NCERT Solutions Class 9 Maths Chapter 4 Exercise 4.2

Understanding Infinitely Many Solutions

A linear equation in two variables is represented in the general form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero. Every point on the line represented by this equation is a solution. Since a straight line consists of an infinite number of points, a linear equation in two variables has infinitely many solutions.
To find these solutions, we can choose any real number as the value of one variable (say, $x$) and substitute it into the equation to find the corresponding value of the other variable (say, $y$). This pair of values, written as coordinates $(x, y)$, represents a unique solution. Every time you change the value of $x$, you will obtain a new, unique value for $y$. Thus, there is no end to the number of solutions you can generate.

Step-by-Step Method to Find Solutions

1. Isolate One Variable — Express one variable in terms of the other to simplify calculation. For example, express $y$ in terms of $x$ to get $y = \frac{-c - ax}{b}$.
2. Select Values for the Independent Variable — Choose convenient integer values for $x$ (such as $0, 1, 2, -1$) to keep calculations straightforward and avoid complicated fractions.
3. Substitute and Solve — Substitute the chosen value of $x$ into your isolated equation and compute the corresponding value of $y$.
4. Write the Coordinate Pair — Express your final answer as an ordered pair $(x, y)$ which represents a point on the coordinate plane.

NCERT Exercise 4.2 Solved Examples

• Example 1: Find four solutions for the equation $2x + y = 7$. Solution: 1. First, express $y$ in terms of $x$: $y = 7 - 2x$. 2. Let $x = 0$: $y = 7 - 2(0) = 7$. First solution is $(0, 7)$. 3. Let $x = 1$: $y = 7 - 2(1) = 5$. Second solution is $(1, 5)$. 4. Let $x = 2$: $y = 7 - 2(2) = 3$. Third solution is $(2, 3)$. 5. Let $x = 3$: $y = 7 - 2(3) = 1$. Fourth solution is $(3, 1)$. Therefore, the four solutions are $(0, 7), (1, 5), (2, 3),$ and $(3, 1)$.
• Example 2: Find the value of $k$ if $x = 2, y = 1$ is a solution of the equation $2x + 3y = k$. Solution: 1. Since $(2, 1)$ is a solution, it must satisfy the given equation. 2. Substitute $x = 2$ and $y = 1$ into $2x + 3y = k$: $2(2) + 3(1) = k$ $4 + 3 = k$ $k = 7$ Therefore, the value of $k$ is $7$.

Avoid Common Mistakes in Board Exams

• Incorrect Coordinate Order: Always write coordinates as $(x, y)$. A very common mistake is writing $(y, x)$, which completely changes the position and value.
• Verifying LHS = RHS: When testing if a point is a solution, substitute both values. If LHS equals RHS, it is a solution. If not, it is not a solution.
• Choose Values Smartly: If your equation has a fraction (like $y = \frac{x}{3}$), select values of $x$ that are multiples of the denominator (like $0, 3, 6, -3$) to keep your answers as integers.

Practice Questions with Solutions

• Q: Write four solutions for the equation $\pi x + y = 9$. A: Step 1: Express $y$ in terms of $x$: $y = 9 - \pi x$. Step 2: Substitute $x = 0$: $y = 9 - \pi(0) = 9$. Solution is $(0, 9)$. Step 3: Substitute $x = 1$: $y = 9 - \pi(1) = 9 - \pi$. Solution is $(1, 9 - \pi)$. Step 4: Substitute $x = 2$: $y = 9 - \pi(2) = 9 - 2\pi$. Solution is $(2, 9 - 2\pi)$. Step 5: Substitute $x = -1$: $y = 9 - \pi(-1) = 9 + \pi$. Solution is $(-1, 9 + \pi)$. Final answer: The four solutions are $(0, 9)$, $(1, 9 - \pi)$, $(2, 9 - 2\pi)$, and $(-1, 9 + \pi)$.
• Q: Check if $(2, 0)$ and $(4, 0)$ are solutions of the equation $x - 2y = 4$. A: Step 1: For $(2, 0)$, substitute $x = 2$ and $y = 0$ into the LHS: $2 - 2(0) = 2$. Since LHS $\neq$ RHS ($2 \neq 4$), $(2, 0)$ is not a solution. Step 2: For $(4, 0)$, substitute $x = 4$ and $y = 0$ into the LHS: $4 - 2(0) = 4$. Since LHS = RHS ($4 = 4$), $(4, 0)$ is a solution. Final answer: $(2, 0)$ is not a solution, but $(4, 0)$ is a solution.
• Q: Find the value of $k$ if $x = 3, y = 2$ is a solution of the equation $5x - 4y = k$. A: Step 1: Substitute $x = 3$ and $y = 2$ into the equation: $5(3) - 4(2) = k$. Step 2: Simplify the values: $15 - 8 = k$. Step 3: Solve for $k$: $k = 7$. Final answer: The value of $k$ is 7.
• Q: Express $y$ in terms of $x$ for the equation $3x + 4y = 12$, and find the value of $y$ when $x = -4$. A: Step 1: Isolate $y$ on the LHS: $4y = 12 - 3x$, which gives $y = \frac{12 - 3x}{4}$. Step 2: Substitute $x = -4$ into this expression: $y = \frac{12 - 3(-4)}{4}$. Step 3: Simplify the numerator: $y = \frac{12 + 12}{4} = \frac{24}{4} = 6$. Final answer: Expressed in terms of $x$, $y = \frac{12 - 3x}{4}$. When $x = -4$, $y = 6$.

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