Rbse Solutions for Class 9 Maths Chapter 4 Exercise 4.1 | Linear Equations in Two Variables
Get step-by-step NCERT Solutions for Class 9 Maths Chapter 4, Exercise 4.1. Learn to represent word problems as linear equations and express equations in the standard form $ax+by+c=0$, identifying coefficients $a, b, \text{ and } c$.}$
1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Let the cost of a notebook be $\mathbf{x}$ and the cost of a pen be $\mathbf{y}$.
The statement says:
$$(\text{Cost of notebook}) = 2 \times (\text{Cost of pen})$$
Substituting the variables:
$$x = 2y$$
To write it in the standard linear equation form ($ax + by + c = 0$):
$$\mathbf{x – 2y + 0 = 0}$$
2. Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of $a, b$ and $c$ in each case:
The standard form is $\mathbf{ax + by + c = 0}$.
(i) $2x + 3y = 9.3\overline{5}$
Move the constant term to the Left Hand Side (LHS):
$$2x + 3y – 9.3\overline{5} = 0$$
$$\mathbf{a = 2, \quad b = 3, \quad c = -9.3\overline{5}}$$
(ii) $x – \frac{y}{5} – 10 = 0$
This is already in the standard form. We can write $x$ as $1x$ and $-\frac{y}{5}$ as $-\frac{1}{5}y$:
$$1x + \left(-\frac{1}{5}\right)y + (-10) = 0$$
$$\mathbf{a = 1, \quad b = -\frac{1}{5}, \quad c = -10}$$
(iii) $-2x + 3y = 6$
Move the constant term to the LHS:
$$-2x + 3y – 6 = 0$$
$$\mathbf{a = -2, \quad b = 3, \quad c = -6}$$
(iv) $x = 3y$
Move the $y$ term to the LHS and write $0$ for the constant:
$$x – 3y + 0 = 0$$
$$\mathbf{a = 1, \quad b = -3, \quad c = 0}$$
(v) $2x = -5y$
Move the $y$ term to the LHS and write $0$ for the constant:
$$2x + 5y + 0 = 0$$
$$\mathbf{a = 2, \quad b = 5, \quad c = 0}$$
(vi) $3x + 2 = 0$
This equation is missing the $y$ term, so its coefficient is $0$:
$$3x + 0y + 2 = 0$$
$$\mathbf{a = 3, \quad b = 0, \quad c = 2}$$
(vii) $y – 2 = 0$
This equation is missing the $x$ term, so its coefficient is $0$:
$$0x + 1y – 2 = 0$$
$$\mathbf{a = 0, \quad b = 1, \quad c = -2}$$
(viii) $5 = 2x$
Rearrange the terms to get $x$ and constants on the LHS (or move 5 to RHS, resulting in $0 = 2x – 5$):
$$2x + 0y – 5 = 0$$
$$\mathbf{a = 2, \quad b = 0, \quad c = -5}$$