CBSE Class 9 Mathematics Linear Equations in two variables Assignment Set 03

Question. State true or false : ax + by + c = 0, represents a closed curve.
Answer: False. A linear equation of the form \( ax + by + c = 0 \) represents a straight line, which is an open curve extending infinitely in both directions, not a closed curve.
In simple words: The equation represents a straight line. Since a line goes on forever in both directions, it is not closed.

Exam Tip: A linear equation in two variables always represents a straight line, which is an open geometric figure.

Question. Find the coefficient of x in the equation { \( \sqrt{a^2 + b^2} \) } x + { \( \sqrt{a^2 - b^2} \) } y = \( \sqrt{a^2 b^2} \).
Answer: The coefficient of \( x \) in the given equation is \( \sqrt{a^2 + b^2} \).
In simple words: The coefficient of \( x \) is simply the value or expression that is multiplied by \( x \). Here, it is \( \sqrt{a^2+b^2} \).

Exam Tip: In any algebraic term, the coefficient of a variable is the coefficient multiplying that variable. Be careful to include any radical signs or brackets that group the terms.

Question. Write the coordinate axis represented by the line y = 0 .
Answer: The line \( y = 0 \) represents the x-axis.
In simple words: On the horizontal axis (x-axis), the value of the y-coordinate is always zero.

Exam Tip: Always remember that the equation of the x-axis is \( y = 0 \), and the equation of the y-axis is \( x = 0 \).

Question. Write the equation \( 12x + 3y = 20 \) in the form of \( ax+by+c= 0 \) and find out the values of a, b and c.
Answer: Rearranging the given equation, we get \( 12x + 3y - 20 = 0 \). Comparing this with the standard form \( ax + by + c = 0 \), we find that \( a = 12 \), \( b = 3 \), and \( c = -20 \).
In simple words: Move the 20 to the left side of the equals sign to make it zero on the right side. Then, look at the numbers in front of \( x \), \( y \pm \), and the constant.

Exam Tip: Don't forget the sign of the constant term \( c \). When transferring \( 20 \) to the left side, it becomes negative, so \( c = -20 \).

Question. Write \( 2x = 3y+5 \) in standard form of equation in two variables.
Answer: Bringing all terms to the left side, the equation \( 2x = 3y + 5 \) can be written in the standard form \( ax + by + c = 0 \) as:
\( 2x - 3y - 5 = 0 \)
In simple words: Shift all terms to one side of the equation to set it equal to zero, which gives \( 2x - 3y - 5 = 0 \).

Exam Tip: Standard linear form is represented as \( ax + by + c = 0 \). Ensure the right side of the equation is zero.

Question. Write the equation of the line parallel to the y-axis.
Answer: The general equation of a line parallel to the y-axis is \( x = a \), where \( a \) is a constant representing its distance from the y-axis.
In simple words: Any line that runs parallel to the vertical y-axis has an equation like \( x = \text{constant} \).

Exam Tip: Remember that a line parallel to the y-axis has a constant \( x \) value and is vertical.

Question. Write the equation of the line parallel to x-axis.
Answer: The general equation of a line parallel to the x-axis is \( y = b \), where \( b \) is a constant representing its distance from the x-axis.
In simple words: Any horizontal line parallel to the x-axis has an equation of the form \( y = \text{constant} \).

Exam Tip: A line parallel to the x-axis is a horizontal line and has a constant \( y \) value.

Question. Write the coordinate axis represented by the line \( x = 0 \).
Answer: The equation \( x = 0 \) represents the y-axis.
In simple words: Every point lying on the vertical y-axis has an x-coordinate of zero.

Exam Tip: Memorize that \( x = 0 \) is the equation for the y-axis, and \( y = 0 \) is the equation for the x-axis.

Question. State true or false: A linear equation in two variables can have only two solutions.
Answer: False. A linear equation in two variables represents a straight line, which contains infinitely many points; thus, it has infinitely many solutions.
In simple words: For any value you choose for \( x \), you can find a corresponding value for \( y \). This means there are endless solutions, not just two.

Exam Tip: A linear equation in one variable has a unique solution, while a linear equation in two variables always has infinitely many solutions.

Question. State true or false: \( x = ay \) is the equation of line passing through origin.
Answer: True. If we substitute the coordinates of the origin \( (0, 0) \) into the equation \( x = ay \), we get \( 0 = a(0) \), which simplifies to \( 0 = 0 \). Since the point satisfies the equation, the line passes through the origin.
In simple words: Since the point \( (0,0) \) works in this equation, the line definitely goes through the center of the graph (the origin).

Exam Tip: Any linear equation that does not have a constant term (i.e., of the form \( y = mx \) or \( x = ay \)) always passes through the origin.

