GSEB Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.3

Question 1. Draw the graph of each of the following linear equations in two variaNes.
(i) x + y = 4
(ii) x - y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Answer:
(i) For the equation \( x + y = 4 \), we can rewrite it as \( y = 4 - x \). Let's prepare a table of values for x and y to plot the graph:

Now, we plot the points \( (0, 4), (1, 3), (2, 2) \) and \( (3, 1) \) on graph paper and connect them using a ruler to form a straight line. This line represents the graph of the equation \( x + y = 4 \).

In simple words: First, find some points that fit the equation. Then, mark these points on the graph and connect them with a straight line. This line visually shows all the possible answers for the equation.

Exam Tip: To draw a graph accurately, calculate at least three points for the linear equation. Two points are enough to define a line, but a third point helps to check for any errors in calculation.

(ii) For the equation \( x - y = 2 \), we can rearrange it as \( x = y + 2 \) or \( y = x - 2 \). Here is a table of points that satisfy this equation:

Next, we plot the points \( (2, 0), (3, 1), (4, 2) \) and \( (1, -1) \) on the graph paper. We then connect these points with a ruler to obtain the straight line representing the graph of \( x - y = 2 \).

In simple words: Arrange the equation to make 'y' the subject. Then, find several pairs of 'x' and 'y' values that fit. Mark these points on graph paper and connect them with a straight line.

Exam Tip: When drawing graphs, always label your axes (X, X', Y, Y') and the origin (O). Also, clearly mark the scale on both axes to ensure readability and accuracy.

(iii) For the equation \( y = 3x \), we can easily find corresponding y values for various x values. Here's a table showing some points for this equation:

Now, plot the points \( (-1, -3), (0, 0), (1, 3) \) and \( (2, 6) \) on graph paper and connect them using a ruler. This straight line displays the graph for the equation \( y = 3x \).

In simple words: Pick some x-values, calculate the y-values using \( y = 3x \), and then draw a line through those points on a graph.

Exam Tip: For equations like \( y = kx \), the line always passes through the origin \( (0,0) \). This can be a useful check when plotting such graphs.

(iv) For the equation \( 3 = 2x + y \), we can rearrange it to make y the subject: \( y = 3 - 2x \). Let's generate a table of values:

Finally, plot the points \( (0, 3), (1, 1), (2, -1) \) and \( (-1, 5) \) on graph paper and join them with a ruler to create the line that represents the graph of \( 3 = 2x + y \).

In simple words: Rewrite the equation to isolate 'y'. Find a few points that fit, mark them on a graph, and draw a straight line through them.

Exam Tip: Always double-check your calculations for at least two points to ensure they accurately represent the given linear equation. This helps prevent plotting errors.

Question 2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer: Two possible equations for lines that pass through the point \( (2, 14) \) are:
1. \( x + y = 16 \) (Since \( 2 + 14 = 16 \))
2. \( 2x + y = 18 \) (Since \( 2(2) + 14 = 4 + 14 = 18 \))
There are an infinite number of lines that can pass through a single given point. This is because a point represents a specific location, and an endless amount of lines can be drawn to cross through that exact spot from different angles and directions.
In simple words: You can find many equations for lines going through \( (2, 14) \). There are actually countless lines that can all pass through one single point, like spokes on a wheel.

Exam Tip: Remember that an infinite number of lines can pass through a single point. To find equations for such lines, substitute the given coordinates into the general linear equation \( Ax + By = C \) and choose arbitrary values for A, B, and C that make the equation true.

Question 3. If the point (3, 4) lies on the graph of the equation \( 3y = ax + 7 \), find the value of a.
Answer: If the point \( (3, 4) \) is on the graph of the equation \( 3y = ax + 7 \), it means that when \( x = 3 \) and \( y = 4 \), the equation will hold true. We can substitute these values into the equation to find 'a':
Given equation: \( 3y = ax + 7 \)
Substitute \( x = 3 \) and \( y = 4 \):
\( 3 \times 4 = a \times 3 + 7 \)
\( 12 = 3a + 7 \)
To solve for 'a', subtract 7 from both sides:
\( 12 - 7 = 3a \)
\( 5 = 3a \)
Divide by 3:
\( a = \frac {5}{3} \)
Therefore, the value of 'a' is \( \frac {5}{3} \).
In simple words: Put the x and y values from the point into the equation. Then, solve the simple math problem to find what 'a' equals.

