NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables — Complete Guide

A solution of a linear equation $ax + by + c = 0$ is an ordered pair $(x_0, y_0)$ such that $ax_0 + by_0 + c = 0$.

Key Fact: A linear equation in two variables has infinitely many solutions. These solutions, when plotted on the Cartesian plane, form a straight line.

For example, consider $x + y = 5$. Some solutions are:
- $(0, 5)$: $0 + 5 = 5$ ✓
- $(1, 4)$: $1 + 4 = 5$ ✓
- $(2, 3)$: $2 + 3 = 5$ ✓
- $(5, 0)$: $5 + 0 = 5$ ✓
- $(-1, 6)$: $-1 + 6 = 5$ ✓
- $(2.5, 2.5)$: $2.5 + 2.5 = 5$ ✓

There are infinitely many such pairs, and they all lie on the line $x + y = 5$.

Method: To find solutions of $ax + by + c = 0$:
1. Choose a convenient value for $x$ (usually $0, 1, 2$, etc.).
2. Substitute into the equation and solve for $y$.
3. Each $(x, y)$ pair is one solution.

Example: Find three solutions of $2x + 3y = 12$.

**When $x = 0$:** $3y = 12 \implies y = 4$. Solution: $(0, 4)$.
**When $x = 3$:** $6 + 3y = 12 \implies y = 2$. Solution: $(3, 2)$.
**When $x = 6$:** $12 + 3y = 12 \implies y = 0$. Solution: $(6, 0)$.

Tip: Choosing $x = 0$ gives the $y$-intercept, and choosing $y = 0$ gives the $x$-intercept. These two points are usually the easiest to plot.

Question: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Solution:
Let the cost of a notebook be $x$ rupees and the cost of a pen be $y$ rupees.

"Cost of notebook is twice the cost of a pen" means:

This is in the form $ax + by + c = 0$ with $a = 1$, $b = -2$, $c = 0$.

Key Insight: Translating word problems into equations is a critical skill. Identify the unknowns, assign variables, and write the relationship.

Question: Express the following linear equations in the form $ax + by + c = 0$ and indicate the values of $a$, $b$, and $c$:
(i) $2x + 3y = 9.\overline{35}$
(ii) $x - \frac{y}{5} - 10 = 0$
(iii) $-2x + 3y = 6$
(iv) $x = 3y$
(v) $2x = -5y$
(vi) $3x + 2 = 0$
(vii) $y - 2 = 0$
(viii) $5 = 2x$

Solution:

Key Observation: When $b = 0$, the equation involves only $x$ — its graph is a vertical line. When $a = 0$, the equation involves only $y$ — its graph is a horizontal line.

Question: Which one of the following options is true, and why?
$y = 3x + 5$ has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions

Solution:
(iii) Infinitely many solutions is correct.

For every real number $x$, we get a corresponding $y = 3x + 5$. Since there are infinitely many real numbers, there are infinitely many solutions.

Examples: $(0, 5)$, $(1, 8)$, $(-1, 2)$, $(\frac{1}{3}, 6)$, $(-\frac{5}{3}, 0)$, ...

Question: Write four solutions for the equation $2x + y = 7$.

Solution:
Rewrite as $y = 7 - 2x$, then substitute:

Four solutions: $(0, 7)$, $(1, 5)$, $(2, 3)$, $(3, 1)$.

Bonus: The $x$-intercept is at $y = 0$: $2x = 7 \implies x = 3.5$, giving $(3.5, 0)$.

Question: Check which of the following are solutions of the equation $x - 2y = 4$:
(i) $(0, 2)$ (ii) $(2, 0)$ (iii) $(4, 0)$ (iv) $(\sqrt{2}, 4\sqrt{2})$ (v) $(1, 1)$

Solution:
Substitute each pair into $x - 2y$. It should equal 4.

(i) $0 - 2(2) = -4 \neq 4$. Not a solution.

(ii) $2 - 2(0) = 2 \neq 4$. Not a solution.

(iii) $4 - 2(0) = 4 = 4$ ✓. Yes, a solution.

(iv) $\sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2} \neq 4$. Not a solution.

(v) $1 - 2(1) = -1 \neq 4$. Not a solution.

Only $(4, 0)$ is a solution.

Question: Find the value of $k$ if $x = 2$, $y = 1$ is a solution of the equation $2x + 3y = k$.

Solution:
Substitute $x = 2$, $y = 1$:

Verification: The equation $2x + 3y = 7$ with $(2, 1)$: $4 + 3 = 7$ ✓.

Step 1: Rewrite the equation in the form $y = mx + c$ (if possible), or find at least two solution pairs by substituting values of $x$.

Step 2: Plot at least two points on the Cartesian plane. (Three points is safer — the third point serves as a check.)

Step 3: Draw a straight line through the plotted points. Extend it in both directions with arrows to indicate that the line continues infinitely.

Step 4: Label the line with its equation.

Finding the two easiest points:
- **$x$-intercept:** Set $y = 0$, solve for $x$. This gives the point where the line crosses the $x$-axis.
- **$y$-intercept:** Set $x = 0$, solve for $y$. This gives the point where the line crosses the $y$-axis.

For example, for $2x + 3y = 6$:
- $x$-intercept: $2x = 6 \implies x = 3$. Point: $(3, 0)$.
- $y$-intercept: $3y = 6 \implies y = 2$. Point: $(0, 2)$.
- Third check point: $x = 1 \implies 2 + 3y = 6 \implies y = \frac{4}{3}$. Point: $(1, \frac{4}{3})$.

Question: Draw the graph of the equation $2x + y = 7$.