Question. State true or false: A linear equation of two variable can have infinitely many solutions.
Answer: True. A linear equation in two variables can be satisfied by infinitely many ordered pairs of values for \( x \) and \( y \), as they correspond to the infinite points on a straight line.
In simple words: You can find an infinite number of coordinate pairs that make the equation true.

Exam Tip: Do not confuse a system of two linear equations (which can have a unique solution) with a single linear equation in two variables, which always has infinitely many solutions.

Question. State true or false: \( x = y + 2 \) represent a line passes through origin.
Answer: False. Substituting the origin \( (0, 0) \) into the equation \( x = y + 2 \) gives \( 0 = 0 + 2 \), which is \( 0 = 2 \) (not true). Hence, the line does not pass through the origin.
In simple words: When you plug in zero for both \( x \) and \( y \), the equation does not work, meaning the line does not go through the origin.

Exam Tip: A line can only pass through the origin if its constant term \( c \) is zero when written in standard form. Here, \( x - y - 2 = 0 \), so \( c = -2 \neq 0 \).

Question. Find the value of k, if \( x=2 \) , \( y=1 \) is a solution of the equation \( 2x+3y=k \).
Answer: Substituting \( x = 2 \) and \( y = 1 \) into the given equation:
\( 2(2) + 3(1) = k \)
\( \implies 4 + 3 = k \)
\( \implies k = 7 \)
Therefore, the value of \( k \) is \( 7 \).
In simple words: Put 2 in place of \( x \) and 1 in place of \( y \) in the equation, then calculate the math to find \( k \).

Exam Tip: When a point is a solution to an equation, its coordinates must satisfy the equation. Always substitute carefully and double-check your arithmetic.

Question. The cost of a notebook is twice as the cost of a pen. Write a linear equation in two variables to represent this statement.
Answer: Let the cost of a notebook be Rs. \( x \) and the cost of a pen be Rs. \( y \). According to the given condition:
Cost of notebook = \( 2 \times \) Cost of pen
\( \implies x = 2y \)
\( \implies x - 2y = 0 \)
This is the required linear equation.
In simple words: If we call the notebook's cost \( x \) and the pen's cost \( y \), then \( x \) equals two times \( y \).

Exam Tip: Always clearly state what variables represent at the beginning. Standardizing the equation to the form \( ax + by + c = 0 \) is recommended.

Question. Find the value of k in the equation \( 2x+3y= k \), where \( x = 2 \) and \( y = 3 \) is the solution of the equation.
Answer: Substituting the given solution \( x = 2 \) and \( y = 3 \) into the equation \( 2x + 3y = k \):
\( 2(2) + 3(3) = k \)
\( \implies 4 + 9 = k \)
\( \implies k = 13 \)
Hence, the value of \( k \) is \( 13 \).
In simple words: Just replace \( x \) with 2 and \( y \) with 3 in the formula, then add the results to find \( k \).

Exam Tip: Keep your calculations simple and ensure you follow the order of operations: multiply first, then add.

Question. Find the two different solutions of the equation \( 2x+y=4 \).
Answer: To find two solutions for \( 2x + y = 4 \), we can substitute arbitrary values for \( x \):
Case 1: Let \( x = 0 \)
\( 2(0) + y = 4 \)
\( \implies y = 4 \)
Thus, \( (0, 4) \) is the first solution.
Case 2: Let \( y = 0 \)
\( 2x + 0 = 4 \)
\( \implies 2x = 4 \)
Thus, \( (2, 0) \) is the second solution.
In simple words: By choosing 0 for \( x \), we find \( y \) is 4. Choosing 0 for \( y \) gives \( x \) as 2. These give two coordinate points on the line.

Exam Tip: Setting \( x = 0 \) and then \( y = 0 \) is the easiest way to find two clean solutions representing the intercepts of the line.

Question. If a = b = 3, then find the value of x from the equation \( \sqrt{a^2 + b^2} x + \sqrt{a^2 - b^2} y = \sqrt{a^2 b^2} \).
Answer: Given that \( a = 3 \) and \( b = 3 \), substitute these values into the equation:
\( \sqrt{3^2 + 3^2} x + \sqrt{3^2 - 3^2} y = \sqrt{3^2 \cdot 3^2} \)
\( \implies \sqrt{9 + 9} x + \sqrt{0} y = \sqrt{81} \)
\( \implies \sqrt{18} x + 0 = 9 \)
\( \implies 3\sqrt{2} x = 9 \)
\( \implies x = \frac{9}{3\sqrt{2}} \)
\( \implies x = \frac{3}{\sqrt{2}} \) (or \( \frac{3\sqrt{2}}{2} \))
In simple words: Put 3 in place of both \( a \) and \( b \). The term with \( y \) becomes zero because \( 3^2 - 3^2 = 0 \). Then solve the remaining equation to get \( x = \frac{3}{\sqrt{2}} \).

