CBSE Class 9 Maths Chapter 4: Linear Equations in Two Variables NCERT Solutions

Let the cost of a notebook be represented by the variable $x$ and the cost of a pen be represented by the variable $y$.

According to the problem statement, the cost of a notebook is twice the cost of a pen. This can be written as:

$$x\; =\; 2y$$

To express this in the standard form of a linear equation in two variables ($ax\; +\; by\; +\; c\; =\; 0$), we can rearrange the equation:

$$x\; -\; 2y\; =\; 0$$

Thus, the linear equation representing the statement is $x\; -\; 2y\; =\; 0$.

1. Given equation: $2x\; +\; 3y\; =\; 9.3\backslash overline\{5\}$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we move the constant term to the left side:

   $$2x\; +\; 3y\; -\; 9.3\backslash overline\{5\}\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 2$, $b\; =\; 3$, $c\; =\; -9.3\backslash overline\{5\}$

2. Given equation: $x\; -\; \backslash frac\{y\}\{5\}\; -\; 10\; =\; 0$

   This equation is already in the form $ax\; +\; by\; +\; c\; =\; 0$.

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 1$, $b\; =\; -\backslash frac\{1\}\{5\}$, $c\; =\; -10$

3. Given equation: $-2x\; +\; 3y\; =\; 6$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we move the constant term to the left side:

   $$-2x\; +\; 3y\; -\; 6\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; -2$, $b\; =\; 3$, $c\; =\; -6$

4. Given equation: $x\; =\; 3y$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we move the $3y$ term to the left side and add $c=0$:

   $$x\; -\; 3y\; +\; 0\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 1$, $b\; =\; -3$, $c\; =\; 0$

5. Given equation: $2x\; =\; -5y$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we move the $-5y$ term to the left side and add $c=0$:

   $$2x\; +\; 5y\; +\; 0\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 2$, $b\; =\; 5$, $c\; =\; 0$

6. Given equation: $3x\; +\; 2\; =\; 0$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we can include the $y$ term with a coefficient of 0:

   $$3x\; +\; 0\; \backslash cdot\; y\; +\; 2\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 3$, $b\; =\; 0$, $c\; =\; 2$

7. Given equation: $y\; -\; 2\; =\; 0$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we can include the $x$ term with a coefficient of 0:

   $$0\; \backslash cdot\; x\; +\; 1\; \backslash cdot\; y\; -\; 2\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; 0$, $b\; =\; 1$, $c\; =\; -2$

8. Given equation: $5\; =\; 2x$

   To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we rearrange the terms:

   $$5\; -\; 2x\; =\; 0$$

   Then, we order the terms and include the $y$ term with a coefficient of 0:

   $$-2x\; +\; 0\; \backslash cdot\; y\; +\; 5\; =\; 0$$

   Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get:

   $a\; =\; -2$, $b\; =\; 0$, $c\; =\; 5$

The correct option is (iii) infinitely many solutions.

A linear equation in two variables, such as $y\; =\; 3x\; +\; 5$, represents a straight line on a graph. A straight line consists of an infinite number of points. Each point on the line corresponds to a pair of (x, y) values that satisfies the equation. Therefore, there are infinitely many solutions to this equation. We can find different solutions by choosing any real value for $x$ and calculating the corresponding value of $y$.

1. For the equation $2x\; +\; y\; =\; 7$:

   We can find solutions by choosing values for $x$ and calculating the corresponding $y$ values.

   Solution 1: Let $x\; =\; 1$. Then, $2(1)\; +\; y\; =\; 7\; \backslash implies\; 2\; +\; y\; =\; 7\; \backslash implies\; y\; =\; 7\; -\; 2\; =\; 5$. So, (1, 5) is a solution.

   Solution 2: Let $x\; =\; 2$. Then, $2(2)\; +\; y\; =\; 7\; \backslash implies\; 4\; +\; y\; =\; 7\; \backslash implies\; y\; =\; 7\; -\; 4\; =\; 3$. So, (2, 3) is a solution.

