# Linear Equations Solving Methods in Class 9 Mathematics

QuietMagicRealism
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## Solving Linear Equations in Class 9th PSEB Mathematics

In your mathematics journey through Class 9 with Punjab School Education Board (PSEB), you'll encounter linear equations — foundational building blocks of algebra and problem-solving skills. In this section, let's explore how to solve these simple yet powerful expressions that describe real life situations using two fundamental methods.

### One Step Method

This method is effective when solving one variable equations like (ax + b = c) where coefficients ((a,\ b,\ c)) involve only whole numbers or integers. Here's how it works:

• If there's a coefficient before the x term, isolate it by dividing both sides by the number, i.e., set [ax = c - b]
• Next, move all terms containing x to one side while placing constants to another side: [x = \frac{c - b}{a}]

Example: Solve (7x - 6= 8.) First, divide both sides by 7: [ x=\frac{8+6}{7} ] Then simplify: [ x=\frac{14}{7}=\frac{2\times 7}{7}=2 ]

### Two Step Method

The two step method allows us to solve more complex linear equations involving fractions, decimals, or multiplication and division operations within parentheses. It involves following these steps:

1. Simplify each term without changing its sign.
2. Perform any operation inside parentheses first.
3. Move all terms containing (x) to one side while keeping their signs intact.
4. Set up the equation so that the coefficient of (x) equals zero. Then, find the constant part divided by the coefficient of (x).

Let's apply our newfound knowledge to some examples:

Solve (3-\dfrac{x}{5}=7.)

Firstly, reverse the order of the fraction term and multiply the entire equation by 5: [ 5\left( 3-\frac{x}{5} \right)=5(7)\quad\Rightarrow\quad 15-x,= ,35 ] Next, add (x) to both sides: [ x=20-15 \quad\Rightarrow \quad x=5 ]

### Applications and Real Life Situations

Linear relationships often represent proportions, rates, slopes of lines, and distance-time graphs. For instance:

• A shopkeeper sells apples at Rs.25 per dozen ((12)); if he earns Rs.300 selling them, determine how many dozens were sold. To solve this, we can create a proportion: (\frac{\text{number of apples}}{12}=\frac{Rs300}{25}.)

• Determine the time taken to travel 40 km at a speed of 30kmph. To do this, formulate a ratio: [\frac{\text{distance}}{\text{speed}}=\left(\frac{40,\text{km}}{30,\text{kmh}^{-1}}\right)=\left(\frac{4}{3},\text{hour}\right)^{-1}=\left(\frac{3}{4},\text{hour}\right),] which indicates 0.75 hours or (45\ minutes.)

As you progress further into Class 9 math, keep practicing problems related to linear equations; they will help you understand higher level concepts such as quadratic equations, systems of linear equations, and graphical representation of data much easier!

## Description

Explore the one-step and two-step methods for solving linear equations in 9th Grade Mathematics with Punjab School Education Board (PSEB). Learn how to manipulate equations involving whole numbers, fractions, and decimals to find solutions. Apply these techniques to real-life scenarios involving proportions, rates, and distance-time graphs.