Exam Tip: When one of the terms simplifies to zero (like the \( y \) term here), the equation reduces to a single-variable equation, which is straightforward to solve.

Question. Let \( y \) varies direct as \( x \). If \( y=14 \), when \( x=7 \), then write a linear equation. What is the value of \( y \) when \( x= -2 \)?
Answer: Since \( y \) varies directly with \( x \), we can write the relation as:
\( y = kx \) (where \( k \) is a constant)
Substituting \( y = 14 \) and \( x = 7 \) to find \( k \):
\( 14 = k(7) \)
\( \implies k = 2 \)
Thus, the linear equation is:
\( y = 2x \)
To find \( y \) when \( x = -2 \):
\( y = 2(-2) \)
\( \implies y = -4 \)
In simple words: Direct variation means if one doubles, the other doubles. Here, \( y \) is always 2 times \( x \). So, when \( x = -2 \), \( y \) will be \( 2 \times (-2) = -4 \).

Exam Tip: "Varies directly" means \( y = kx \). Find the constant \( k \) first, formulate the equation, and then substitute the new value to solve.

Question. How many solution(s) of the equation \( 5x-3=3x+5 \) are there on the:
(i) Number line
(ii) Cartesian plane.
Answer: First, solve the linear equation:
\( 5x - 3 = 3x + 5 \)
\( \implies 5x - 3x = 5 + 3 \)
\( \implies 2x = 8 \)
\( \implies x = 4 \)
(i) On a number line, \( x = 4 \) is represented as a single unique point. Hence, there is **one solution**.
(ii) On the Cartesian plane, the equation \( x = 4 \) represents a vertical straight line parallel to the y-axis. Any point on this line is a solution, so there are **infinitely many solutions**.
In simple words: On a 1D line, \( x = 4 \) is just one spot. In a 2D plane, \( x = 4 \) is a whole vertical line, meaning any point on that line works.

Exam Tip: Remember that a single variable equation has a unique solution in one dimension, but representing it in two dimensions transforms it into a line with infinite solutions.

Question. For what value of k the point \( (k,5) \) lies on the line \( 4x-5y=10 \) ?
Answer: Since the point \( (k, 5) \) lies on the line \( 4x - 5y = 10 \), it must satisfy the equation:
\( 4(k) - 5(5) = 10 \)
\( \implies 4k - 25 = 10 \)
\( \implies 4k = 35 \)
\( \implies k = \frac{35}{4} \) (or \( 8.75 \))
In simple words: Substitute \( k \) for \( x \) and 5 for \( y \) in the line's equation, then solve for \( k \).

Exam Tip: When a coordinate point lies on a given line, plugging its coordinates into the line's equation must produce a true statement.

Question. If \( x = 1 \) and \( y = 1 \) is the solution of the equation \( 5x + 2ay = 3a \), find the value of a.
Answer: Since \( x = 1 \) and \( y = 1 \) is a solution, substituting these coordinates into \( 5x + 2ay = 3a \) gives:
\( 5(1) + 2a(1) = 3a \)
\( \implies 5 + 2a = 3a \)
\( \implies 3a - 2a = 5 \)
\( \implies a = 5 \)
In simple words: Substitute 1 for both \( x \) and \( y \), then group the terms containing \( a \) to solve for \( a \).

Exam Tip: Group variables on one side of the equation carefully to prevent sign errors when isolating \( a \).

Question. Solve the equation \( \frac{p^2 - 16}{p + 4} = 5 \quad (p \neq -4) \).
Answer: We are given:
\[ \frac{p^2 - 16}{p + 4} = 5 \quad (p \neq -4) \]
Factoring the numerator as a difference of squares:
\[ \frac{(p - 4)(p + 4)}{p + 4} = 5 \]
Since \( p \neq -4 \), we can divide both numerator and denominator by \( (p + 4) \):
\( p - 4 = 5 \)
\( \implies p = 9 \)
In simple words: The numerator is a difference of squares, so it splits into \( (p-4)(p+4) \). The \( p+4 \) terms cancel out, leaving \( p - 4 = 5 \), which gives \( p = 9 \).

Exam Tip: Factoring expressions is a crucial skill. Always look for algebraic identities like \( a^2 - b^2 = (a-b)(a+b) \) to simplify rational expressions.