   Solution 3: Let $x\; =\; 3$. Then, $2(3)\; +\; y\; =\; 7\; \backslash implies\; 6\; +\; y\; =\; 7\; \backslash implies\; y\; =\; 7\; -\; 6\; =\; 1$. So, (3, 1) is a solution.

   Solution 4: Let $x\; =\; 4$. Then, $2(4)\; +\; y\; =\; 7\; \backslash implies\; 8\; +\; y\; =\; 7\; \backslash implies\; y\; =\; 7\; -\; 8\; =\; -1$. So, (4, -1) is a solution.

   Therefore, four solutions are (1, 5), (2, 3), (3, 1), and (4, -1).

2. For the equation $\backslash pi\; x\; +\; y\; =\; 9$:

   We can find solutions by choosing values for $x$ and calculating the corresponding $y$ values.

   Solution 1: Let $x\; =\; \backslash frac\{1\}\{\backslash pi\}$. Then, $\backslash pi\; \backslash left(\backslash frac\{1\}\{\backslash pi\}\backslash right)\; +\; y\; =\; 9\; \backslash implies\; 1\; +\; y\; =\; 9\; \backslash implies\; y\; =\; 9\; -\; 1\; =\; 8$. So, $\backslash left(\backslash frac\{1\}\{\backslash pi\},\; 8\backslash right)$ is a solution.

   Solution 2: Let $x\; =\; \backslash frac\{2\}\{\backslash pi\}$. Then, $\backslash pi\; \backslash left(\backslash frac\{2\}\{\backslash pi\}\backslash right)\; +\; y\; =\; 9\; \backslash implies\; 2\; +\; y\; =\; 9\; \backslash implies\; y\; =\; 9\; -\; 2\; =\; 7$. So, $\backslash left(\backslash frac\{2\}\{\backslash pi\},\; 7\backslash right)$ is a solution.

   Solution 3: Let $x\; =\; \backslash frac\{3\}\{\backslash pi\}$. Then, $\backslash pi\; \backslash left(\backslash frac\{3\}\{\backslash pi\}\backslash right)\; +\; y\; =\; 9\; \backslash implies\; 3\; +\; y\; =\; 9\; \backslash implies\; y\; =\; 9\; -\; 3\; =\; 6$. So, $\backslash left(\backslash frac\{3\}\{\backslash pi\},\; 6\backslash right)$ is a solution.

   Solution 4: Let $x\; =\; 0$. Then, $\backslash pi(0)\; +\; y\; =\; 9\; \backslash implies\; 0\; +\; y\; =\; 9\; \backslash implies\; y\; =\; 9$. So, (0, 9) is a solution.

   Therefore, four solutions are $\backslash left(\backslash frac\{1\}\{\backslash pi\},\; 8\backslash right)$, $\backslash left(\backslash frac\{2\}\{\backslash pi\},\; 7\backslash right)$, $\backslash left(\backslash frac\{3\}\{\backslash pi\},\; 6\backslash right)$, and (0, 9).

3. For the equation $x\; =\; 4y$:

   We can find solutions by choosing values for $y$ and calculating the corresponding $x$ values.

   Solution 1: Let $y\; =\; 1$. Then, $x\; =\; 4(1)\; \backslash implies\; x\; =\; 4$. So, (4, 1) is a solution.

   Solution 2: Let $y\; =\; 2$. Then, $x\; =\; 4(2)\; \backslash implies\; x\; =\; 8$. So, (8, 2) is a solution.

   Solution 3: Let $y\; =\; 0$. Then, $x\; =\; 4(0)\; \backslash implies\; x\; =\; 0$. So, (0, 0) is a solution.

   Solution 4: Let $y\; =\; -1$. Then, $x\; =\; 4(-1)\; \backslash implies\; x\; =\; -4$. So, (-4, -1) is a solution.

   Therefore, four solutions are (4, 1), (8, 2), (0, 0), and (-4, -1).