes the boat upstream ($u - v$). Thus, downstream speed is always greater than upstream speed.

Linear Equations: Geometric Classifications & Algebraic Solving Matrix

Chapter Summary & 10 Key Takeaways

Takeaway 1

A linear equation in one variable $ax + b = 0$ ($a \neq 0$) has a unique root $x = -b/a$.

Takeaway 2

A linear equation in two variables $ax + by + c = 0$ represents a straight line and has infinitely many solutions $(x, y)$.

Takeaway 3

A pair of linear equations is consistent if it has at least one solution, and inconsistent if it has no solution.

Takeaway 4

Intersecting Lines: $a_1/a_2 \neq b_1/b_2$ gives a unique point of intersection (Consistent).

Takeaway 5

Coincident Lines: $a_1/a_2 = b_1/b_2 = c_1/c_2$ gives infinitely many overlapping solutions (Consistent & Dependent).

Takeaway 6

Parallel Lines: $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ gives no solution (Inconsistent).

Takeaway 7

Substitution Method isolates one variable and substitutes it into the other equation.

Takeaway 8

Elimination Method equalizes coefficients of one variable by multiplication, then eliminates it via addition or subtraction.

Takeaway 9

Cross-Multiplication Method uses the circular 2312 determinant sequence to directly compute $x$ and $y$.

Takeaway 10

Applied word problems translate real-world conditions (ages, two-digit numbers $10x + y$, upstream/downstream relative speeds $u \pm v$) into simultaneous linear systems.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

For what value of $k$ will the following system of linear equations have infinitely many solutions?
$$kx + 3y - (k - 3) = 0$$
$$12x + ky - k = 0$$

Reveal Answer & Explanation

Answer: Step 1: Write down the condition for infinitely many solutions:
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
Here $a_1 = k, b_1 = 3, c_1 = -(k - 3)$ and $a_2 = 12, b_2 = k, c_2 = -k$.
$$\frac{k}{12} = \frac{3}{k} = \frac{-(k - 3)}{-k} = \frac{k - 3}{k}$$
Step 2: Solve the first pair:
$$\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$$
Step 3: Test $k = +6$ and $k = -6$ in the second ratio:
• If $k = 6$: $\frac{3}{6} = \frac{1}{2}$ and $\frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}$. Both equal $\frac{6}{12} = \frac{1}{2}$. Satisfied!
• If $k = -6$: $\frac{3}{-6} = -\frac{1}{2}$, but $\frac{-6 - 3}{-6} = \frac{-9}{-6} = \frac{3}{2}$. Mismatch!
$$\mathbf{\text{Answer: } k = 6}$$
Equate $k/12 = 3/k = (k-3)/k$. Note that $k = -6$ does not satisfy the second equality.

Solve the following system by elimination:
$$0.2x + 0.3y = 1.3$$
$$0.4x + 0.5y = 2.3$$

Reveal Answer & Explanation

Answer: Step 1: Multiply both equations by 10 to clear decimals:
$$2x + 3y = 13 \quad \text{--- (1)}$$
$$4x + 5y = 23 \quad \text{--- (2)}$$
Step 2: Multiply equation (1) by 2:
$$4x + 6y = 26 \quad \text{--- (3)}$$
Step 3: Subtract equation (2) from equation (3):
$$(4x + 6y) - (4x + 5y) = 26 - 23 \implies y = \mathbf{3}$$
Step 4: Substitute $y = 3$ into equation (1):
$$2x + 3(3) = 13 \implies 2x + 9 = 13 \implies 2x = 4 \implies x = \mathbf{2}$$
$$\mathbf{\text{Answer: } x = 2, \quad y = 3}$$
Multiply by 10 first to convert into integer coefficients.

Determine graphically or algebraically whether the pair of equations $2x - 3y = 8$ and $4x - 6y = 9$ is consistent or inconsistent.