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 kilometres and total fare as rupees, write a linear equation for this information.
Answer: Let the distance covered be \( x \) km and the total fare be Rs. \( y \). According to the given details:
Fare for the first kilometer = Rs. 8
Subsequent distance = \( (x - 1) \) km
Fare for the subsequent distance = Rs. \( 5(x - 1) \)
Thus, the total fare is:
\( y = 8 + 5(x - 1) \)
\( \implies y = 8 + 5x - 5 \)
\( \implies y = 5x + 3 \)
Or in standard form:
\( 5x - y + 3 = 0 \)
In simple words: We pay Rs. 8 for the first km. For the rest of the distance \( (x-1) \), we pay Rs. 5 per km. Adding these together gives the total fare, which is \( y = 5x + 3 \).

Exam Tip: When representing real-world scenarios with linear equations, always break down the fixed charge and the variable rate clearly before combining them.

Question. Find four different solutions of the equation \( x+2y=6 \).
Answer: We can find four distinct solutions by substituting different values for one variable:
1. Let \( x = 0 \)
\( 0 + 2y = 6 \)
\( \implies 2y = 6 \)
\( \implies y = 3 \).
First solution: \( (0, 3) \).
2. Let \( y = 0 \)
\( x + 2(0) = 6 \)
\( \implies x = 6 \).
Second solution: \( (6, 0) \).
3. Let \( x = 2 \)
\( 2 + 2y = 6 \)
\( \implies 2y = 4 \)
\( \implies y = 2 \).
Third solution: \( (2, 2) \).
4. Let \( x = 4 \)
\( 4 + 2y = 6 \)
\( \implies 2y = 2 \)
\( \implies y = 1 \).
Fourth solution: \( (4, 1) \).
In simple words: Plug in different numbers for \( x \) (like 0, 6, 2, and 4) and calculate the corresponding \( y \) values to get four separate coordinates.

Exam Tip: Choosing even numbers for \( x \) ensures that \( y \) evaluates to an integer, making calculations simpler and cleaner.

Question. Write each of the following equations in the form \( ax+by+c=0 \) and indicate the values of a, b and c in each case:
(i) \( 2x + 3y = 4.37 \)
(ii) \( x-4 = 3y \)
(iii) \( 4= 5x-3y \)
Answer:
(i) For \( 2x + 3y = 4.37 \):
Rearranging gives: \( 2x + 3y - 4.37 = 0 \)
Comparing with \( ax + by + c = 0 \):
\( a = 2, b = 3, c = -4.37 \)
(ii) For \( x - 4 = 3y \):
Rearranging gives: \( x - 3y - 4 = 0 \)
Comparing with \( ax + by + c = 0 \):
\( a = 1, b = -3, c = -4 \)
(iii) For \( 4 = 5x - 3y \):
Rearranging gives: \( 5x - 3y - 4 = 0 \)
Comparing with \( ax + by + c = 0 \):
\( a = 5, b = -3, c = -4 \)
In simple words: Move all elements of the equation to one side to get zero on the other side. Then identify \( a \), \( b \), and \( c \).

Exam Tip: Pay close attention to the signs (+ or -) of each constant. A negative sign belongs to the coefficient.

Question. Solve the equation \( 2x+1=x-3 \), and represent the solution(s) on
(i) the number line.
(ii) the Cartesian plane.
Answer: Solving the given equation:
\( 2x + 1 = x - 3 \)
\( \implies 2x - x = -3 - 1 \)
\( \implies x = -4 \)
(i) On the number line, \( x = -4 \) represents a unique point:

(ii) On the Cartesian plane, \( x = -4 \) represents a straight vertical line parallel to the y-axis, passing through the point \( (-4, 0) \):

In simple words: The solution to the equation is \( x = -4 \). On a basic 1D number line, this is just a single point. On a 2D grid, it becomes a vertical line where \( x \) is always \( -4 \) for any \( y \) value.

Exam Tip: For representation on a Cartesian plane, write the equation as \( 1\cdot x + 0\cdot y = -4 \) to understand that \( y \) can take any real value while \( x \) remains \( -4 \).

Question. Evaluate: \( (5x+1)(x+3)-8=5(x+1)(x+2) \).
Answer: Expand both sides of the equation:
Left-Hand Side (LHS):
\( (5x + 1)(x + 3) - 8 = 5x^2 + 15x + x + 3 - 8 \)
\( \implies 5x^2 + 16x - 5 \)
Right-Hand Side (RHS):
\( 5(x + 1)(x + 2) = 5(x^2 + 2x + x + 2) \)
\( \implies 5(x^2 + 3x + 2) \)
\( \implies 5x^2 + 15x + 10 \)
Now, set LHS equal to RHS:
\( 5x^2 + 16x - 5 = 5x^2 + 15x + 10 \)
Subtracting \( 5x^2 \) from both sides:
\( 16x - 5 = 15x + 10 \)
\( \implies 16x - 15x = 10 + 5 \)
\( \implies x = 15 \)
In simple words: Expand the multiplication on both sides. The \( 5x^2 \) terms on both sides cancel out, allowing us to find that \( x = 15 \).