Exam Tip: When a point 'lies on the graph' of an equation, it means its coordinates satisfy that equation. Always substitute the x-coordinate for x and the y-coordinate for y carefully to solve for any unknown variables.

Question 4. The taxi fare in a city is as follows for the first kilometre, the fare is 8 and for the subsequent distance, it is 5 per km. Taking the distance covered as x km and total fare as y. Write a linear equation for this information and draw its graph.
Answer: Let's define the variables:
Total distance covered = \( x \) km
Total fare = Rs. \( y \)

Now, let's break down the fare components:
Fare for the first kilometre = Rs. 8
Remaining distance (subsequent distance) = \( (x - 1) \) km
Fare for the subsequent distance = Rs. \( 5(x - 1) \)

According to the information given in the problem, the total fare \( y \) is the sum of the fare for the first kilometre and the fare for the subsequent distance:
\( y = 8 + 5(x - 1) \)
\( y = 8 + 5x - 5 \)

Combining the constant terms, we get the linear equation:
\( \implies y = 5x + 3 \)

Now, let's prepare a table of values for plotting the graph of \( y = 5x + 3 \):

We now plot the points \( (0, 3), (-1, -2), (-2, -7) \) and \( (1, 8) \) on the graph paper and join them with a ruler. This line will show the graph of the equation \( y = 5x + 3 \).

In simple words: First, create an equation that shows how the total taxi fare is calculated based on the distance. Then, find some points from that equation, mark them on a graph, and draw a line.

Exam Tip: When formulating linear equations from word problems, clearly define your variables. Carefully identify the fixed costs (like the first km fare) and variable costs (like the fare per subsequent km) to construct the equation correctly.

Question 5. From the choice given below, choose the equation whose graphs are given in figure 1 and fig. 2 for fig. 1 for fig. 2
(i) y = x       (i) y = x+2
(ii) x + y = 0    (ii) y = x - 2
(iii) y = 2x     (iii) y = -x + 2
(iv) 2 + 3y = 7   (iv) x + 2y = 6
Answer:
Let's analyze each figure and match it with the correct equation based on the points provided in the graph:

For Figure 1:
The graph passes through the points \( (-1, 1), (0, 0), \) and \( (1, -1) \). Let's test these points with the given equations for Figure 1 (left column):
(i) \( y = x \): For \( (-1, 1) \), \( 1 \neq -1 \). So, this is incorrect.
(ii) \( x + y = 0 \):
   - For \( (-1, 1) \): \( -1 + 1 = 0 \). This is correct.
   - For \( (0, 0) \): \( 0 + 0 = 0 \). This is correct.
   - For \( (1, -1) \): \( 1 + (-1) = 0 \). This is correct.
(iii) \( y = 2x \): For \( (-1, 1) \), \( 1 \neq 2(-1) \). So, this is incorrect.
(iv) \( 2 + 3y = 7 \): For \( (0, 0) \), \( 2 + 3(0) = 2 \neq 7 \). So, this is incorrect.

Therefore, the correct equation for Figure 1 is \( \textbf{(ii) x + y = 0} \).

For Figure 2:
The graph passes through the points \( (-1, 3), (0, 2), \) and \( (2, 0) \). Let's test these points with the given equations for Figure 2 (right column):
(i) \( y = x+2 \): For \( (2, 0) \), \( 0 \neq 2+2 \). So, this is incorrect.
(ii) \( y = x - 2 \): For \( (-1, 3) \), \( 3 \neq -1-2 \). So, this is incorrect.
(iii) \( y = -x + 2 \):
   - For \( (-1, 3) \): \( 3 = -(-1) + 2 = 1 + 2 = 3 \). This is correct.
   - For \( (0, 2) \): \( 2 = -(0) + 2 = 2 \). This is correct.
   - For \( (2, 0) \): \( 0 = -(2) + 2 = 0 \). This is correct.
(iv) \( x + 2y = 6 \): For \( (-1, 3) \), \( -1 + 2(3) = -1 + 6 = 5 \neq 6 \). So, this is incorrect.

