CBSE Class 9 Mathematics Linear Equations in two variables Assignment Set 05

Key Concepts

• An equation written in the form \( ax + by + c = 0 \), where \( a, b, \text{ and } c \) are real numbers and both \( a \) and \( b \) are not zero, is known as a linear equation in two variables.
• A set of values for \( x \) and \( y \) that makes the equation true when substituted is called a solution.
• A linear equation with two variables possesses an infinite number of solutions.
• The graphical representation of any linear equation in two variables forms a straight line.
• The equation representing the x-axis is \( y = 0 \), while the equation for the y-axis is \( x = 0 \).
• The graph of \( x = a \) is a straight line that is parallel to the y-axis.
• The graph of \( y = a \) is a straight line that is parallel to the x-axis.
• An equation of the form \( y = mx \) represents a straight line that goes directly through the origin.

Question. The point (a, a) always lies on the line
(a) \( y = x \)
(b) y - axis
(c) x - axis
(d) \( x + y = 0 \)
Answer: (a) \( y = x \)
In simple words: Since both the x-coordinate and the y-coordinate are equal to \( a \), the point always satisfies the relationship where \( y \) is equal to \( x \).

Exam Tip: If both coordinates of a point are identical, the point must lie on the line \( y = x \). Substituting the coordinates into the equation is a quick way to verify.

Question. The point (m, -m) always lies on the line.
(a) \( x = -y \)
(b) \( y = -x \)
(c) \( x + y = 0 \)
(d) \( x = y \)
Answer: (c) \( x + y = 0 \)
In simple words: If you add the coordinates \( m \) and \( -m \) together, they cancel each other out to make 0, so the point is on the line \( x + y = 0 \).

Exam Tip: When the coordinates are exact opposites of each other, their sum is zero. Look for the option \( x + y = 0 \) in such cases.

Question. If \( x = -2 \) and \( y = 3 \) is a solution of the equation \( 3x - 5y = a \), then value of a is
(a) 19
(b) -21
(c) -9
(d) -18
Answer: (b) -21
In simple words: By putting the given numbers into the equation, we find that \( a \) must be equal to -21.

Exam Tip: Make sure you pay close attention to positive and negative signs when substituting values. A simple sign mistake can lead to incorrect options.

Question. \( x = 3, y = -2 \) is a solution of the equation.
(a) \( x + y = 5 \)
(b) \( 3x - 2y = 11 \)
(c) \( 4x - 3y = 18 \)
(d) \( 3x + y = 5 \)
Answer: (c) \( 4x - 3y = 18 \)
In simple words: When we insert these values into the third equation, the left side simplifies to 18, which is equal to the right side.

Exam Tip: Instead of solving, systematically check the options by substituting the coordinates. This approach is highly efficient for multiple-choice questions.

Question. \( x = -5 \) can be written in the form of equation in two variable as
(a) \( x + 0\cdot y + 5 = 0 \)
(b) \( 0\cdot x + y = -5 \)
(c) \( 0\cdot x + 0\cdot y = -5 \)
(d) \( 0\cdot x + 0\cdot y = +5 \)
Answer: (a) \( x + 0\cdot y + 5 = 0 \)
In simple words: We can include the letter \( y \) by multiplying it by 0, which does not change the equation, and move -5 to the other side to get \( x + 0\cdot y + 5 = 0 \).

Exam Tip: When a variable is missing in a linear equation, write it with a coefficient of 0, like \( 0\cdot y \) or \( 0\cdot x \), to convert it to standard two-variable form.

Question. The linear equation \( 3x - 2y = 5 \) has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions.
Answer: (d) infinitely many solutions.
In simple words: Because we can choose any number for \( x \) and find a matching number for \( y \), there is no limit to the number of solutions.

Exam Tip: Remember that a single linear equation in two variables always has infinitely many solutions, whereas a system of two equations can have a unique, infinite, or no solution.

Question. The equation of x-axis is
(a) \( x = k \)
(b) \( y = 0 \)
(c) \( x = 0 \)
(d) \( y = k \)
Answer: (b) \( y = 0 \)
In simple words: Every point on the horizontal axis has a height of 0, so the equation is simply \( y = 0 \).

Exam Tip: Remember that the equation of the x-axis is \( y = 0 \) and the equation of the y-axis is \( x = 0 \). Do not get these two mixed up.

Question. Any point on the y-axis is of the form
(a) \( (x, y) \)
(b) \( (x, x) \)
(c) \( (0, y) \)
(d) \( (x, 0) \)
Answer: (c) \( (0, y) \)
In simple words: Since points on the vertical axis do not move left or right, their first number is always 0.