Exam Tip: Don't let quadratic terms confuse you; when they appear with the same coefficient on both sides, they subtract out, leaving a linear equation.

Question. If the point \( (-1,-5) \) lies on the graphs of \( 3x=ay+7 \) and \( y=bx+7 \), find the value of a and b.
Answer: Substitute the point \( (-1, -5) \) into both equations.
For the first equation \( 3x = ay + 7 \):
\( 3(-1) = a(-5) + 7 \)
\( \implies -3 = -5a + 7 \)
\( \implies -5a = -10 \)
\( \implies a = 2 \)
For the second equation \( y = bx + 7 \):
\( -5 = b(-1) + 7 \)
\( \implies -5 = -b + 7 \)
\( \implies -b = -12 \)
\( \implies b = 12 \)
Thus, the values are \( a = 2 \) and \( b = 12 \).
In simple words: Put \( -1 \) for \( x \) and \( -5 \) for \( y \) in both equations. Solving them separately gives \( a = 2 \) and \( b = 12 \).

Exam Tip: Treat each equation as an independent substitution problem to find the value of each variable separately.

Question. At what point does the graph of the linear equation \( 2x+3y=9 \) meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis.
Answer: First, determine the equation of the line parallel to the y-axis:
A line parallel to the y-axis and at a distance of 4 units to the right of the y-axis has the equation:
\( x = 4 \)
To find the intersection point, substitute \( x = 4 \) into the given linear equation \( 2x + 3y = 9 \):
\( 2(4) + 3y = 9 \)
\( \implies 8 + 3y = 9 \)
\( \implies 3y = 1 \)
\( \implies y = \frac{1}{3} \)
Therefore, the graph meets the line at the point \( \left(4, \frac{1}{3}\right) \).
In simple words: A line 4 units to the right of the y-axis is \( x = 4 \). Put \( x = 4 \) into the equation to get the y-value, which is \( \frac{1}{3} \).

Exam Tip: Carefully translate geometric descriptions (like "parallel to the y-axis, at a distance of 4 units") into equations like \( x = 4 \).

Question. If the point \( (4,3) \) lies on the graph of the equation \( 3x-ay=6 \), find whether \( (-2,-6) \) also lies on the same graph.
Answer: First, find the value of \( a \) by substituting \( (4, 3) \) into the equation:
\( 3(4) - a(3) = 6 \)
\( \implies 12 - 3a = 6 \)
\( \implies 3a = 6 \)
\( \implies a = 2 \)
Thus, the equation of the line is:
\( 3x - 2y = 6 \)
Now, check if \( (-2, -6) \) satisfies this equation by substituting \( x = -2 \) and \( y = -6 \):
LHS = \( 3(-2) - 2(-6) \)
\( \implies -6 + 12 = 6 \)
Since LHS = RHS = 6, the point \( (-2, -6) \) also lies on the same graph.
In simple words: Use the first point to find that \( a = 2 \). This makes our equation \( 3x - 2y = 6 \). When we plug in \( (-2, -6) \), both sides match, so it does lie on the graph.

Exam Tip: When solving such two-part problems, first use the known point to find the constant, rewrite the complete equation, and then verify the second point.

Question. Give the equations of two lines passing through \( (-2,-4) \). How many more such lines are there, and why?
Answer: Two simple lines that pass through the point \( (-2, -4) \) are:
1. \( y = 2x \) (since \( -4 = 2(-2) \) is true)
2. \( x + y = -6 \) (since \( -2 + (-4) = -6 \) is true)
There are **infinitely many** such lines because an infinite number of lines can pass through any single given point in a plane.
In simple words: Equations like \( y = 2x \) and \( x+y = -6 \) pass through \( (-2, -4) \). There are infinitely many such lines because you can draw endless lines through any single dot.

Exam Tip: A single point has infinite degrees of freedom, meaning you can write an infinite number of equations in the form \( ax + by + c = 0 \) that satisfy the point.

Question. In countries like USA and Canada, 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: \( F=(9/5)C+32 \) Draw the graph of linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
Answer: Let us find two points to plot the linear equation \( F = \frac{9}{5}C + 32 \), treating Celsius (\( C \)) on the x-axis and Fahrenheit (\( F \)) on the y-axis:
1. When \( C = 0 \):
\( F = \frac{9}{5}(0) + 32 = 32 \)
Point: \( (0, 32) \)
2. When \( C = -40 \):
\( F = \frac{9}{5}(-40) + 32 = -72 + 32 = -40 \)
Point: \( (-40, -40) \)
Using these coordinates, we can plot the straight line:

In simple words: We can find two key temperature readings, like \( 0^\circ\text{C} = 32^\circ\text{F} \) and \( -40^\circ\text{C} = -40^\circ\text{F} \), and draw a line passing through these two points on the graph paper.

