CBSE Class 9 Mathematics Linear Equations in two variables Assignment Set 04

Question. State true or false : ax + by + c = 0, represents a closed curve.
Answer: False. The equation \( ax + by + c = 0 \) is the standard form of a linear equation in two variables, which represents a straight line. A straight line is not a closed curve because it extends infinitely in both directions.
In simple words: False. This equation represents a straight line, which goes on forever in both directions and does not close back on itself.
Exam Tip: Remember that 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 \( \{ \sqrt{a^2+b^2} \} x + \{ \sqrt{a^2-b^2} \} y = \sqrt{a^2b^2} \).
Answer: The coefficient of \( x \) in the given equation is \( \sqrt{a^2+b^2} \).
In simple words: The number or expression multiplying \( x \) is \( \sqrt{a^2+b^2} \).

Exam Tip: The coefficient is the entire constant expression multiplying the variable, including any square roots or brackets.

Question. Write the coordinate axis represented by the line y = 0 .
Answer: The line \( y = 0 \) represents the x-axis.
In simple words: The line \( y = 0 \) is the x-axis, because every point on the x-axis has a y-coordinate of zero.

Exam Tip: The equation of the x-axis is always \( y = 0 \), and the equation of the y-axis is always \( x = 0 \).

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 equation to the form \( ax + by + c = 0 \): \( 12x + 3y - 20 = 0 \) By comparing this with the standard form, we get the values of the constants: \( a = 12 \) \( b = 3 \) \( c = -20 \)
In simple words: Move all terms to the left side so that the right side is zero, then match the numbers with \( a \), \( b \), and \( c \).

Exam Tip: Be careful with the sign of the constant term \( c \); since the standard form has \( +c \), the negative sign of \( -20 \) must be included in the value of \( c \).

Question. Write 2x = 3y+5 in standard form of equation in two variables.
Answer: The given equation can be rewritten in the standard form \( ax + by + c = 0 \) by bringing all terms to one side: \( 2x - 3y - 5 = 0 \)
In simple words: To write the equation in standard form, shift \( 3y \) and \( 5 \) to the left side of the equals sign.

Exam Tip: When you transpose a term to the other side of an equation, always remember to reverse its sign.

Question. Write the equation of the line parallel to the y-axis.
Answer: The equation of any straight line parallel to the y-axis is of the form: \( x = a \) where \( a \) is any real number representing the constant distance of the line from the y-axis.
In simple words: Any line parallel to the vertical axis has the equation \( x = a \), where \( a \) is a constant number.

Exam Tip: Remember that a vertical line has a constant x-value, so its equation is always \( x = \text{constant} \).

Question. Write the equation of the line parallel to x-axis.
Answer: The equation of any straight line parallel to the x-axis is represented in the form: \( y = b \) where \( b \) is any constant value representing the distance of the line from the x-axis.
In simple words: Any line parallel to the horizontal axis has the equation \( y = b \), where \( b \) is a constant.

Exam Tip: A horizontal line has a constant y-value, so its equation is always of the form \( y = \text{constant} \).

Question. Write the coordinate axis represented by the line x = 0 .
Answer: The line \( x = 0 \) represents the y-axis.
In simple words: The y-axis is defined by the equation \( x = 0 \) because the x-coordinate of every point on this vertical line is zero.

Exam Tip: Keep in mind that the y-axis is represented by \( x = 0 \) and the x-axis is represented by \( y = 0 \).

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 in the coordinate plane. Since a straight line contains infinitely many points, the equation has infinitely many solutions.
In simple words: False. There are infinitely many pairs of values that can satisfy a linear equation in two variables.

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

Question. State true or false: x = ay is the equation of line passing through origin.
Answer: True. If we substitute \( x = 0 \) and \( y = 0 \) (the coordinates of the origin) into the equation \( x = ay \), we get \( 0 = a(0) \), which is \( 0 = 0 \). Since the coordinates of the origin satisfy the equation, the line passes through the origin.
In simple words: True. If you plug in \( 0 \) for both \( x \) and \( y \), the equation works, which means the line goes through the point \( (0,0) \).