Exam Tip: For points on coordinate axes, one of the coordinates is always zero. Use this fact to eliminate incorrect options quickly.

Question. Draw the graph of the equation \( x - 2y = 0 \)
Answer: To draw the graph of \( x - 2y = 0 \), we rewrite it as \( x = 2y \). Let us find a few coordinate pairs that satisfy this equation:
- If \( y = 0 \implies x = 2(0) = 0 \). Point: \( (0, 0) \)
- If \( y = 1 \implies x = 2(1) = 2 \). Point: \( (2, 1) \)
- If \( y = -1 \implies x = 2(-1) = -2 \). Point: \( (-2, -1) \)
Plotting these points on a cartesian plane and joining them with a straight line gives the required graph.
In simple words: We find a few points like \( (0,0) \) and \( (2,1) \) by plugging in numbers, plot them on a grid, and draw a straight line through them.

Exam Tip: Ensure that you plot at least three points when graphing a line. This helps verify that your calculations are correct and that the points align perfectly.

Question. The cost of a pen is four times the cost of a pencil express the statement as a linear equation in two variables.
Answer: Let the cost of a pen be Rs. \( x \) and the cost of a pencil be Rs. \( y \).
According to the given condition:
Cost of pen = \( 4 \times \) Cost of pencil
\( \implies x = 4y \)
\( \implies x - 4y = 0 \)
This is the required linear equation.
In simple words: If a pen costs \( x \) and a pencil costs \( y \), then \( x \) is equal to 4 times \( y \), which we can write as \( x - 4y = 0 \).

Exam Tip: Always start by clearly defining your variables, specifying what \( x \) and \( y \) represent, to score full marks in descriptive problems.

Question. Write any four solutions for each of the following equations.
(a) \( 5x - 2 = 0 \)
(b) \( 3x + y = 7 \)
Answer: Let us find four solutions for each equation:
(a) \( 5x - 2 = 0 \)
We can express this in two variables as \( 5x + 0\cdot y - 2 = 0 \). Solving for \( x \), we get \( x = \frac{2}{5} \). Since the coefficient of \( y \) is 0, \( x \) will always be \( \frac{2}{5} \) regardless of the value of \( y \). Thus, four solutions are:
1. \( \left(\frac{2}{5}, 0\right) \)
2. \( \left(\frac{2}{5}, 1\right) \)
3. \( \left(\frac{2}{5}, 2\right) \)
4. \( \left(\frac{2}{5}, -1\right) \)

(b) \( 3x + y = 7 \)
We can write this as \( y = 7 - 3x \). Let us choose different values for \( x \) to find corresponding values of \( y \):
1. For \( x = 0 \implies y = 7 - 3(0) = 7 \). Solution: \( (0, 7) \)
2. For \( x = 1 \implies y = 7 - 3(1) = 4 \). Solution: \( (1, 4) \)
3. For \( x = 2 \implies y = 7 - 3(2) = 1 \). Solution: \( (2, 1) \)
4. For \( x = 3 \implies y = 7 - 3(3) = -2 \). Solution: \( (3, -2) \)
In simple words: To find solutions, pick any number for one variable and calculate the other. For the first equation, \( x \) is always \( 2/5 \). For the second, different values of \( x \) give different values for \( y \).

Exam Tip: When one variable is missing, its value remains constant. You can pick any arbitrary values for the missing variable to generate solutions.

Question. Find the value of a if (-1, 1) is a solution of the equation \( 3x - ay = 5 \)
Answer: Since \( (-1, 1) \) is a solution of the equation \( 3x - ay = 5 \), these coordinates must satisfy it. Substituting \( x = -1 \) and \( y = 1 \) into the equation:
\( 3(-1) - a(1) = 5 \)
\( \implies -3 - a = 5 \)
\( \implies -a = 5 + 3 \)
\( \implies -a = 8 \)
\( \implies a = -8 \)
In simple words: By putting the numbers \( -1 \) and \( 1 \) in place of \( x \) and \( y \), we solve the equation to find that \( a \) is \( -8 \).

Exam Tip: Always substitute carefully and verify your signs, as negative signs are where students frequently make algebraic errors.

Question. If (3,1) is a solution of the equation \( 3x + 2y = k \), find the value of k.
Answer: Since \( (3, 1) \) is a solution of the equation \( 3x + 2y = k \), substituting \( x = 3 \) and \( y = 1 \) gives:
\( 3(3) + 2(1) = k \)
\( \implies 9 + 2 = k \)
\( \implies k = 11 \)
In simple words: We substitute 3 for \( x \) and 1 for \( y \) into the equation, which directly gives \( k = 11 \).

Exam Tip: These types of problems are highly scoring and frequently appear in exams. Just substitute the values and solve for the unknown.