Exam Tip: Label your axes clearly with the physical units (Celsius on the horizontal axis, Fahrenheit on the vertical axis) and mark at least two points on the line.

Question. Write each of the following as an equation in two variables:
(i) \( x=-5 \)
(ii) \( y=2 \)
(iii) \( 2x=3 \)
(iv) \( 5y=2 \)
Answer: We can express a single-variable equation in two variables by including the other variable with a coefficient of 0:
(i) \( x = -5 \) is written as:
\( 1\cdot x + 0\cdot y = -5 \) (or \( 1\cdot x + 0\cdot y + 5 = 0 \))
(ii) \( y = 2 \) is written as:
\( 0\cdot x + 1\cdot y = 2 \) (or \( 0\cdot x + 1\cdot y - 2 = 0 \))
(iii) \( 2x = 3 \) is written as:
\( 2x + 0\cdot y = 3 \) (or \( 2x + 0\cdot y - 3 = 0 \))
(iv) \( 5y = 2 \) is written as:
\( 0\cdot x + 5y = 2 \) (or \( 0\cdot x + 5y - 2 = 0 \))
In simple words: To write a one-variable equation with two variables, we add the missing variable multiplied by 0.

Exam Tip: Writing standard forms with \( 0\cdot y \) or \( 0\cdot x \) shows that the missing variable does not affect the equation's value but allows it to be plotted in 2D.

Question. The taxi fare in a city is as follows: For the first kilometer, the fare is Rs.20 and for the subsequent distance it is Rs. 6 per km. Taking \( x \) km as the distance covered and Rs. \( y \) as the total fare, write 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 given details:
Fare for the first kilometer = Rs. 20
Subsequent distance = \( (x - 1) \) km
Fare for the subsequent distance = Rs. \( 6(x - 1) \)
Thus, the total fare equation is:
\( y = 20 + 6(x - 1) \)
\( \implies y = 20 + 6x - 6 \)
\( \implies y = 6x + 14 \)
To plot the graph of this equation, let's find a few coordinate pairs:
- For \( x = 1 \): \( y = 6(1) + 14 = 20 \). Point is \( (1, 20) \).
- For \( x = 2 \): \( y = 6(2) + 14 = 26 \). Point is \( (2, 26) \).
Plotting these points, we get the graph:

In simple words: We pay a fixed Rs. 20 for the first km and Rs. 6 for each additional km. This gives us the equation \( y = 6x + 14 \), which we can plot on a graph by finding points like \( (1, 20) \) and \( (2, 26) \).

Exam Tip: Since distance and fare cannot be negative, the graph of this function only makes sense in the first quadrant of the Cartesian plane.

Question. The cost of a box is Rs.25. Taking x as the number of boxes and y, the total cost in rupees, construct a linear equation. Also, draw the graph.
Answer: Let the number of boxes be \( x \) and the total cost in rupees be \( y \). Since one box costs Rs. 25, the cost of \( x \) boxes will be \( 25x \). This gives the linear equation:
\( y = 25x \)
In standard form, we write this as:
\( 25x - y = 0 \)
To plot the graph of this equation, we determine a few coordinate points:
- When \( x = 0 \), \( y = 25(0) = 0 \). Point: \( (0, 0) \)
- When \( x = 1 \), \( y = 25(1) = 25 \). Point: \( (1, 25) \)
- When \( x = 2 \), \( y = 25(2) = 50 \). Point: \( (2, 50) \)
We plot these points and join them with a straight line passing through the origin.

In simple words: The total cost depends directly on how many boxes you buy. One box is Rs. 25, so multiplying 25 by the number of boxes gives you the total bill.
Exam Tip: Remember to clearly write the scale chosen for both axes on your graph sheet to score full presentation marks.

Question. The ratio of hydrogen and oxygen in water is 2:1. Set up an equation between hydrogen and oxygen and draw its graph. From the graph read the hydrogen if oxygen is 6 gram.
Answer: Let the quantity of oxygen in water be \( x \) grams and the quantity of hydrogen be \( y \) grams. Given that the ratio of hydrogen to oxygen is \( 2:1 \), we have:
\( \frac{y}{x} = \frac{2}{1} \implies y = 2x \)
We find a few coordinate points to draw the graph:
- If \( x = 0 \), then \( y = 0 \). Point: \( (0, 0) \)
- If \( x = 2 \), then \( y = 4 \). Point: \( (2, 4) \)
- If \( x = 4 \), then \( y = 8 \). Point: \( (4, 8) \)
- If \( x = 6 \), then \( y = 12 \). Point: \( (6, 12) \)
Reading from the graph, when oxygen \( x = 6 \) grams, the corresponding value of hydrogen \( y \) is \( 12 \) grams.

