NCERT Solutions for Class 9 Maths Exercise 4.2 Question 3

NCERT Solutions for Class 9 Maths Exercise 4.2 Question 3

NCERT Solutions for Class 9 Maths Exercise 4.2 Question 3

ncert-solutions-for-class-9-maths-exercise-4-2-question-3

Understanding the Question 🧐

This question asks us to verify whether certain points are “solutions” to a given linear equation. A point is a solution if its coordinates make the equation true. The method is to substitute the &&x&& and &&y&& values from each point into the equation and check if the Left Hand Side (LHS) equals the Right Hand Side (RHS). Let’s go through these ncert solutions step-by-step.

Check which of the following are solutions of the equation &&x – 2y = 4&& and which are not:
(i) &&(0, 2)&& (ii) &&(2, 0)&& (iii) &&(4, 0)&& (iv) &&(\sqrt{2}, 4\sqrt{2})&& (v) &&(1, 1)&&

The given equation is &&x – 2y = 4&&.
Here, LHS = &&x – 2y&& and RHS = &&4&&.

Part (i): Checking the point &&(0, 2)&& 📝

For the point &&(0, 2)&&, we have &&x = 0&& and &&y = 2&&.

Substitute these values into the LHS:

LHS = &&x – 2y = (0) – 2(2) = 0 – 4 = -4&&

Now, compare LHS with RHS:

We have LHS = &&-4&& and RHS = &&4&&. Since &&-4 \neq 4&&, the LHS is not equal to the RHS.

Conclusion: ❌ The point &&(0, 2)&& is not a solution.


Part (ii): Checking the point &&(2, 0)&& 📝

For the point &&(2, 0)&&, we have &&x = 2&& and &&y = 0&&.

Substitute these values into the LHS:

LHS = &&x – 2y = (2) – 2(0) = 2 – 0 = 2&&

Now, compare LHS with RHS:

We have LHS = &&2&& and RHS = &&4&&. Since &&2 \neq 4&&, the LHS is not equal to the RHS.

Conclusion: ❌ The point &&(2, 0)&& is not a solution.


Part (iii): Checking the point &&(4, 0)&& 📝

For the point &&(4, 0)&&, we have &&x = 4&& and &&y = 0&&.

Substitute these values into the LHS:

LHS = &&x – 2y = (4) – 2(0) = 4 – 0 = 4&&

Now, compare LHS with RHS:

We have LHS = &&4&& and RHS = &&4&&. Since LHS = RHS, the equation is satisfied.

Conclusion: ✅ The point &&(4, 0)&& is a solution.


Part (iv): Checking the point &&(\sqrt{2}, 4\sqrt{2})&& 📝

For the point &&(\sqrt{2}, 4\sqrt{2})&&, we have &&x = \sqrt{2}&& and &&y = 4\sqrt{2}&&.

Substitute these values into the LHS:

LHS = &&x – 2y = (\sqrt{2}) – 2(4\sqrt{2}) = \sqrt{2} – 8\sqrt{2} = -7\sqrt{2}&&

Now, compare LHS with RHS:

We have LHS = &&-7\sqrt{2}&& and RHS = &&4&&. Since &&-7\sqrt{2} \neq 4&&, the LHS is not equal to the RHS.

Conclusion: ❌ The point &&(\sqrt{2}, 4\sqrt{2})&& is not a solution.


Part (v): Checking the point &&(1, 1)&& 📝

For the point &&(1, 1)&&, we have &&x = 1&& and &&y = 1&&.

Substitute these values into the LHS:

LHS = &&x – 2y = (1) – 2(1) = 1 – 2 = -1&&

Now, compare LHS with RHS:

We have LHS = &&-1&& and RHS = &&4&&. Since &&-1 \neq 4&&, the LHS is not equal to the RHS.

Conclusion: ❌ The point &&(1, 1)&& is not a solution.

Final Summary ✅

After checking all the given points, only one point satisfies the equation &&x – 2y = 4&&.

  • The only solution is &&(4, 0)&&.

Watch Out!

A common mistake is mixing up &&x&& and &&y&&! Always remember that in an ordered pair &&(x, y)&&, the first number is for &&x&& and the second is for &&y&&. Forgetting this is a frequent reason for incorrect answers.

Key Concepts

  • To verify a solution, substitute the values of &&x&& and &&y&& into the equation.
  • If the Left Hand Side (LHS) equals the Right Hand Side (RHS), the point is a solution.
  • If LHS is not equal to RHS, the point is not a solution.

FAQ (Frequently Asked Questions)

Q: What does it mean for a point to be a ‘solution’ of an equation?

A: A point &&(x, y)&& is a solution to an equation if, when you substitute the &&x&& and &&y&& values from the point into the equation, it results in a true statement (i.e., the Left Hand Side equals the Right Hand Side).

Q: What are LHS and RHS in an equation?

A: LHS stands for Left Hand Side, and RHS stands for Right Hand Side. They are the expressions on either side of the equals sign. For the equation &&x – 2y = 4&&, the LHS is ‘&&x – 2y&&’ and the RHS is ‘&&4&&’.

Q: In the point &&(\sqrt{2}, 4\sqrt{2})&&, which value is &&x&& and which is &&y&&?

A: In any ordered pair &&(a, b)&&, the first value is always for &&x&& and the second value is for &&y&&. So, &&x = \sqrt{2}&& and &&y = 4\sqrt{2}&&.

Further Reading

For more information on linear equations and to access the official textbook, you can visit the NCERT website. Official NCERT Website.

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