NCERT Solutions for Class 9 Maths Exercise 4.2 Question 2
Understanding the Question 🧐
This question asks us to find four different solutions for three separate linear equations. A “solution” is an ordered pair &&(x, y)&& that makes the equation true. Since these are linear equations in two variables, they have an infinite number of solutions. Our job is to find any four of them for each equation. The method is simple: we choose a value for one variable (like &&x&&) and then solve the equation to find the value of the other variable (&&y&&). These ncert solutions will guide you through the process.
Write four solutions for each of the following equations:
Part (i): &&2x + y = 7&& 📝
To make it easier to find solutions, let’s first rearrange the equation to isolate &&y&&.
&&y = 7 – 2x&&
Now, we can choose any value for &&x&& and easily calculate &&y&&.
-
Let &&x = 0&&:
&&y = 7 – 2(0) = 7 – 0 = 7&&.
The solution is &&(0, 7)&&. -
Let &&x = 1&&:
&&y = 7 – 2(1) = 7 – 2 = 5&&.
The solution is &&(1, 5)&&. -
Let &&x = 2&&:
&&y = 7 – 2(2) = 7 – 4 = 3&&.
The solution is &&(2, 3)&&. -
Let &&x = 3&&:
&&y = 7 – 2(3) = 7 – 6 = 1&&.
The solution is &&(3, 1)&&.
So, four solutions for &&2x + y = 7&& are &&(0, 7), (1, 5), (2, 3),&& and &&(3, 1)&&.
Part (ii): &&\pi x + y = 9&& 📝
Don’t be intimidated by &&\pi&& (pi)! Just treat it as a constant number. Again, let’s isolate &&y&&.
&&y = 9 – \pi x&&
Now, let’s substitute values for &&x&&.
-
Let &&x = 0&&:
&&y = 9 – \pi(0) = 9 – 0 = 9&&.
The solution is &&(0, 9)&&. -
Let &&x = 1&&:
&&y = 9 – \pi(1) = 9 – \pi&&.
The solution is &&(1, 9 – \pi)&&. -
Let &&x = 2&&:
&&y = 9 – \pi(2) = 9 – 2\pi&&.
The solution is &&(2, 9 – 2\pi)&&. -
Let &&x = -1&&:
&&y = 9 – \pi(-1) = 9 + \pi&&.
The solution is &&(-1, 9 + \pi)&&.
So, four solutions for &&\pi x + y = 9&& are &&(0, 9), (1, 9 – \pi), (2, 9 – 2\pi),&& and &&(-1, 9 + \pi)&&.
Part (iii): &&x = 4y&& 📝
In this equation, &&x&& is already isolated. So, it’s much easier to choose values for &&y&& and then find &&x&&.
-
Let &&y = 0&&:
&&x = 4(0) = 0&&.
The solution is &&(0, 0)&&. -
Let &&y = 1&&:
&&x = 4(1) = 4&&.
The solution is &&(4, 1)&&. -
Let &&y = 2&&:
&&x = 4(2) = 8&&.
The solution is &&(8, 2)&&. -
Let &&y = -1&&:
&&x = 4(-1) = -4&&.
The solution is &&(-4, -1)&&.
So, four solutions for &&x = 4y&& are &&(0, 0), (4, 1), (8, 2),&& and &&(-4, -1)&&.
Quick Tip!
The easiest values to start with are almost always &&x=0&& and &&y=0&&. Finding the points where the line crosses the x and y axes (the intercepts) often gives you two very quick solutions!Key Concepts
- A solution must be written as an ordered pair &&(x, y)&&. The &&x&& value always comes first.
- There are infinitely many correct answers to this question. The solutions shown here are just examples.
- Rearranging the equation to isolate one variable first (like getting &&y = …&&) can save a lot of work.
FAQ (Frequently Asked Questions)
Q: How many solutions does a linear equation in two variables have?
A: A linear equation in two variables has infinitely many solutions. This question just asks for any four of them as examples.
Q: Is there only one correct set of four solutions for these equations?
A: No, there are infinite correct solutions. The solutions provided are just examples. You can find different solutions by picking different initial values for &&x&& or &&y&&.
Q: What does &&\pi&& (pi) represent in the equation &&\pi x + y = 9&&?
A: In this equation, &&\pi&& (pi) is treated as a constant, just like any other number. It is an irrational number approximately equal to 3.14159. The solutions will simply have &&\pi&& in them, and you don’t need to substitute its decimal value.
Q: For the equation &&x = 4y&&, why is it easier to choose values for &&y&& first?
A: Since the equation is already solved for &&x&& (&&x = 4y&&), it’s much faster to choose a value for &&y&& and multiply by 4 to find &&x&&. If you chose &&x&& first, you would have to divide by 4, which might lead to fractions (e.g., if you picked &&x=1&&, then &&y = 1/4&&).
Further Reading
For more information on linear equations and to access the official textbook, you can visit the NCERT website. Official NCERT Website.
