Linear Equations: Concept, Formulas, And Practice Problems

Understanding Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations are called "linear" because they represent straight lines when graphed on a coordinate plane.

The general form of a linear equation in one variable is:

$$ ax + b = 0 $$

where:

• \( a \) and \( b \) are constants,
• \( x \) is the variable.

In two variables, a linear equation is represented as:

$$ ax + by = c $$

where:

• \( a \), \( b \), and \( c \) are constants,
• \( x \) and \( y \) are variables.

Types of Linear Equations

1. Linear Equations in One Variable: These equations involve only one variable and can be solved to get a single value for that variable.

   • Example: \( 3x + 5 = 11 \)

2. Linear Equations in Two Variables: These equations involve two variables and can have multiple solutions that form a line when plotted on a graph.

   • Example: \( 2x + 3y = 6 \)

Solving Linear Equations in One Variable

To solve linear equations, the goal is to isolate the variable on one side of the equation. Here’s a simple approach:

1. Move all variable terms to one side of the equation.
2. Move constant terms to the opposite side.
3. Isolate the variable by dividing or multiplying if necessary.

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Example Problems and Solutions

Problem 1: Linear Equation in One Variable

Solve for \( x \):

$$ 3x - 7 = 11 $$

Solution:

1. Move the constant term to the right: $$ 3x = 11 + 7 $$
2. Simplify: $$ 3x = 18 $$
3. Divide by 3 to isolate \( x \): $$ x = \frac{18}{3} = 6 $$

Answer: \( x = 6 \)

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Problem 2: Linear Equation in Two Variables

Solve the system of equations:

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

Solution: We can solve this using substitution or elimination. Let’s use substitution.

1. Solve the second equation for \( x \): $$ x = y + 2 $$
2. Substitute \( x = y + 2 \) into the first equation: $$ 2(y + 2) + 3y = 12 $$
3. Distribute and simplify: $$ 2y + 4 + 3y = 12 $$ $$ 5y = 8 $$ $$ y = \frac{8}{5} = 1.6 $$
4. Substitute \( y = 1.6 \) back into \( x = y + 2 \): $$ x = 1.6 + 2 = 3.6 $$

Answer: \( x = 3.6 \), \( y = 1.6 \)

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Problem 3: Word Problem Involving Linear Equations

Problem: A school has 5 times as many students as teachers. If the total number of students and teachers is 600, how many students and teachers are there?

Solution:

1. Let \( x \) be the number of teachers.
2. Then, the number of students is \( 5x \).
3. According to the problem: $$ x + 5x = 600 $$
4. Combine like terms: $$ 6x = 600 $$
5. Divide by 6: $$ x = \frac{600}{6} = 100 $$

So, there are \( 100 \) teachers and \( 5 \times 100 = 500 \) students.

Answer: 100 teachers and 500 students.

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Graphing Linear Equations

Graphing linear equations in two variables (e.g., \( y = 2x + 3 \)) helps visualize solutions as points on a line. The slope-intercept form \( y = mx + c \) is often used, where:

• \( m \) is the slope (rise over run),
• \( c \) is the y-intercept.

For example, to graph \( y = 2x + 3 \):

• Start at the y-intercept (0, 3).
• Use the slope \( m = 2 \) (rise 2 units, run 1 unit right) to plot additional points.

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Conclusion

Linear equations form the foundation of algebra and have broad applications in problem-solving, graphing, and real-world situations. Mastering these equations helps with more advanced topics and builds a strong mathematical foundation.