Linear Equations in Two Variables: A Comprehensive Guide

 Linear Equations in Two Variables: A Comprehensive Guide

Linear equations in two variables form the cornerstone of algebra and are essential for solving real-life problems involving two unknown quantities. This blog will provide a detailed explanation of the concept, methods of solving these equations, and illustrative examples.

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What is a Linear Equation in Two Variables?

A linear equation in two variables is an equation that can be written in the standard form:
$ax + by + c = 0$
where:

• $a$, $b$, and $c$ are constants, and
• $x$ and $y$ are the variables.

For example:

1. $2x + 3y = 6$
2. $x - y = 2$

Each solution of the equation is a pair of values $(x, y)$ that satisfies the equation.

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Solving Pairs of Linear Equations

To solve two linear equations in two variables, we aim to find the values of $x$ and $y$ that satisfy both equations simultaneously. The general methods include:

1. Graphical Method

The graphical method involves plotting both equations on a graph and finding their point of intersection.

Steps:

1. Rewrite each equation in the slope-intercept form $y = mx + c$.
2. Plot the graph for each equation.
3. The intersection point gives the solution.

Example:
Solve:

1. $x + y = 5$
2. $x - y = 1$

For $x + y = 5$:

• If $x = 0$, then $y = 5$.
• If $y = 0$, then $x = 5$.

For $x - y = 1$:

• If $x = 0$, then $y = -1$.
• If $y = 0$, then $x = 1$.

Plot these points on a graph. The lines intersect at $(3, 2)$, which is the solution.

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2. Substitution Method

In this method, we solve one equation for one variable and substitute it into the other equation.

Steps:

1. Solve one equation for $x$ or $y$.
2. Substitute this expression into the second equation.
3. Solve for the remaining variable.
4. Substitute the value back into the first equation to find the other variable.

Example:
Solve:

1. $x + y = 5$
2. $x - y = 1$

From $x + y = 5$, $x = 5 - y$.
Substitute $x = 5 - y$ into $x - y = 1$:
$(5 - y) - y = 1$
$5 - 2y = 1$
$-2y = -4$
$y = 2$

Substitute $y = 2$ into $x + y = 5$:
$x + 2 = 5$
$x = 3$.

Solution: $(x, y) = (3, 2)$.

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3. Elimination Method

This method eliminates one variable by adding or subtracting the equations.

Steps:

1. Align both equations.
2. Multiply one or both equations to make the coefficients of one variable equal.
3. Add or subtract the equations to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.

Example:
Solve:

1. $2x + 3y = 12$
2. $x - y = 3$

Multiply the second equation by 2:

1. $2x + 3y = 12$
2. $2x - 2y = 6$

Subtract the second equation from the first:
$(2x + 3y) - (2x - 2y) = 12 - 6$
$5y = 6$
$y = 6/5$.

Substitute $y = 6/5$ into $x - y = 3$:
$x - 6/5 = 3$
$x = 3 + 6/5 = 15/5 + 6/5 = 21/5$.

Solution: $(x, y) = (21/5, 6/5)$.

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4. Cross-Multiplication Method

This method uses a formulaic approach to solve the equations directly.

Formula:
For equations:
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$

The solution is:

$$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \quad y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}.$$

Example:
Solve:

1. $3x + 2y - 5 = 0$
2. $2x - y - 4 = 0$

Using the formula:

$$x = \frac{(2)(-4) - (-1)(-5)}{(3)(-1) - (2)(2)} = \frac{-8 - 5}{-3 - 4} = \frac{-13}{-7} = \frac{13}{7},$$ $$y = \frac{(5)(2) - (-4)(3)}{(3)(-1) - (2)(2)} = \frac{10 + 12}{-7} = \frac{22}{-7} = \frac{-22}{7}.$$

Solution: $(x, y) = \left(\frac{13}{7}, -\frac{22}{7}\right)$.

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Applications of Linear Equations in Two Variables

• Business: Cost-revenue analysis and profit prediction.
• Science: Chemical equations balancing and physics problems.
• Daily Life: Solving problems involving distances, mixtures, or investments.

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Conclusion

Mastering linear equations in two variables equips students with a critical tool for solving mathematical and real-world problems. Each method has its advantages, and understanding all of them ensures flexibility and efficiency in different scenarios. Practice regularly to gain confidence and proficiency!