Question. Verify that x = 2, y = -1, is a solution of the linear equation \( 7x + 3y = 11 \)
Answer: To verify, let us substitute \( x = 2 \) and \( y = -1 \) into the Left Hand Side (LHS) of the equation:
LHS = \( 7(2) + 3(-1) \)
\( \implies \text{LHS} = 14 - 3 \)
\( \implies \text{LHS} = 11 \)
Since LHS is equal to the Right Hand Side (RHS), which is 11, the given values are indeed a solution.
In simple words: When we plug 2 and -1 into the left side of the equation, it works out to 11, which matches the right side.

Exam Tip: For verification questions, clearly show LHS and RHS separately to demonstrate a complete and structured proof.

Question. Write one solution of each of the following equations
(a) \( 4x - 3y = 0 \)
(b) \( 2y - y = 3 \)
Answer: Let us find one solution for each equation:
(a) \( 4x - 3y = 0 \)
If we choose \( x = 0 \), then:
\( 4(0) - 3y = 0 \implies -3y = 0 \implies y = 0 \).
Thus, one solution is \( (0, 0) \).

(b) \( 2y - y = 3 \)
Simplifying the equation, we get \( y = 3 \). In two-variable form, this is \( 0\cdot x + y = 3 \). Since \( y \) must always be 3, we can pick any value for \( x \). Let us choose \( x = 1 \).
Thus, one solution is \( (1, 3) \).
In simple words: For the first equation, \( (0,0) \) is an easy solution. For the second equation, \( y \) is always 3, so \( (1,3) \) works perfectly.

Exam Tip: Whenever the constant term on the RHS is 0, the origin \( (0,0) \) is always a solution.

Question. The cost of 2 pencils is same as the cost of 5 erasers. Express the statement as a linear equation in two variables.
Answer: Let the cost of a pencil be Rs. \( x \) and the cost of an eraser be Rs. \( y \).
According to the problem, the cost of 2 pencils equals the cost of 5 erasers:
\( 2x = 5y \)
\( \implies 2x - 5y = 0 \)
This is the required linear equation.
In simple words: If one pencil costs \( x \) and one eraser costs \( y \), then 2 pencils cost \( 2x \) and 5 erasers cost \( 5y \). Putting them equal gives \( 2x - 5y = 0 \).

Exam Tip: Always clearly mention the units, such as 'Rs.', when defining variables for cost-related problems.

Question. Give the geometrical representation of the equation y = 3 as an equation.
(i) In one variable
(ii) In two variables
Answer: (i) In one variable, \( y = 3 \) is represented as a point on a number line:

(ii) In two variables, the equation is written as \( 0\cdot x + y = 3 \), which is represented as a straight horizontal line parallel to the x-axis:
In simple words: On a single line, \( y = 3 \) is just one point. On a 2D grid, it becomes a flat line that stays at a height of 3.

Exam Tip: Be sure to draw coordinate axes cleanly with arrows and labeled units when showing 2D geometrical representations.

Question. Ramesh is driving his car with a uniform speed of 80 km/hr. Draw the time distance graph. Form the graph find the distance travelled by him in.
(i) \( 1\frac{1}{2} \) hr
(ii) 3 hours
Answer: Let the time taken be \( x \) hours and the distance travelled be \( y \) km.
Since distance = speed \( \times \) time, the relationship is:
\( y = 80x \)
Let us find a few coordinates to plot this graph:
- For \( x = 0 \implies y = 0 \). Point: \( (0, 0) \)
- For \( x = 1 \implies y = 80 \). Point: \( (1, 80) \)
- For \( x = 2 \implies y = 160 \). Point: \( (2, 160) \)
- For \( x = 3 \implies y = 240 \). Point: \( (3, 240) \)
Plotting these points on a coordinate plane gives our time-distance graph.

From the graph, we can find the distance for the given times:
(i) At \( x = 1\frac{1}{2} = 1.5 \) hours, we move vertically to the line and read the Y-value, which is 120 km.
(ii) At \( x = 3 \) hours, the corresponding Y-value is 240 km.
In simple words: We draw a line where the distance goes up by 80 km for every hour. Looking at the graph, 1.5 hours corresponds to 120 km, and 3 hours corresponds to 240 km.

Exam Tip: When drawing a time-distance graph, always label the X-axis with 'Time (hours)' and the Y-axis with 'Distance (km)' to avoid losing marks.