Exam Tip: Any linear equation that does not have a non-zero constant term (i.e., of the form \( ax + by = 0 \)) will always pass 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 corresponds to a line on a graph, and since there are infinitely many points on any line, the equation has an infinite number of solutions.
In simple words: True. You can choose any value for \( x \) and find a matching value for \( y \), so there are endless solutions.

Exam Tip: Remember that any point lying on the line representing a linear equation is a solution to that equation.

Question. State true or false: x = y + 2 represent a line passes through origin.
Answer: False. Substituting the coordinates of the origin \( (0, 0) \) into the equation \( x = y + 2 \), we get \( 0 = 0 + 2 \), which simplifies to \( 0 = 2 \). Since this is not true, the line does not pass through the origin.
In simple words: False. If we plug in \( 0 \) for \( x \) and \( y \), we get \( 0 = 2 \), which is wrong, so the line does not go through \( (0,0) \).

Exam Tip: If a linear equation has a constant term other than zero when written as \( ax + by = c \), it cannot pass through the origin.

Question. Find the value of k, if x=2 , y=1 is a solution of the equation 2x+3y=k.
Answer: Since \( x = 2 \) and \( y = 1 \) is a solution of the equation \( 2x + 3y = k \), substituting these values into the equation gives: \( 2(2) + 3(1) = k \)
\( \implies 4 + 3 = k \)
\( \implies k = 7 \) Therefore, the value of \( k \) is \( 7 \).
In simple words: Put \( x = 2 \) and \( y = 1 \) into the equation to calculate that \( k \) is \( 7 \).

Exam Tip: To verify if a point is a solution, substitute its coordinates into the equation and check if the left-hand side equals the right-hand side.

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, the cost of the notebook is twice the cost of the pen. This statement can be written as: \( x = 2y \) Or in standard form: \( x - 2y = 0 \)
In simple words: If a notebook costs \( x \) and a pen costs \( y \), then because the notebook is twice as expensive, we write \( x = 2y \).

Exam Tip: Clearly define your variables (with units if applicable) before formulating the equation to avoid confusion.

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: Given that \( x = 2 \) and \( y = 3 \) is a solution to the equation \( 2x + 3y = k \), we can substitute these coordinates into the equation: \( 2(2) + 3(3) = k \)
\( \implies 4 + 9 = k \)
\( \implies k = 13 \) Hence, the value of \( k \) is \( 13 \).
In simple words: Substituting \( 2 \) for \( x \) and \( 3 \) for \( y \) in the equation gives \( 4 + 9 \), which means \( k = 13 \).

Exam Tip: Always double-check basic arithmetic when substituting values into equations to prevent silly errors.

Question. Find the two different solutions of the equation 2x+y=4.
Answer: To find two solutions for \( 2x + y = 4 \), we can choose arbitrary values for one of the variables and solve for the other:
(i) Let \( x = 0 \): \( 2(0) + y = 4 \)
\( \implies y = 4 \) So, \( (0, 4) \) is a solution.
(ii) Let \( y = 0 \): \( 2x + 0 = 4 \)
\( \implies 2x = 4 \)
\( \implies x = 2 \) So, \( (2, 0) \) is a solution. Two different solutions of the equation are \( (0, 4) \) and \( (2, 0) \).
In simple words: You can find two points by setting \( x = 0 \) to get \( y = 4 \), and setting \( y = 0 \) to get \( x = 2 \).

Exam Tip: Setting \( x = 0 \) and \( y = 0 \) is the easiest way to find two simple, clean solutions for any linear equation.