Solution:
Rewrite: $y = 7 - 2x$.

Plot these points and draw a straight line through them. The line has a negative slope (going downward from left to right), $y$-intercept at $(0, 7)$, and $x$-intercept at $(3.5, 0)$.

Question: Draw the graph of the equation $x + y = 0$.

Solution:
$y = -x$.

This line passes through the origin and makes a $135°$ angle with the positive $x$-axis (or $45°$ below the horizontal going right). It passes through Quadrants II and IV.

Question: Draw the graph of $y = 3x$.

Solution:

This line passes through the origin with a steep positive slope. It passes through Quadrants I and III.

Key Observation: Any equation of the form $y = mx$ (no constant term) passes through the origin.

Question: The taxi fare in a city is as follows: for the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as $x$ km and total fare as Rs $y$, write a linear equation for this information, and draw its graph.

Solution:
For $x \geq 1$:
- First km costs Rs 8.
- Remaining $(x - 1)$ km costs Rs $5(x - 1)$.

Linear equation: $y = 5x + 3$ (for $x \geq 1$).

Plot these points and draw a line. Since $x$ represents distance, only $x \geq 1$ is meaningful.

From the graph: For a distance of 5 km, the fare = $5(5) + 3 =$ Rs 28.

The equation $y = k$ (where $k$ is a constant) represents a horizontal line parallel to the $x$-axis, passing through all points with $y$-coordinate equal to $k$.

• $y = 3$ is a horizontal line 3 units above the $x$-axis.
  - $y = -2$ is a horizontal line 2 units below the $x$-axis.
  - $y = 0$ is the $x$-axis itself.

In standard form: $0 \cdot x + 1 \cdot y - k = 0$, i.e., $a = 0$.

The equation $x = k$ (where $k$ is a constant) represents a vertical line parallel to the $y$-axis, passing through all points with $x$-coordinate equal to $k$.

• $x = 4$ is a vertical line 4 units to the right of the $y$-axis.
  - $x = -1$ is a vertical line 1 unit to the left of the $y$-axis.
  - $x = 0$ is the $y$-axis itself.

In standard form: $1 \cdot x + 0 \cdot y - k = 0$, i.e., $b = 0$.

Question: Give the geometric representations of $y = 3$ as an equation (i) in one variable and (ii) in two variables.

Solution:

(i) In one variable: $y = 3$ represents a single point on the number line (the point 3 on the $y$-axis).

(ii) In two variables: $y = 3$ (or $0 \cdot x + y - 3 = 0$) represents a straight line in the Cartesian plane. This line is parallel to the $x$-axis, passing through all points of the form $(a, 3)$ where $a$ is any real number.

Some points on this line: $(-2, 3)$, $(0, 3)$, $(1, 3)$, $(5, 3)$.

Question: Give the geometric representations of $2x + 9 = 0$ as an equation in two variables.

Solution:
$2x + 9 = 0 \implies x = -\frac{9}{2} = -4.5$

In two variables, this is $2x + 0 \cdot y + 9 = 0$, representing a vertical line at $x = -4.5$.

This line is parallel to the $y$-axis, passing through $(-4.5, 0)$, $(-4.5, 1)$, $(-4.5, -3)$, etc.

Question: Find the slope and $y$-intercept of $3x + 2y = 8$.

Solution:
Rewrite in slope-intercept form ($y = mx + c$):

Slope $m = -\frac{3}{2}$ (negative, so line goes downward).
$y$-intercept $c = 4$ (the line crosses the $y$-axis at $(0, 4)$).

Question: Find the slope of the line passing through $(1, 3)$ and $(4, 9)$.

Solution:

The line rises 2 units for every 1 unit it moves to the right.

Question: Find the points where the line $3x - 4y = 12$ crosses the $x$-axis and $y$-axis.

Solution:
**$x$-intercept** (set $y = 0$): $3x = 12 \implies x = 4$. Point: $(4, 0)$.
**$y$-intercept** (set $x = 0$): $-4y = 12 \implies y = -3$. Point: $(0, -3)$.

The line crosses the $x$-axis at $(4, 0)$ and the $y$-axis at $(0, -3)$.

Question: Write the equation of a line that passes through the origin and has a slope of 2.

Solution:
A line through the origin has $y$-intercept $= 0$. With slope $m = 2$:

Or in standard form: $2x - y = 0$.

Question: Write the equation of a line passing through $(1, 3)$ and $(3, 7)$.

Solution:
Slope $= \frac{7-3}{3-1} = \frac{4}{2} = 2$.

Using point-slope form with $(1, 3)$:

Standard form: $2x - y + 1 = 0$.

Verification: $(3, 7)$: $2(3) - 7 + 1 = 0$ ✓.

Question: The temperature in Fahrenheit ($F$) and Celsius ($C$) are related by $F = \frac{9}{5}C + 32$. Draw the graph and find:
(i) The temperature when both scales read the same.
(ii) The Fahrenheit equivalent of $30°C$.

Solution:

(i) When $F = C$: $C = \frac{9}{5}C + 32 \implies C - \frac{9}{5}C = 32 \implies -\frac{4}{5}C = 32 \implies C = -40$.
So $F = C = -40°$. Both scales read the same at $-40°$!

(ii) $F = \frac{9}{5}(30) + 32 = 54 + 32 = 86°F$.

Question: The perimeter of a rectangle is 40 cm. Express the length $l$ in terms of the breadth $b$. Draw the graph and find the dimensions when $l = b$.

Solution:

When $l = b$: $b = 20 - b \implies 2b = 20 \implies b = 10$, $l = 10$.

The rectangle is a square with side 10 cm.