Class IX Mathematics worksheet on linear equations in two variables, featuring problems on finding values, writing equations, and graphing.

Problem: Solve the given problems from the image and explain the solutions.

#### Section A (1 Mark)

1. Find \( m \), if the point \( (7, -3) \) lies on the equation \( y = \frac{3}{7}m(x - \frac{2}{7}) \).

- Solution:
The point \( (7, -3) \) satisfies the given equation. Substitute \( x = 7 \) and \( y = -3 \) into the equation:
\[
-3 = \frac{3}{7}m\left(7 - \frac{2}{7}\right)
\]
Simplify the term inside the parentheses:
\[
7 - \frac{2}{7} = \frac{49}{7} - \frac{2}{7} = \frac{47}{7}
\]
So the equation becomes:
\[
-3 = \frac{3}{7}m \cdot \frac{47}{7}
\]
Simplify the right-hand side:
\[
-3 = \frac{3 \cdot 47}{49}m = \frac{141}{49}m
\]
Solve for \( m \):
\[
m = -3 \cdot \frac{49}{141} = -\frac{147}{141} = -\frac{49}{47}
\]
Therefore, the value of \( m \) is:
\[
\boxed{-\frac{49}{47}}
\]

2. Find the value of \( \alpha \) in the equation \( \alpha x + y = 5 \) if \( x = 2 \) and \( y = 3 \).

- Solution:
Substitute \( x = 2 \) and \( y = 3 \) into the equation \( \alpha x + y = 5 \):
\[
\alpha \cdot 2 + 3 = 5
\]
Simplify:
\[
2\alpha + 3 = 5
\]
Subtract 3 from both sides:
\[
2\alpha = 2
\]
Divide by 2:
\[
\alpha = 1
\]
Therefore, the value of \( \alpha \) is:
\[
\boxed{1}
\]

3. If \( x - 4 = \sqrt{3}y \) is written in the standard form \( ax + by + c = 0 \), then find the values of \( a \), \( b \), and \( c \).

- Solution:
The given equation is \( x - 4 = \sqrt{3}y \). Rearrange it to the standard form \( ax + by + c = 0 \):
\[
x - \sqrt{3}y - 4 = 0
\]
Comparing this with \( ax + by + c = 0 \), we get:
\[
a = 1, \quad b = -\sqrt{3}, \quad c = -4
\]
Therefore, the values are:
\[
\boxed{a = 1, b = -\sqrt{3}, c = -4}
\]

#### Section B (2 Marks)

4. Represent an equation of a straight line which is parallel to the x-axis and at a distance of 2.5 units below it.

- Solution:
A line parallel to the x-axis has a constant y-coordinate. If the line is 2.5 units below the x-axis, its y-coordinate is \(-2.5\). Therefore, the equation of the line is:
\[
y = -2.5
\]
or equivalently:
\[
y + 2.5 = 0
\]
Therefore, the equation is:
\[
\boxed{y + 2.5 = 0}
\]

5. For the first km, the fare is Rs 15 and for the successive distance it is Rs 8 per km. Taking distance covered as \( x \) (km) and the total fare as \( y \) (Rs), represent a linear equation in two variables.

- Solution:
The fare for the first kilometer is Rs 15. For the remaining distance \( (x - 1) \) km, the fare is Rs 8 per km. Therefore, the total fare \( y \) can be expressed as:
\[
y = 15 + 8(x - 1)
\]
Simplify the equation:
\[
y = 15 + 8x - 8 = 8x + 7
\]
Therefore, the linear equation is:
\[
\boxed{y = 8x + 7}
\]

6. If \( (2, 3) \) and \( (4, 0) \) lie on the graph of the equation \( ax + by = 1 \), then find \( a \) and \( b \).

- Solution:
Substitute the points \( (2, 3) \) and \( (4, 0) \) into the equation \( ax + by = 1 \).

For \( (2, 3) \):
\[
a \cdot 2 + b \cdot 3 = 1 \implies 2a + 3b = 1 \quad \text{(1)}
\]

For \( (4, 0) \):
\[
a \cdot 4 + b \cdot 0 = 1 \implies 4a = 1 \implies a = \frac{1}{4} \quad \text{(2)}
\]

Substitute \( a = \frac{1}{4} \) into equation (1):
\[
2 \left(\frac{1}{4}\right) + 3b = 1 \implies \frac{1}{2} + 3b = 1 \implies 3b = 1 - \frac{1}{2} \implies 3b = \frac{1}{2} \implies b = \frac{1}{6}
\]
Therefore, the values are:
\[
\boxed{a = \frac{1}{4}, b = \frac{1}{6}}
\]

7. Find the coordinates of the points where the graph of the equation \( 7x - 3y = 4 \) cuts the x-axis and y-axis.

- Solution:
To find the x-intercept, set \( y = 0 \) in the equation \( 7x - 3y = 4 \):
\[
7x - 3 \cdot 0 = 4 \implies 7x = 4 \implies x = \frac{4}{7}
\]
So, the x-intercept is \( \left( \frac{4}{7}, 0 \right) \).

To find the y-intercept, set \( x = 0 \) in the equation \( 7x - 3y = 4 \):
\[
7 \cdot 0 - 3y = 4 \implies -3y = 4 \implies y = -\frac{4}{3}
\]
So, the y-intercept is \( \left( 0, -\frac{4}{3} \right) \).