In simple words: For every part of oxygen, there are two parts of hydrogen. So, if we have 6 grams of oxygen, we must have double that amount of hydrogen, which is 12 grams.
Exam Tip: Use dashed lines on the graph to show how you traced from \( x = 6 \) on the horizontal axis up to the line and then over to \( y = 12 \) on the vertical axis.

Question. The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is (i) 5 m/sec\(^2\) , (ii) 6 m/sec\(^2\).
Answer: Let the acceleration produced be \( x \) m/sec\(^2\) and the force exerted be \( y \) Newtons. Since force is directly proportional to acceleration and the mass is the constant of proportionality (\( m = 6 \) kg), we write the equation using Newton's second law (\( F = ma \)):
\( y = 6x \)
Let us obtain coordinates to plot this line:
- If \( x = 0 \), then \( y = 0 \). Point: \( (0, 0) \)
- If \( x = 5 \), then \( y = 30 \). Point: \( (5, 30) \)
- If \( x = 6 \), then \( y = 36 \). Point: \( (6, 36) \)
Now, reading from our plotted graph:
(i) When the acceleration \( x = 5 \) m/sec\(^2\), the force required \( y = 30 \) Newtons.
(ii) When the acceleration \( x = 6 \) m/sec\(^2\), the force required \( y = 36 \) Newtons.

In simple words: Force equals mass times acceleration. Since the mass is 6, you just multiply the acceleration value by 6 to find the force.
Exam Tip: Be sure to write the units (Newtons for force, m/sec² for acceleration) when writing down final values read from the graph.

Question. The parking charges of a car at certain place in Delhi is Rs.50 for first one hour and Rs. 10 for subsequent hours. (a) Write down the equation and draw the graph for the data.(b) Read the charges from the graph (i) for 2 hours (ii) for 8 hours.
Answer: Let the total parking duration be \( x \) hours and the total charges be Rs. \( y \).
The charge for the very first hour is Rs. 50. The remaining duration is \( (x - 1) \) hours, charged at Rs. 10 per hour.
Thus, the linear equation representing this situation is:
\( y = 50 + 10(x - 1) \)
\( y = 50 + 10x - 10 \)
\( y = 10x + 40 \) (valid for \( x \ge 1 \))
Let us find some coordinates to draw the graph:
- For \( x = 1 \): \( y = 10(1) + 40 = 50 \). Point: \( (1, 50) \)
- For \( x = 2 \): \( y = 10(2) + 40 = 60 \). Point: \( (2, 60) \)
- For \( x = 8 \): \( y = 10(8) + 40 = 120 \). Point: \( (8, 120) \)
We plot these coordinates and connect them with a straight line starting from \( x = 1 \).
Reading from the graph:
(i) Parking charges for 2 hours are Rs. 60.
(ii) Parking charges for 8 hours are Rs. 120.

In simple words: You pay a flat rate of Rs. 50 for the first hour. After that, you pay an extra Rs. 10 for each extra hour.
Exam Tip: This equation is only valid for \( x \ge 1 \). Avoid extending the line on your graph to values where \( x < 1 \) as negative or zero hours do not make sense in this context.

Question. Two players A and B together scored 40 runs in a cricket match. If there is no extra run scored in their partnership, then represent this information in the form of linear equation in two variables. Draw graph of the linear equation. From the graph, find the runs recorded by player A if run scored by player B is 10.
Answer: Let the runs scored by player A be \( x \) and the runs scored by player B be \( y \). Since their combined score is 40 runs, we get the following linear equation:
\( x + y = 40 \)
Let us determine some coordinate points to plot this line:
- If \( x = 40 \), then \( y = 0 \). Point: \( (40, 0) \)
- If \( x = 0 \), then \( y = 40 \). Point: \( (0, 40) \)
- If \( x = 30 \), then \( y = 10 \). Point: \( (30, 10) \)
Plotting these points on a grid and joining them gives a straight line. If player B scores \( y = 10 \) runs, we locate \( 10 \) on the vertical axis, trace to our line, and find that player A scored \( x = 30 \) runs.

In simple words: The scores of both players must add up to 40. If one player makes 10 runs, the other must have made 30 runs to reach the total of 40.
Exam Tip: Since runs cannot be negative, keep your graph line confined to the first quadrant only.

