Linear Equations in Two Variables: Solved Examples

Yaar, ever felt that moment in a math exam where you see a question and think, 'Ugh, not this again!'? Especially when it comes to solving those tricky equations with 'x' and 'y'? You're not alone, bilkul!

Many of us struggle with Linear Equations in Two Variables, even though it's super important for Class 9 and 10. But what if I told you it's actually one of the easiest topics to master once you get the hang of it?

This article is your personal guide to cracking these problems. We'll go through solved examples, step-by-step, using methods like substitution, elimination, and even cross-multiplication. So, grab your notebook, and let's conquer this!

Linear equations in two variables, like $ax + by + c = 0$, are fundamental building blocks in mathematics. They show up everywhere, from physics problems to economics models. For Class 9, you're introduced to them, and in Class 10 (NCERT Chapter 3 for CBSE, or Selina Concise Chapter 5 for ICSE), you dive deep into pairs of linear equations.

Accha, did you know that 'Pair of Linear Equations in Two Variables' carries a significant weightage in your board exams? For CBSE Class 10, it's often 6-8 marks, and it's a core part of the ICSE syllabus too, tested thoroughly in Section A and B.

The challenge often isn't the concept itself, but applying the right method efficiently and avoiding silly calculation mistakes. That's where practice with solved examples really helps, suno!

To solve a pair of linear equations in two variables, you essentially need to find values for 'x' and 'y' that satisfy both equations simultaneously. There are three main algebraic methods we use:

1. Substitution Method: Express one variable in terms of the other from one equation, then substitute it into the second equation.
2. Elimination Method: Make the coefficients of one variable equal in both equations (by multiplying them by suitable numbers), then add or subtract to eliminate that variable.
3. Cross-Multiplication Method: A direct formula-based approach that's super quick once you memorise it, especially useful for competitive exams or when you're short on time.

We'll explore each with examples, so you understand when to use which method effectively.

Let's kick things off with a classic using the substitution method. This is often the most intuitive method for many students.

Problem: Solve the following pair of linear equations by the substitution method:

Solution:

Step 1: Express one variable in terms of the other.
From Equation 1, we can write $y = 14 - x$. Let's call this Equation 3.

Step 2: Substitute this expression into the other equation.
Substitute $y = 14 - x$ into Equation 2:
$x - (14 - x) = 4$
$x - 14 + x = 4$
$2x - 14 = 4$

Step 3: Solve for the single variable.
$2x = 4 + 14$
$2x = 18$
$x = \frac{18}{2}$
$x = 9$

Step 4: Substitute the value back to find the other variable.
Substitute $x = 9$ into Equation 3 ($y = 14 - x$):
$y = 14 - 9$
$y = 5$

Answer: So, the solution is $x = 9$ and $y = 5$. You can always verify your answer by plugging these values back into the original equations!

Now, let's try the elimination method. This is often faster, especially when coefficients are easy to match.

Problem: Solve the following pair of linear equations by the elimination method:

Solution:

Step 1: Make the coefficients of one variable equal.
Look at the 'y' coefficients: 4 and -2. If we multiply Equation 2 by 2, the 'y' coefficient becomes -4, which is easy to eliminate with +4y from Equation 1.

Multiply Equation 2 by 2:
$2 \times (2x - 2y) = 2 \times 2$
$4x - 4y = 4 \quad \text{(Equation 3)}$

Step 2: Add or subtract the equations to eliminate a variable.
Now, add Equation 1 and Equation 3:
$(3x + 4y) + (4x - 4y) = 10 + 4$
$3x + 4x + 4y - 4y = 14$
$7x = 14$

Step 3: Solve for the single variable.
$x = \frac{14}{7}$
$x = 2$

Step 4: Substitute the value back to find the other variable.
Substitute $x = 2$ into Equation 1 ($3x + 4y = 10$):
$3(2) + 4y = 10$
$6 + 4y = 10$
$4y = 10 - 6$
$4y = 4$
$y = \frac{4}{4}$
$y = 1$

Answer: The solution is $x = 2$ and $y = 1$. See how quick that was?

Word problems are where the real fun (and challenge) begins! They test your ability to translate real-life scenarios into mathematical equations. These are super common in CBSE Class 10 (from NCERT Exercise 3.3/3.4) and ICSE exams.

Problem: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

Solution:

Step 1: Define variables.
Let the present age of Jacob be $x$ years.
Let the present age of his son be $y$ years.

Step 2: Formulate equations from the given conditions.

Condition 1: Five years hence (i.e., in 5 years):
Jacob's age will be $x + 5$.
Son's age will be $y + 5$.
According to the problem: $x + 5 = 3(y + 5)$
$x + 5 = 3y + 15$
$x - 3y = 10 \quad \text{(Equation 1)}$

Condition 2: Five years ago:
Jacob's age was $x - 5$.
Son's age was $y - 5$.
According to the problem: $x - 5 = 7(y - 5)$
$x - 5 = 7y - 35$
$x - 7y = -30 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations (using elimination here).
We have:
$x - 3y = 10 \quad \text{(Equation 1)}$
$x - 7y = -30 \quad \text{(Equation 2)}$

Notice that the coefficient of 'x' is already 1 in both equations. We can subtract Equation 2 from Equation 1 to eliminate 'x'.

$(x - 3y) - (x - 7y) = 10 - (-30)$
$x - 3y - x + 7y = 10 + 30$
$4y = 40$
$y = \frac{40}{4}$
$y = 10$

Step 4: Substitute the value back to find the other variable.
Substitute $y = 10$ into Equation 1 ($x - 3y = 10$):
$x - 3(10) = 10$
$x - 30 = 10$
$x = 10 + 30$
$x = 40$

Answer: Jacob's present age is 40 years, and his son's present age is 10 years. Always write your final answer in the context of the word problem!

It's easy to think math is just about textbooks, right? But linear equations are everywhere! From figuring out how much fuel a rocket needs (okay, maybe not your daily life, but still!), to calculating costs for a business, they are super useful.

Think about setting up a budget: you have fixed expenses and variable ones, forming a linear relationship. Or even in sports, calculating speed and distance. In computer programming and data science, linear equations are fundamental for building predictive models and machine learning algorithms. In fact, NASSCOM projects India's AI market to reach $17 billion by 2027, and 73% of data science job postings require proficiency in statistics and linear algebra, guess what linear algebra is built on? You got it, linear equations!

So, mastering this isn't just for your board exams; it's building a foundation for a future where logical thinking and problem-solving are key!