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 the equation: \( \sqrt{a^2 + b^2} x + \sqrt{a^2 - b^2} y = \sqrt{a^2b^2} \) Substituting \( a = 3 \) and \( b = 3 \) into this equation: \( \sqrt{3^2 + 3^2} x + \sqrt{3^2 - 3^2} y = \sqrt{3^2 \cdot 3^2} \)
\( \implies \sqrt{9 + 9} x + \sqrt{9 - 9} y = \sqrt{81} \)
\( \implies \sqrt{18} x + 0 \cdot y = 9 \)
\( \implies 3\sqrt{2} x = 9 \)
\( \implies x = \frac{9}{3\sqrt{2}} \)
\( \implies x = \frac{3}{\sqrt{2}} \) Multiplying the numerator and denominator by \( \sqrt{2} \), we can also write: \( x = \frac{3\sqrt{2}}{2} \)
In simple words: Since \( a = b = 3 \), the \( y \) term disappears because \( a^2 - b^2 = 0 \). Solving the remaining part gives \( x = \frac{3}{\sqrt{2}} \).

Exam Tip: Notice how the difference of squares \( a^2 - b^2 \) equals zero when \( a = b \); recognizing this immediately simplifies the problem by eliminating the \( y \) variable.

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 as \( x \), we can express this relationship as: \( y = kx \) where \( k \) is the constant of variation. Using the given values \( y = 14 \) and \( x = 7 \): \( 14 = k(7) \)
\( \implies k = 2 \) Thus, the linear equation representing this relationship is: \( y = 2x \) To find the value of \( y \) when \( x = -2 \), we substitute this into our equation: \( y = 2(-2) \)
\( \implies y = -4 \)
In simple words: Since \( y \) is always double \( x \), the equation is \( y = 2x \). When \( x = -2 \), \( y \) must be \( -4 \).

Exam Tip: Direct variation means the ratio \( y/x \) remains constant. First find this constant \( k \) using the given pair, then use it to find any other values.

Question. How many solution(s) of the equation 5x-3=3x+5 are there on the:
(i) Number line
(ii) Cartesian plane.
Answer: First, let us simplify the given equation: \( 5x - 3 = 3x + 5 \)
\( \implies 5x - 3x = 5 + 3 \)
\( \implies 2x = 8 \)
\( \implies x = 4 \) (i) On the number line: The equation \( x = 4 \) represents a single, unique point. Thus, there is **one solution**. (ii) On the Cartesian plane: The equation \( x = 4 \) can be written as a linear equation in two variables: \( 1x + 0y = 4 \). This represents a straight line parallel to the y-axis. Since any point on this line satisfies the equation, there are **infinitely many solutions** of the form \( (4, y) \), where \( y \) is any real number.
In simple words: On a single number line, \( x = 4 \) is just one point. On a flat grid, it represents a vertical line, which contains infinitely many points.

Exam Tip: Understand the difference in representation: an equation in one variable corresponds to a single point on a 1D line, but on a 2D coordinate plane, it represents an entire line with infinitely many coordinate points.

Question. For what value of k the point (k,5) lies on the line 4x-5y=10 ?
Answer: If the point \( (k, 5) \) lies on the line \( 4x - 5y = 10 \), its coordinates must satisfy the equation of the line. Substituting \( x = k \) and \( y = 5 \) into the equation: \( 4(k) - 5(5) = 10 \)
\( \implies 4k - 25 = 10 \)
\( \implies 4k = 10 + 25 \)
\( \implies 4k = 35 \)
\( \implies k = \frac{35}{4} \text{ (or } 8.75)\)
In simple words: Put \( k \) for \( x \) and \( 5 \) for \( y \) into the equation, then solve to get \( k = 8.75 \).

Exam Tip: When a point lies on a line, always plug its coordinates into the equation to find any unknown constants.

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 of the equation \( 5x + 2ay = 3a \), we substitute these values into the equation: \( 5(1) + 2a(1) = 3a \)
\( \implies 5 + 2a = 3a \) Subtracting \( 2a \) from both sides: \( 3a - 2a = 5 \)
\( \implies a = 5 \) Therefore, the value of \( a \) is \( 5 \).
In simple words: Substitute \( x = 1 \) and \( y = 1 \) into the equation to get \( 5 + 2a = 3a \), which simplifies to \( a = 5 \).

Exam Tip: Collect all terms involving the variable you are solving for (in this case, \( a \)) on one side of the equation to simplify.