Question. If \( \frac{3x+6}{8} - \frac{11x-8}{24} + \frac{x}{3} = \frac{3x}{4} - \frac{x+7}{24} \), then the value of x is
Answer: Let us solve the given equation:
\( \frac{3x+6}{8} - \frac{11x-8}{24} + \frac{x}{3} = \frac{3x}{4} - \frac{x+7}{24} \)
Multiply every term on both sides by 24 (the LCM of the denominators 8, 24, 3, and 4) to eliminate the fractions:
\( 3(3x + 6) - (11x - 8) + 8(x) = 6(3x) - (x + 7) \)
Now expand and simplify both sides:
\( 9x + 18 - 11x + 8 + 8x = 18x - x - 7 \)
Group the terms together:
\( (9x - 11x + 8x) + (18 + 8) = 17x - 7 \)
\( 6x + 26 = 17x - 7 \)
Rearrange to solve for \( x \):
\( 26 + 7 = 17x - 6x \)
\( 33 = 11x \)
\( x = 3 \)
Thus, the value of \( x \) is 3.
In simple words: First multiply the whole equation by 24 to get rid of the denominators. After that, simplify the terms on both sides to find that x equals 3.
Exam Tip: Be very careful with the negative signs in front of fractions, such as \( -\frac{11x-8}{24} \). The negative sign distributes to both terms in the numerator, changing \( -8 \) into \( +8 \).

Question. If one-fourth of the sum of a number and seven is four less than three times the number, find the number.
Answer: Let the unknown number be \( x \).
According to the statement:
- The sum of the number and seven is \( (x + 7) \).
- One-fourth of this sum is \( \frac{1}{4}(x + 7) \).
- Three times the number is \( 3x \).
- Four less than three times the number is \( (3x - 4) \).
We set up the equation as:
\( \frac{1}{4}(x + 7) = 3x - 4 \)
Multiply both sides by 4 to clear the fraction:
\( x + 7 = 4(3x - 4) \)
\( x + 7 = 12x - 16 \)
Collect the variables on one side and the constants on the other:
\( 7 + 16 = 12x - x \)
\( 23 = 11x \)
\( x = \frac{23}{11} \)
So, the number is \( \frac{23}{11} \).
In simple words: Write out the word problem as an algebraic equation. Multiply by 4 to remove the fraction, and solve for x to get 23 divided by 11.
Exam Tip: Verify your answer by plugging \( \frac{23}{11} \) back into the original statement to check if both sides evaluate to the same value.

Question. Solve the equation 2x + 1 = x - 3, and represent the solution(s) on
(i) the number line,
(ii) the Cartesian plane.
Answer: Let us first solve the linear equation:
\( 2x + 1 = x - 3 \)
\( 2x - x = -3 - 1 \)
\( x = -4 \)
Now we represent this solution in both formats:
(i) **On a number line:** The equation \( x = -4 \) represents a single point. We place a solid dot at the position \( -4 \) on the number line.

(ii) **On the Cartesian plane:** In a two-dimensional coordinate system, the equation \( x = -4 \) is written as \( x + 0y = -4 \). This represents a vertical line parallel to the y-axis, passing through the point \( (-4, 0) \).

In simple words: The solution to the equation is x = -4. On a single line, it is just one dot. On a flat 2D grid, it is a vertical line that crosses the horizontal axis at -4.
Exam Tip: Remember that a single variable equation has a point solution in 1D, but represents a line in a 2D Cartesian plane.

Question. Find two solutions for each of the following equations:
(i) 4x + 3y = 12
(ii) 2x + 5y = 0
(iii) 3y + 4 = 0
Answer: Let us find two solutions for each equation by substituting arbitrary values:

(i) \( 4x + 3y = 12 \)
- Let \( x = 0 \): then \( 4(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \). First solution: \( (0, 4) \).
- Let \( y = 0 \): then \( 4x + 3(0) = 12 \implies 4x = 12 \implies x = 3 \). Second solution: \( (3, 0) \).

(ii) \( 2x + 5y = 0 \)
- Let \( x = 0 \): then \( 2(0) + 5y = 0 \implies 5y = 0 \implies y = 0 \). First solution: \( (0, 0) \).
- Let \( x = 5 \): then \( 2(5) + 5y = 0 \implies 10 + 5y = 0 \implies 5y = -10 \implies y = -2 \). Second solution: \( (5, -2) \).

(iii) \( 3y + 4 = 0 \)
This equation can be rewritten as \( y = -\frac{4}{3} \). Since \( x \) does not affect the equation, \( x \) can be any real number while \( y \) remains constant at \( -\frac{4}{3} \).
- Let \( x = 0 \): then \( y = -\frac{4}{3} \). First solution: \( (0, -\frac{4}{3}) \).
- Let \( x = 1 \): then \( y = -\frac{4}{3} \). Second solution: \( (1, -\frac{4}{3}) \).
In simple words: Choose any number for one of the variables and solve for the other to find points that satisfy the equation.
Exam Tip: Setting \( x = 0 \) and \( y = 0 \) is usually the easiest way to quickly find two distinct points for linear equations.