Therefore, the coordinates are:
\[
\boxed{\left( \frac{4}{7}, 0 \right), \left( 0, -\frac{4}{3} \right)}
\]

#### Section C (3 Marks)

8. Solve \( \frac{3x + 2}{7} + \frac{4(x + 1)}{5} = \frac{2(2x + 1)}{3} \).

- Solution:
First, find a common denominator for the fractions. The least common multiple of 7, 5, and 3 is 105. Rewrite each fraction with the common denominator:
\[
\frac{3x + 2}{7} = \frac{15(3x + 2)}{105} = \frac{45x + 30}{105}
\]
\[
\frac{4(x + 1)}{5} = \frac{21 \cdot 4(x + 1)}{105} = \frac{84x + 84}{105}
\]
\[
\frac{2(2x + 1)}{3} = \frac{35 \cdot 2(2x + 1)}{105} = \frac{140x + 70}{105}
\]
Substitute these into the equation:
\[
\frac{45x + 30}{105} + \frac{84x + 84}{105} = \frac{140x + 70}{105}
\]
Combine the fractions on the left-hand side:
\[
\frac{45x + 30 + 84x + 84}{105} = \frac{140x + 70}{105}
\]
Simplify the numerator on the left-hand side:
\[
\frac{129x + 114}{105} = \frac{140x + 70}{105}
\]
Since the denominators are the same, equate the numerators:
\[
129x + 114 = 140x + 70
\]
Rearrange to solve for \( x \):
\[
114 - 70 = 140x - 129x \implies 44 = 11x \implies x = 4
\]
Therefore, the solution is:
\[
\boxed{4}
\]

9. Draw the graph of the linear equations \( y = x \) and \( y = -x \) on the same Cartesian plane. What do you observe?

- Solution:
The equation \( y = x \) represents a straight line passing through the origin with a slope of 1. The equation \( y = -x \) represents a straight line passing through the origin with a slope of \(-1\). When these two lines are drawn on the same Cartesian plane, they intersect at the origin \((0, 0)\) and are perpendicular to each other.

Observation: The two lines intersect at the origin and are perpendicular to each other.

Therefore, the observation is:
\[
\boxed{\text{Point of intersection is the origin}}
\]

10. Draw the graph of the equation \( 2x + 3y = 12 \) and check whether the points \( (4.5, 1) \) and \( (1.5, 3) \) lie on the graph.

- Solution:
To draw the graph, find the x-intercept and y-intercept of the equation \( 2x + 3y = 12 \).

For the x-intercept, set \( y = 0 \):
\[
2x + 3 \cdot 0 = 12 \implies 2x = 12 \implies x = 6
\]
So, the x-intercept is \( (6, 0) \).

For the y-intercept, set \( x = 0 \):
\[
2 \cdot 0 + 3y = 12 \implies 3y = 12 \implies y = 4
\]
So, the y-intercept is \( (0, 4) \).

Plot the points \( (6, 0) \) and \( (0, 4) \) and draw a straight line through them.

Now, check if the points \( (4.5, 1) \) and \( (1.5, 3) \) lie on the graph by substituting them into the equation \( 2x + 3y = 12 \).

For \( (4.5, 1) \):
\[
2(4.5) + 3(1) = 9 + 3 = 12
\]
Since the equation is satisfied, the point \( (4.5, 1) \) lies on the graph.

For \( (1.5, 3) \):
\[
2(1.5) + 3(3) = 3 + 9 = 12
\]
Since the equation is satisfied, the point \( (1.5, 3) \) lies on the graph.

Therefore, the points are:
\[
\boxed{(4.5, 1) \text{ and } (1.5, 3)}
\]

11. Give the geometrical interpretation of \( 5x + 3 = 3x - 7 \) as an equation:
- i) In one variable
- ii) In two variables

- Solution:
i) In one variable:
The equation \( 5x + 3 = 3x - 7 \) is a linear equation in one variable. Solving for \( x \):
\[
5x - 3x = -7 - 3 \implies 2x = -10 \implies x = -5
\]
Geometrically, this represents a single point on the number line at \( x = -5 \).

ii) In two variables:
Rewrite the equation \( 5x + 3 = 3x - 7 \) in the form \( ax + by + c = 0 \):
\[
5x + 3 - 3x + 7 = 0 \implies 2x + 10 = 0 \implies 2x + 0y + 10 = 0
\]
This is a linear equation in two variables. Geometrically, it represents a vertical line in the Cartesian plane at \( x = -5 \).

Therefore, the interpretations are:
\[
\boxed{\text{i) A single point on the number line at } x = -5, \text{ ii) A vertical line in the Cartesian plane at } x = -5}
\]

Final Answer:

\[
\boxed{-\frac{49}{47}, 1, a = 1, b = -\sqrt{3}, c = -4, y + 2.5 = 0, y = 8x + 7, a = \frac{1}{4}, b = \frac{1}{6}, \left( \frac{4}{7}, 0 \right), \left( 0, -\frac{4}{3} \right), 4, \text{Point of intersection is the origin}, (4.5, 1) \text{ and } (1.5, 3), \text{i) A single point on the number line at } x = -5, \text{ ii) A vertical line in the Cartesian plane at } x = -5}
\]