Question. Solve the equation \( \frac{p^2 - 16}{p + 4} = 5 \) (\( p \neq -4 \)).
Answer: Given equation: \( \frac{p^2 - 16}{p + 4} = 5 \) We can factor the numerator \( p^2 - 16 \) using the identity \( a^2 - b^2 = (a - b)(a + b) \): \( \frac{(p - 4)(p + 4)}{p + 4} = 5 \) Since we are given that \( p \neq -4 \), the term \( p + 4 \) is non-zero, and we can cancel it from the numerator and denominator: \( p - 4 = 5 \)
\( \implies p = 5 + 4 \)
\( \implies p = 9 \)
In simple words: Factor the top part into \( (p - 4)(p + 4) \), cancel out the \( p + 4 \) terms, and solve \( p - 4 = 5 \) to get \( p = 9 \).

Exam Tip: Always check the given conditions, like \( p \neq -4 \), which ensure that you are not dividing by zero when canceling terms.

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 us define the variables as follows: - Total distance covered = \( x \) km - Total taxi fare = Rs. \( y \) According to the given fare structure: - Fare for the first kilometer = Rs. 8 - Remaining distance after the first kilometer = \( (x - 1) \) km - Fare for this remaining distance = Rs. \( 5(x - 1) \) The total fare \( y \) is the sum of these two fares: \( y = 8 + 5(x - 1) \)
\( \implies y = 8 + 5x - 5 \)
\( \implies y = 5x + 3 \) This can also be written in the standard form: \( 5x - y + 3 = 0 \)
In simple words: The fare is Rs. 8 for the first km and Rs. 5 for each extra km. So the equation is \( y = 5x + 3 \).

Exam Tip: Break down the total distance into two parts (the first kilometer and the remaining kilometers) to construct the correct equation.

Question. Find four different solutions of the equation x+2y=6.
Answer: To find four different solutions for the equation \( x + 2y = 6 \), we can substitute different values for either \( x \) or \( y \) and solve for the other variable:
(i) Setting \( x = 0 \): \( 0 + 2y = 6 \)
\( \implies 2y = 6 \)
\( \implies y = 3 \) So, \( (0, 3) \) is the first solution.
(ii) Setting \( y = 0 \): \( x + 2(0) = 6 \)
\( \implies x = 6 \) So, \( (6, 0) \) is the second solution.
(iii) Setting \( x = 2 \): \( 2 + 2y = 6 \)
\( \implies 2y = 6 - 2 \)
\( \implies 2y = 4 \)
\( \implies y = 2 \) So, \( (2, 2) \) is the third solution.
(iv) Setting \( x = 4 \): \( 4 + 2y = 6 \)
\( \implies 2y = 6 - 4 \)
\( \implies 2y = 2 \)
\( \implies y = 1 \) So, \( (4, 1) \) is the fourth solution. The four solutions are \( (0, 3) \), \( (6, 0) \), \( (2, 2) \), and \( (4, 1) \).
In simple words: Choose different values for one variable and calculate the other. For example, if you set \( x = 0, 6, 2, 4 \), you get the pairs \( (0,3), (6,0), (2,2), (4,1) \).

Exam Tip: To make calculation easier, substitute values for the variable that has a coefficient of 2 (in this case, \( y \)) or choose even numbers for \( x \) to avoid fractions.

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: We rewrite each equation in the general form \( ax + by + c = 0 \) and determine the values of \( a \), \( b \), and \( c \):
(i) \( 2x + 3y = 4.37 \) Rearranging to the standard form: \( 2x + 3y - 4.37 = 0 \) Comparing with \( ax + by + c = 0 \): \( a = 2 \) \( b = 3 \) \( c = -4.37 \)
(ii) \( x - 4 = 3y \) Rearranging to the standard form: \( 1x - 3y - 4 = 0 \) Comparing with \( ax + by + c = 0 \): \( a = 1 \) \( b = -3 \) \( c = -4 \)
(iii) \( 4 = 5x - 3y \) Rearranging to the standard form: \( 5x - 3y - 4 = 0 \) Comparing with \( ax + by + c = 0 \): \( a = 5 \) \( b = -3 \) \( c = -4 \)
In simple words: Shift all terms to the left side so the right side is 0, then identify the coefficient of \( x \) as \( a \), the coefficient of \( y \) as \( b \), and the constant as \( c \).