Therefore, the correct equation for Figure 2 is \( \textbf{(iii) y = -x + 2} \).
In simple words: Look at the points shown on each graph. Then, try putting these x and y values into each equation until you find the equation that works for all the points on that graph.

Exam Tip: To verify if a point lies on a graph, substitute its x and y coordinates into the equation. If both sides of the equation are equal, the point is on the graph. Otherwise, it is not.

Question 6. If the work done by a body on application of a constant force is directly proportional to Travelled by the body, express this in the form of an equation in two variables anu araw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is (i) 2 units (ii) 0 unit
Answer: Let's define our variables:
Distance travelled by the body = \( x \) units
Work done by the constant force = \( y \) units
Given constant force = 5 units

The formula for work done is: Work done = Force \( \times \) Displacement
Substituting our variables and the given force:
\( y = 5 \times x \)
\( y = 5x \)

This is the linear equation in two variables representing the given information. Now, we will prepare a table of values to draw its graph:

We plot the points \( (0, 0), (1, 5) \) and \( (-1, -5) \) on graph paper and connect them using a ruler to form a straight line. This line represents the graph of the equation \( y = 5x \).

Now, let's read the work done from the graph for the given distances:
(i) To find the work done when the distance travelled is 2 units:
   - Locate \( x = 2 \) on the x-axis (point A).
   - Draw a vertical line from \( A(2,0) \) parallel to the y-axis. This line intersects the graph of \( y = 5x \) at point B.
   - From point B, draw a horizontal line parallel to the x-axis. This line intersects the y-axis at point C \( (0, 10) \).
   - Therefore, when the distance travelled is 2 units, the work done is 10 units.
(ii) To find the work done when the distance travelled is 0 units:
   - From the graph, we can clearly see that when \( x = 0 \), the corresponding \( y \) value is \( 0 \).
   - Thus, when the distance travelled is 0 units, the work done is 0 units.
In simple words: Write the equation 'Work = Force x Distance'. Since the force is 5, it becomes 'y = 5x'. Then, draw the line for this equation on a graph. To find work for a certain distance, find that distance on the x-axis, go up to the line, and then across to the y-axis to read the work.

Exam Tip: When reading values from a graph, always start from the given axis (e.g., x-axis for distance), move perpendicularly to the line, and then move perpendicularly to the other axis (e.g., y-axis for work done) to find the corresponding value. Show these steps clearly if required.

Question 7. Yamini and Fatima, two students of class IX of a school, together contributed loo towards the Prime minister's relief fund to help the earthquake victims. Write a linear equation which satisfies this data. (You take their contributions as x and y). Draw the graph of the same.
Answer: Let's define the contributions:
Contribution of Yamini = Rs. \( x \)
Contribution of Fatima = Rs. \( y \)

According to the problem, their combined contribution is Rs. 100. So, the linear equation representing this situation is:
\( x + y = 100 \)
We can rearrange this equation to make \( y \) the subject for easier plotting:
\( y = 100 - x \)

Now, let's create a table of values to plot the graph:

Finally, plot the points \( (0, 100), (25, 75), (50, 50), (75, 25) \) and \( (100, 0) \) on graph paper. Connect these points with a ruler to get the straight line that is the graph of \( x + y = 100 \).

In simple words: Add their contributions, \( x \) and \( y \), to get the total of 100. Then, draw a graph by finding points that make \( x + y = 100 \) true.

Exam Tip: In word problems involving contributions, make sure to form the linear equation by adding the individual contributions and equating them to the total amount. Always ensure the graph accurately represents positive contributions only, as negative contributions don't make sense in this context.

