NCERT Solutions for Class 9 Maths Exercise 4.1 Question 2
Understanding the Question 🧐
In this question, we are given several linear equations. Our task is to rewrite each of them into the standard form of a linear equation in two variables, which is &&ax + by + c = 0&&. After rewriting them, we need to identify the values of the coefficients &&a&&, &&b&&, and the constant &&c&&. This is a great exercise from the ncert solutions to practice algebraic manipulation.
Express the following linear equations in the form &&ax + by + c = 0&& and indicate the values of &&a, b&& and &&c&& in each case.
Part (i): &&2x + 3y = 9.3\bar{5}&& 📝
First, we need to move all terms to the left-hand side to make the right-hand side zero.
Given equation: &&2x + 3y = 9.3\bar{5}&&
Subtract &&9.3\bar{5}&& from both sides:
&&2x + 3y – 9.3\bar{5} = 0&&
Now, we compare this with the standard form &&ax + by + c = 0&&.
On comparing, we get: &&a = 2&&, &&b = 3&&, &&c = -9.3\bar{5}&&
Part (ii): &&x – \frac{y}{5} – 10 = 0&& 📝
This equation is already in the standard form &&ax + by + c = 0&&, so we can directly compare it.
Given equation: &&x – \frac{y}{5} – 10 = 0&&
Let’s write the term &&-\frac{y}{5}&& as &&(-\frac{1}{5})y&& to make the coefficient clear.
So, the equation is &&(1)x + (-\frac{1}{5})y + (-10) = 0&&.
Comparing with &&ax + by + c = 0&&:
On comparing, we get: &&a = 1&&, &&b = -\frac{1}{5}&&, &&c = -10&&
Part (iii): &&-2x + 3y = 6&& 📝
We need to move the constant term &&6&& to the left-hand side.
Given equation: &&-2x + 3y = 6&&
Subtract &&6&& from both sides:
&&-2x + 3y – 6 = 0&&
Comparing this with &&ax + by + c = 0&&:
On comparing, we get: &&a = -2&&, &&b = 3&&, &&c = -6&&
Part (iv): &&x = 3y&& 📝
Move the term &&3y&& to the left-hand side. Notice that there is no constant term, which means &&c=0&&.
Given equation: &&x = 3y&&
Subtract &&3y&& from both sides:
&&x – 3y = 0&&
To match the standard form exactly, we can write it as &&x – 3y + 0 = 0&&.
Comparing this with &&ax + by + c = 0&&:
On comparing, we get: &&a = 1&&, &&b = -3&&, &&c = 0&&
Part (v): &&2x = -5y&& 📝
Move the term &&-5y&& to the left-hand side. When it moves, its sign will change from negative to positive.
Given equation: &&2x = -5y&&
Add &&5y&& to both sides:
&&2x + 5y = 0&&
We can write this as &&2x + 5y + 0 = 0&&. There is no constant term, so &&c=0&&.
Comparing with &&ax + by + c = 0&&:
On comparing, we get: &&a = 2&&, &&b = 5&&, &&c = 0&&
Part (vi): &&3x + 2 = 0&& 📝
This equation has an &&x&& term and a constant, but no &&y&& term. This means the coefficient of &&y&& is zero.
Given equation: &&3x + 2 = 0&&
We can write this as &&3x + 0y + 2 = 0&& to match the standard form.
Comparing with &&ax + by + c = 0&&:
On comparing, we get: &&a = 3&&, &&b = 0&&, &&c = 2&&
Part (vii): &&y – 2 = 0&& 📝
Similarly, this equation has a &&y&& term and a constant, but no &&x&& term. This means the coefficient of &&x&& is zero.
Given equation: &&y – 2 = 0&&
We can write this as &&0x + 1y – 2 = 0&&.
Comparing with &&ax + by + c = 0&&:
On comparing, we get: &&a = 0&&, &&b = 1&&, &&c = -2&&
Part (viii): &&5 = 2x&& 📝
Let’s move the &&2x&& term to the left-hand side to make the RHS zero.
Given equation: &&5 = 2x&&
Subtract &&2x&& from both sides:
&&5 – 2x = 0&&
To match the standard order, we can write it as &&-2x + 5 = 0&&. There is no &&y&& term, so its coefficient is zero.
The equation is &&-2x + 0y + 5 = 0&&.
Comparing with &&ax + by + c = 0&&:
On comparing, we get: &&a = -2&&, &&b = 0&&, &&c = 5&&
Conclusion and Key Points ✅
This exercise shows that any linear equation can be rearranged into the standard form &&ax + by + c = 0&&. This standard form is very useful because it allows us to consistently identify the key characteristics of the equation through its coefficients &&a&& and &&b&& and the constant term &&c&&.
Trick to Remember
A handy trick: If a variable (like &&x&& or &&y&&) is missing from the equation, its coefficient is simply &&0&&! For example, in &&3x + 2 = 0&&, there is no &&y&& term, so we can write it as &&3x + 0y + 2 = 0&&, which makes it clear that &&b=0&&.Points to Remember
- The standard form is always &&ax + by + c = 0&&.
- Always ensure the right-hand side of the equation is zero before you identify &&a&&, &&b&&, and &&c&&.
- Pay close attention to the signs (+ or -) of the coefficients. For example, in &&x – 3y = 0&&, the value of &&b&& is &&-3&&, not &&3&&.
FAQ (Frequently Asked Questions)
Q: What is the standard form of a linear equation in two variables?
A: The standard form is &&ax + by + c = 0&&, where &&x&& and &&y&& are variables, and &&a&&, &&b&&, and &&c&& are real numbers, with the condition that ‘&&a&&’ and ‘&&b&&’ are not both zero.
Q: Why do we need to make the right-hand side of the equation zero?
A: Making the right-hand side zero puts the equation into its standard form, &&ax + by + c = 0&&. This makes it easy and consistent to identify the coefficients &&a&&, &&b&&, and the constant term &&c&&.
Q: What if the ‘y’ term is missing in an equation? What is the value of ‘b’?
A: If a variable term is missing, its coefficient is &&0&&. For example, in the equation &&3x + 2 = 0&&, there is no ‘&&y&&’ term, so we can think of it as &&3x + 0y + 2 = 0&&. Therefore, the value of ‘&&b&&’ is &&0&&.
Q: In the equation &&x – \frac{y}{5} – 10 = 0&&, why is &&b = -\frac{1}{5}&&?
A: The term with ‘&&y&&’ is &&-\frac{y}{5}&&. This can be written as &&(-\frac{1}{5}) \times y&&. When we compare this to the ‘&&by&&’ term in the standard form, we can see that the coefficient ‘&&b&&’ is &&-\frac{1}{5}&&.
Q: Can the values of a, b, or c be fractions or decimals?
A: Yes, &&a&&, &&b&&, and &&c&& can be any real numbers, which includes integers, fractions, and decimals.
Further Reading
For more information on linear equations and to access the official textbook, you can visit the NCERT website. Official NCERT Website.