Exam Tip: Be careful with the signs of \( a \), \( b \), and \( c \). If a term is subtracted, its coefficient is negative.

Question. Solve the equation 2x+1=x-3, and represent the solution(s) on
(i) the number line.
(ii) the Cartesian plane.
Answer: First, solve the linear equation: \( 2x + 1 = x - 3 \) Subtracting \( x \) from both sides and \( 1 \) from both sides: \( 2x - x = -3 - 1 \)
\( \implies x = -4 \) (i) On the number line: The solution is a unique point at \( x = -4 \). (ii) On the Cartesian plane: The equation is expressed in two variables as \( 1x + 0y = -4 \). This represents a vertical line parallel to the y-axis, situated 4 units to the left of the y-axis. In simple words: First solve the equation to get \( x = -4 \). On a number line, it is just a point. On a Cartesian plane, it is a vertical line passing through \( -4 \).

Exam Tip: To plot a 1-variable equation on a Cartesian plane, think of it as having a coefficient of 0 for the other variable, which gives a line parallel to one of the axes.

Question. Evaluate: (5x+1)(x+3)-8=5(x+1)(x+2).
Answer: Let us expand both sides of the given equation: LHS: \( (5x + 1)(x + 3) - 8 \) \( = 5x^2 + 15x + x + 3 - 8 \) \( = 5x^2 + 16x - 5 \) RHS: \( 5(x + 1)(x + 2) \) \( = 5(x^2 + 2x + x + 2) \) \( = 5(x^2 + 3x + 2) \) \( = 5x^2 + 15x + 10 \) Now, equate LHS and RHS: \( 5x^2 + 16x - 5 = 5x^2 + 15x + 10 \) Since \( 5x^2 \) is present on both sides, it cancels out: \( 16x - 5 = 15x + 10 \)
\( \implies 16x - 15x = 10 + 5 \)
\( \implies x = 15 \)
In simple words: Multiply the terms on both sides of the equals sign. The \( 5x^2 \) parts cancel each other out, leaving \( 16x - 5 = 15x + 10 \), which simplifies to \( x = 15 \).

Exam Tip: Expand brackets carefully step-by-step and group like terms before trying to isolate the variable.

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: Since the point \( (-1, -5) \) lies on the graphs of both equations, its coordinates \( x = -1 \) and \( y = -5 \) must satisfy each equation. 1. Substitute \( x = -1 \) and \( y = -5 \) into \( 3x = ay + 7 \): \( 3(-1) = a(-5) + 7 \)
\( \implies -3 = -5a + 7 \) Rearranging the equation: \( 5a = 7 + 3 \)
\( \implies 5a = 10 \)
\( \implies a = 2 \) 2. Substitute \( x = -1 \) and \( y = -5 \) into \( y = bx + 7 \): \( -5 = b(-1) + 7 \)
\( \implies -5 = -b + 7 \) Rearranging the equation: \( b = 7 + 5 \)
\( \implies b = 12 \) Thus, the values are \( a = 2 \) and \( b = 12 \).
In simple words: Plug in \( x = -1 \) and \( y = -5 \) into both equations. Solving the first equation gives \( a = 2 \), and solving the second gives \( b = 12 \).

Exam Tip: Treat each equation as an independent substitution problem to find the two unknown constants one by one.

