CBSE Class 9 Mathematics: Linear Equations in Two Variables NCERT Solutions

Solution:

Let the cost of a notebook be represented by the variable $x$. Let the cost of a pen be represented by the variable $y$. According to the problem statement, the cost of the notebook is twice the cost of the pen. This can be written as: $$x\; =\; 2y$$ To express this as a linear equation in the standard form $ax\; +\; by\; +\; c\; =\; 0$, we can rearrange the terms: $$x\; -\; 2y\; =\; 0$$ Here, $a\; =\; 1$, $b\; =\; -2$, and $c\; =\; 0$. Thus, the linear equation representing the statement is $x\; -\; 2y\; =\; 0$.

Solution:

We need to rewrite each equation in the standard form $ax\; +\; by\; +\; c\; =\; 0$ and identify the coefficients $a$, $b$, and $c$.

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

   To get the form $ax\; +\; by\; +\; c\; =\; 0$, 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: $-2x\; +\; 3y\; =\; 6$

   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: $x\; =\; 3y$

   Rearrange the terms to get the standard form:

   $$x\; -\; 3y\; =\; 0$$ To explicitly include the $c$ term, we can write it as: $$x\; -\; 3y\; +\; 0\; =\; 0$$ Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get: $a\; =\; 1$, $b\; =\; -3$, $c\; =\; 0$

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

   Rearrange the terms to get the standard form:

   $$2x\; +\; 5y\; =\; 0$$ Including the $c$ term: $$2x\; +\; 5y\; +\; 0\; =\; 0$$ Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get: $a\; =\; 2$, $b\; =\; 5$, $c\; =\; 0$

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

   This equation only has an $x$ term and a constant. To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we 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: $y\; -\; 2\; =\; 0$

   This equation only has a $y$ term and a constant. To write it in the form $ax\; +\; by\; +\; c\; =\; 0$, we 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: $5\; =\; 2x$

   Rearrange the terms to get the standard form:

   $$5\; -\; 2x\; =\; 0$$ Rewrite it with $x$ term first and then $y$ term (with coefficient 0): $$-2x\; +\; 0\; \backslash cdot\; y\; +\; 5\; =\; 0$$ Comparing this with $ax\; +\; by\; +\; c\; =\; 0$, we get: $a\; =\; -2$, $b\; =\; 0$, $c\; =\; 5$

Solution:

The correct option is (iii) infinitely many solutions.

Reasoning: 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, a linear equation in two variables has infinitely many solutions. We can find these solutions by choosing any real value for $x$ and then calculating the corresponding value of $y$, or vice versa.

For example:

• If $x\; =\; 0$, then $y\; =\; 3(0)\; +\; 5\; =\; 5$. So, $(0,\; 5)$ is a solution.
• If $x\; =\; 1$, then $y\; =\; 3(1)\; +\; 5\; =\; 8$. So, $(1,\; 8)$ is a solution.
• If $x\; =\; -1$, then $y\; =\; 3(-1)\; +\; 5\; =\; -3\; +\; 5\; =\; 2$. So, $(-1,\; 2)$ is a solution.

Since we can choose infinitely many values for $x$, we can find infinitely many corresponding values for $y$.

Solution:

To find solutions for a linear equation in two variables, we can assign any real value to one variable and then solve for the other variable.

1. Equation: $2x\; +\; y\; =\; 7$

   We can rewrite this as $y\; =\; 7\; -\; 2x$.

   • Let $x\; =\; 0$. Then $y\; =\; 7\; -\; 2(0)\; =\; 7$. Solution: $(0,\; 7)$.
   • Let $x\; =\; 1$. Then $y\; =\; 7\; -\; 2(1)\; =\; 7\; -\; 2\; =\; 5$. Solution: $(1,\; 5)$.
   • Let $x\; =\; 2$. Then $y\; =\; 7\; -\; 2(2)\; =\; 7\; -\; 4\; =\; 3$. Solution: $(2,\; 3)$.
   • Let $x\; =\; 3$. Then $y\; =\; 7\; -\; 2(3)\; =\; 7\; -\; 6\; =\; 1$. Solution: $(3,\; 1)$.
   Thus, four solutions are $(0,\; 7)$, $(1,\; 5)$, $(2,\; 3)$, and $(3,\; 1)$.

2. Equation: $\backslash pi\; x\; +\; y\; =\; 9$

   We can rewrite this as $y\; =\; 9\; -\; \backslash pi\; x$.

   • Let $x\; =\; 0$. Then $y\; =\; 9\; -\; \backslash pi(0)\; =\; 9$. Solution: $(0,\; 9)$.
   • Let $x\; =\; \backslash frac\{1\}\{\backslash pi\}$. Then $y\; =\; 9\; -\; \backslash pi\backslash left(\backslash frac\{1\}\{\backslash pi\}\backslash right)\; =\; 9\; -\; 1\; =\; 8$. Solution: $\backslash left(\backslash frac\{1\}\{\backslash pi\},\; 8\backslash right)$.
   • Let $x\; =\; \backslash frac\{2\}\{\backslash pi\}$. Then $y\; =\; 9\; -\; \backslash pi\backslash left(\backslash frac\{2\}\{\backslash pi\}\backslash right)\; =\; 9\; -\; 2\; =\; 7$. Solution: $\backslash left(\backslash frac\{2\}\{\backslash pi\},\; 7\backslash right)$.
   • Let $x\; =\; \backslash frac\{9\}\{\backslash pi\}$. Then $y\; =\; 9\; -\; \backslash pi\backslash left(\backslash frac\{9\}\{\backslash pi\}\backslash right)\; =\; 9\; -\; 9\; =\; 0$. Solution: $\backslash left(\backslash frac\{9\}\{\backslash pi\},\; 0\backslash right)$.
   Thus, four solutions are $(0,\; 9)$, $\backslash left(\backslash frac\{1\}\{\backslash pi\},\; 8\backslash right)$, $\backslash left(\backslash frac\{2\}\{\backslash pi\},\; 7\backslash right)$, and $\backslash left(\backslash frac\{9\}\{\backslash pi\},\; 0\backslash right)$.

3. Equation: $x\; =\; 4y$

   We can rewrite this as $y\; =\; \backslash frac\{x\}\{4\}$.

   • Let $y\; =\; 0$. Then $x\; =\; 4(0)\; =\; 0$. Solution: $(0,\; 0)$.
   • Let $y\; =\; 1$. Then $x\; =\; 4(1)\; =\; 4$. Solution: $(4,\; 1)$.
   • Let $y\; =\; 2$. Then $x\; =\; 4(2)\; =\; 8$. Solution: $(8,\; 2)$.
   • Let $y\; =\; -1$. Then $x\; =\; 4(-1)\; =\; -4$. Solution: $(-4,\; -1)$.
   Thus, four solutions are $(0,\; 0)$, $(4,\; 1)$, $(8,\; 2)$, and $(-4,\; -1)$.