Question 8. In countries like USA and Canada, the temperature is measured in Fahrenheit. Whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius.
\( F = \frac {9}{5} C + 32 \)
Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F. What is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer:
(i) To draw the graph of the equation \( F = \frac {9}{5} C + 32 \), we need to find some points \( (C, F) \) that satisfy it. We will use Celsius for the x-axis and Fahrenheit for the y-axis. Here's a table of values:

Now, we plot these points (0, 32), (5, 41), (10, 50), and (15, 59) on the graph paper and join them with a ruler to create the line that represents the equation \( F = \frac {9}{5} C + 32 \).

(ii) If the temperature is 30°C, we need to find the equivalent temperature in Fahrenheit.
Given: \( C = 30^\circ C \)
Using the conversion formula: \( F = \frac {9}{5} C + 32 \)
Substitute \( C = 30 \):
\( F = \frac {9}{5} \times 30 + 32 \)
\( F = 9 \times 6 + 32 \)
\( F = 54 + 32 \)
\( \implies F = 86^\circ F \)
Thus, the required temperature in Fahrenheit is \( 86^\circ F \).

(iii) If the temperature is 95°F, we need to find the equivalent temperature in Celsius.
Given: \( F = 95^\circ F \)
Using the conversion formula: \( F = \frac {9}{5} C + 32 \)
Substitute \( F = 95 \):
\( 95 = \frac {9}{5} C + 32 \)
Subtract 32 from both sides:
\( 95 - 32 = \frac {9}{5} C \)
\( 63 = \frac {9}{5} C \)
Multiply by \( \frac {5}{9} \):
\( C = 63 \times \frac {5}{9} \)
\( C = 7 \times 5 \)
\( \implies C = 35^\circ C \)
Thus, the required temperature in Celsius is \( 35^\circ C \).

(iv) If the temperature is 0°C, we find the temperature in Fahrenheit. And if the temperature is 0°F, we find the temperature in Celsius.
Case 1: When \( C = 0^\circ C \)
Using the conversion formula: \( F = \frac {9}{5} C + 32 \)
Substitute \( C = 0 \):
\( F = \frac {9}{5} \times 0 + 32 \)
\( \implies F = 32^\circ F \)
Thus, the temperature is \( 32^\circ F \) when it is \( 0^\circ C \).

Case 2: When \( F = 0^\circ F \)
Using the conversion formula: \( F = \frac {9}{5} C + 32 \)
Substitute \( F = 0 \):
\( 0 = \frac {9}{5} C + 32 \)
Subtract 32 from both sides:
\( -32 = \frac {9}{5} C \)
Multiply by \( \frac {5}{9} \):
\( C = -32 \times \frac {5}{9} \)
\( \implies C = \frac {-160}{9} ^\circ C \)
The required temperature is \( \frac {-160}{9} ^\circ C \).

(v) To find if there's a temperature that is numerically the same in both Fahrenheit and Celsius, let's assume this temperature is \( x \). So, \( F = x \) and \( C = x \).
Using the conversion formula: \( F = \frac {9}{5} C + 32 \)
Substitute \( F = x \) and \( C = x \):
\( x = \frac {9}{5} x + 32 \)
Subtract \( \frac {9}{5} x \) from both sides:
\( x - \frac {9}{5} x = 32 \)
Find a common denominator to subtract the terms with \( x \):
\( \frac {5x - 9x}{5} = 32 \)
\( \frac {-4x}{5} = 32 \)
Multiply both sides by 5:
\( -4x = 32 \times 5 \)
\( -4x = 160 \)
Divide by -4:
\( x = \frac {160}{-4} \)
\( \implies x = -40 \)
Therefore, yes, there is a temperature that is numerically the same in both Fahrenheit and Celsius, which is \( -40^\circ C \) or \( -40^\circ F \).
In simple words: To change Celsius to Fahrenheit, multiply Celsius by \( \frac{9}{5} \) and add 32. To change Fahrenheit to Celsius, subtract 32 from Fahrenheit, then multiply by \( \frac{5}{9} \). We found a special temperature, -40 degrees, where both scales show the same number.

Exam Tip: Remember the Fahrenheit to Celsius conversion formula \( F = \frac{9}{5} C + 32 \). For questions involving conversions or finding common points, careful substitution and algebraic manipulation are key to accurate solutions. Pay close attention to positive and negative signs during calculations.