Reveal Answer & Explanation

Answer: Write both in standard form:
$2x - 3y - 8 = 0 \implies a_1 = 2, b_1 = -3, c_1 = -8$
$4x - 6y - 9 = 0 \implies a_2 = 4, b_2 = -6, c_2 = -9$
Compare the ratios:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$$
$$\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$$
$$\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are strictly parallel.
$$\mathbf{\text{Answer: The system has NO solution and is INCONSISTENT.}}$$
Check if $a_1/a_2 = b_1/b_2 \neq c_1/c_2$.

5 years ago, Nuri was thrice as old as Sonu. 10 years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu today?

Reveal Answer & Explanation

Answer: Let present age of Nuri $= x$ years, and Sonu $= y$ years.
Condition 1 (5 years ago):
$$(x - 5) = 3(y - 5) \implies x - 5 = 3y - 15 \implies x - 3y = -10 \quad \text{--- (1)}$$
Condition 2 (10 years later):
$$(x + 10) = 2(y + 10) \implies x + 10 = 2y + 20 \implies x - 2y = 10 \quad \text{--- (2)}$$
Subtract (1) from (2):
$$(x - 2y) - (x - 3y) = 10 - (-10) \implies y = \mathbf{20}$$
Substitute $y = 20$ into (2):
$$x - 2(20) = 10 \implies x - 40 = 10 \implies x = \mathbf{50}$$
$$\mathbf{\text{Answer: Nuri is 50 years old and Sonu is 20 years old.}}$$
Ages 5 yrs ago: $(x-5)=3(y-5)$; ages 10 yrs later: $(x+10)=2(y+10)$.

Solve for $x$ and $y$ (Equations Reducible to Linear Form):
$$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$$
$$\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$$

Reveal Answer & Explanation

Answer: Let $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$:
$$2u + 3v = 2 \quad \text{--- (1)}$$
$$4u - 9v = -1 \quad \text{--- (2)}$$
Multiply equation (1) by 3:
$$6u + 9v = 6 \quad \text{--- (3)}$$
Add (2) and (3):
$$10u = 5 \implies u = \frac{5}{10} = \frac{1}{2}$$
Substitute $u = \frac{1}{2}$ into (1):
$$2\left(\frac{1}{2}\right) + 3v = 2 \implies 1 + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3}$$
Since $u = \frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 2^2 = \mathbf{4}$.
Since $v = \frac{1}{\sqrt{y}} = \frac{1}{3} \implies \sqrt{y} = 3 \implies y = 3^2 = \mathbf{9}$.
$$\mathbf{\text{Answer: } x = 4, \quad y = 9}$$
Substitute $u = 1/\sqrt{x}$ and $v = 1/\sqrt{y}$, solve for $u, v$, then square.

A fraction becomes $\frac{9}{11}$ if 2 is added to both numerator and denominator. If 3 is added to both, it becomes $\frac{5}{6}$. Find the fraction.

Reveal Answer & Explanation

Answer: Let the numerator be $x$ and denominator be $y$. Fraction $= \frac{x}{y}$.
Condition 1: $\frac{x + 2}{y + 2} = \frac{9}{11} \implies 11(x + 2) = 9(y + 2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4 \quad \text{--- (1)}$
Condition 2: $\frac{x + 3}{y + 3} = \frac{5}{6} \implies 6(x + 3) = 5(y + 3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3 \quad \text{--- (2)}$
Multiply (1) by 5 and (2) by 9:
$$55x - 45y = -20 \quad \text{--- (3)}$$
$$54x - 45y = -27 \quad \text{--- (4)}$$
Subtract (4) from (3):
$$x = -20 - (-27) = \mathbf{7}$$
Substitute $x = 7$ into (2):
$$6(7) - 5y = -3 \implies 42 - 5y = -3 \implies 5y = 45 \implies y = \mathbf{9}$$
$$\mathbf{\text{Answer: The required fraction is } \frac{7}{9}}$$
Cross-multiply $(x+2)/(y+2)=9/11$ and $(x+3)/(y+3)=5/6$ to form two linear equations.

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