Question. Draw the graph of each of the equations \( 2x - 3y + 5 = 0 \) and \( 5x + 4y + 1 = 0 \) and find the coordinates of the point where the lines meet.
Answer: Let us find coordinates for both equations to plot them:
For \( 2x - 3y + 5 = 0 \implies 2x = 3y - 5 \implies x = \frac{3y - 5}{2} \):
- If \( y = 1 \implies x = -1 \). Point: \( (-1, 1) \)
- If \( y = 3 \implies x = 2 \). Point: \( (2, 3) \)
- If \( y = -1 \implies x = -4 \). Point: \( (-4, -1) \)

For \( 5x + 4y + 1 = 0 \implies 5x = -4y - 1 \implies x = \frac{-4y - 1}{5} \):
- If \( y = 1 \implies x = -1 \). Point: \( (-1, 1) \)
- If \( y = -4 \implies x = 3 \). Point: \( (3, -4) \)
- If \( y = 6 \implies x = -5 \). Point: \( (-5, 6) \)
Plotting both lines on the same coordinate grid:

By observing the graph, we can see that both lines intersect at the point \( (-1, 1) \).
In simple words: We plot both lines on the same grid. They cross each other exactly at the point \( (-1, 1) \).

Exam Tip: Clearly write down the final coordinates of the intersection point, as this is the primary answer the examiner looks for.

Question. Draw the graph of the equation \( 5x + 6y - 28 = 0 \) and check whether the point (2,3) lies on the line.
Answer: To draw the graph of \( 5x + 6y - 28 = 0 \), we can express \( x \) as:
\( x = \frac{28 - 6y}{5} \)
Let us find three solutions:
- If \( y = 3 \implies x = \frac{28 - 18}{5} = 2 \). Point: \( (2, 3) \)
- If \( y = 8 \implies x = \frac{28 - 48}{5} = -4 \). Point: \( (-4, 8) \)
- If \( y = -2 \implies x = \frac{28 + 12}{5} = 8 \). Point: \( (8, -2) \)
Let us plot these points and draw the line:

To verify if \( (2, 3) \) lies on the line, we substitute \( x = 2 \) and \( y = 3 \) into the equation:
LHS = \( 5(2) + 6(3) - 28 = 10 + 18 - 28 = 0 \)
Since LHS = RHS, the point \( (2, 3) \) lies on the line.
In simple words: We find points, plot the line, and plug \( (2,3) \) back into the equation. It makes the equation true, so the point lies on the line.

Exam Tip: Always use algebraic verification alongside the graph to double-check your coordinates.

Question. The taxi fare in a city is as follows: For the first kilometer, 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, writes a linear equation for this information, and draw its graph.
Answer: Let the total distance covered be \( x \) km and the total fare be Rs. \( y \).
According to the question:
Fare for the first km = Rs. 8
Remaining distance = \( (x - 1) \) km
Fare for subsequent distance = Rs. \( 5(x - 1) \)
So, total fare is:
\( y = 8 + 5(x - 1) \)
\( \implies y = 8 + 5x - 5 \)
\( \implies y = 5x + 3 \)
Which can be written as:
\( 5x - y + 3 = 0 \)
Let us find some points to plot this graph:
- For \( x = 1 \implies y = 5(1) + 3 = 8 \). Point: \( (1, 8) \)
- For \( x = 2 \implies y = 5(2) + 3 = 13 \). Point: \( (2, 13) \)

In simple words: The base fare is Rs. 8, plus Rs. 5 for every extra kilometer. This gives us the equation \( y = 5x + 3 \), which we can plot as a straight line.

Exam Tip: Since distance and fare cannot be negative, only plot the graph for positive values of \( x \) and \( y \) (first quadrant).

Question. Write three solutions for the equation \( 7x - 8y = 13 \)
Answer: Let us express \( x \) in terms of \( y \):
\( 7x = 8y + 13 \implies x = \frac{8y + 13}{7} \)
Let us choose values for \( y \) that result in integer values for \( x \):
1. If we choose \( y = 1 \):
\( x = \frac{8(1) + 13}{7} = \frac{21}{7} = 3 \). Solution: \( (3, 1) \)
2. If we choose \( y = 8 \):
\( x = \frac{8(8) + 13}{7} = \frac{77}{7} = 11 \). Solution: \( (11, 8) \)
3. If we choose \( y = -6 \):
\( x = \frac{8(-6) + 13}{7} = \frac{-35}{7} = -5 \). Solution: \( (-5, -6) \)
Thus, three solutions are \( (3, 1) \), \( (11, 8) \), and \( (-5, -6) \).
In simple words: We can find solutions by choosing values for \( y \) that make \( x \) come out as whole numbers. Three examples are \( (3,1) \), \( (11,8) \), and \( (-5,-6) \).

Exam Tip: When generating solutions, try to find integer values if possible, as they are much easier to work with and graph than fractions.