In-Depth: Linear Equations

CLAT Application & Relevance

Importance: Medium. While you won't see complex algebraic equations, 'basic algebra' as mentioned in the CLAT syllabus implies the ability to form and solve simple linear equations. This skill is crucial when dealing with caselets where quantities are unknown and their relationships are described textually (e.g., "X is twice Y plus 5," "total of A and B is Z").

How it's tested: Deriving one or two linear equations from a text-based problem and solving them to find unknown values, typically as part of a larger Data Interpretation set.

Section 1: Core Concepts & Solving Methods

A linear equation is an algebraic equation in which each term has an exponent of 1, and the graphing of the equation results in a straight line. They involve variables (like x, y) and constants, connected by arithmetic operations.

Key Definitions

Variable: A symbol (usually a letter like x, y, z) that represents an unknown value.
Constant: A fixed numerical value.
Coefficient: A number that multiplies a variable (e.g., in 3x, 3 is the coefficient).
Linear Equation in One Variable: An equation with one variable (e.g., 2x + 5 = 15).
Linear Equation in Two Variables: An equation with two variables (e.g., 3x + 2y = 10). To solve for two variables, you typically need two independent equations (a system of equations).

Solving Methods for Linear Equations

For one variable: Isolate the variable using inverse operations.

For two variables (Systems of Equations):

Substitution Method:
1. Solve one equation for one variable (e.g., express 'y' in terms of 'x').
2. Substitute this expression into the other equation.
3. Solve the resulting one-variable equation.
4. Substitute the found value back into one of the original equations to find the other variable.
Elimination Method:
1. Multiply one or both equations by constants so that the coefficients of one variable become opposites (e.g., +3x and -3x).
2. Add the two equations together to eliminate that variable.
3. Solve the resulting one-variable equation.
4. Substitute the found value back into one of the original equations to find the other variable.

Section 2: Solved CLAT-Style Examples

Example 1: Linear Equation in One Variable (Caselet)

Passage Context: "A legal services company charges a flat fee plus an hourly rate for its consulting services. For a recent project, the flat fee was ₹5,000. The company worked for 'H' hours and billed a total of ₹25,000. If the hourly rate is constant."

Question: "If the hourly rate charged is ₹500, for how many hours (H) did the company work on this project?"

Detailed Solution:
1. Identify Knowns and Unknown: Flat Fee = ₹5,000 Total Bill = ₹25,000 Hourly Rate = ₹500 Unknown = H (number of hours)
2. Formulate the Equation: Total Bill = Flat Fee + (Hourly Rate × Hours) 25000 = 5000 + (500 * H)
3. Solve for H: 25000 - 5000 = 500 * H 20000 = 500 * H H = 20000 / 500 H = 40.
Answer: The company worked for 40 hours on this project.

Example 2: System of Linear Equations (Caselet)

Passage Context: "A legal research agency employs two types of researchers: Senior Researchers and Junior Researchers. The total number of researchers is 70. Each Senior Researcher is paid ₹1,500 per task, and each Junior Researcher is paid ₹800 per task. If the agency paid a total of ₹70,000 for a particular research task."

Question: "How many Senior Researchers and Junior Researchers does the agency employ?"

Detailed Solution (using Substitution Method):
1. Define Variables: Let 'S' be the number of Senior Researchers. Let 'J' be the number of Junior Researchers.
2. Formulate Equations: Equation 1 (Total Researchers): S + J = 70 Equation 2 (Total Payment): 1500S + 800J = 70000
3. Simplify Equation 2 (divide by 100): 15S + 8J = 700
4. From Equation 1, express J in terms of S: J = 70 - S
5. Substitute into Simplified Equation 2: 15S + 8(70 - S) = 700 15S + 560 - 8S = 700 7S = 700 - 560 7S = 140 S = 140 / 7 = 20.
6. Find J using S: J = 70 - S = 70 - 20 = 50.
Answer: The agency employs 20 Senior Researchers and 50 Junior Researchers.