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: A line parallel to the y-axis at a distance of 4 units to the right of the y-axis has the equation: \( x = 4 \) To find the point of intersection, substitute \( x = 4 \) into the given linear equation \( 2x + 3y = 9 \): \( 2(4) + 3y = 9 \)
\( \implies 8 + 3y = 9 \)
\( \implies 3y = 9 - 8 \)
\( \implies 3y = 1 \)
\( \implies y = \frac{1}{3} \) Therefore, the point where the two lines meet is \( \left(4, \frac{1}{3}\right) \).
In simple words: The vertical line 4 units to the right of the y-axis is \( x = 4 \). Putting \( x = 4 \) into the equation \( 2x + 3y = 9 \) gives \( y = \frac{1}{3} \).

Exam Tip: First translate the description of the second line into its mathematical equation (\( x = 4 \)), then solve the system of equations.

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, since the point \( (4, 3) \) lies on the graph of the equation \( 3x - ay = 6 \), we substitute \( x = 4 \) and \( y = 3 \) to find \( a \): \( 3(4) - a(3) = 6 \)
\( \implies 12 - 3a = 6 \)
\( \implies 3a = 12 - 6 \)
\( \implies 3a = 6 \)
\( \implies a = 2 \) So, the equation of the line is: \( 3x - 2y = 6 \) Now, to check if the point \( (-2, -6) \) lies on this line, substitute \( x = -2 \) and \( y = -6 \) into the Left Hand Side (LHS) of the equation: LHS \( = 3(-2) - 2(-6) \) \( = -6 + 12 \) \( = 6 \) Since LHS is equal to the Right Hand Side (RHS), the coordinates satisfy the equation. Thus, the point \( (-2, -6) \) also lies on the same graph.
In simple words: Substitute \( (4,3) \) into the equation to find that \( a = 2 \), making the equation \( 3x - 2y = 6 \). Plugging \( (-2,-6) \) into this equation works, so this point is also on the line.

Exam Tip: This is a two-step problem: first solve for the unknown constant, and then use that completed equation to test the second point.

Question. Give the equations of two lines passing through (-2,-4). How many more such lines are there, and why?
Answer: We need to find equations of lines that are satisfied by \( x = -2 \) and \( y = -4 \). Two such equations are: 1. \( x - y = 2 \) (Checking: \( -2 - (-4) = -2 + 4 = 2 \), which is true) 2. \( y = 2x \) (Checking: \( -4 = 2(-2) \), which is true) There are **infinitely many** more such lines. This is because \( (-2, -4) \) represents a single point on the Cartesian plane, and through any single point, an infinite number of unique straight lines can be drawn in different directions.
In simple words: Two equations that work for \( (-2, -4) \) are \( x - y = 2 \) and \( y = 2x \). There are infinitely many more such lines because you can draw endless lines through a single point.

Exam Tip: To create equations passing through a point \( (p, q) \), just pick any simple relationship like \( x + y \) or \( x - y \) and calculate the constant on the RHS using the values of \( p \) and \( q \).

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: To draw the graph, we can find a few points on the line by choosing values for Celsius (\( C \)) on the x-axis and calculating Fahrenheit (\( F \)) on the y-axis: 1. Let \( C = 0 \): \( F = \frac{9}{5}(0) + 32 = 32 \) So, the point is \( (0, 32) \). 2. Let \( C = 10 \): \( F = \frac{9}{5}(10) + 32 = 18 + 32 = 50 \) So, the point is \( (10, 50) \). 3. Let \( C = -40 \): \( F = \frac{9}{5}(-40) + 32 = -72 + 32 = -40 \) So, the point is \( (-40, -40) \). We can plot these points on the graph and draw a line passing through them. In simple words: Find three temperature points like \( (0,32) \), \( (10,50) \), and \( (-40,-40) \) where Celsius is on the horizontal axis and Fahrenheit is on the vertical axis, and connect them with a straight line.

Exam Tip: Choosing multiples of 5 for Celsius makes it easier to calculate because it cancels out the denominator of the fraction \( \frac{9}{5} \).

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: To write a single-variable equation as an equation in two variables \( x \) and \( y \), we express the missing variable with a coefficient of \( 0 \):
(i) \( 1x + 0y + 5 = 0 \)
(ii) \( 0x + 1y - 2 = 0 \)
(iii) \( 2x + 0y - 3 = 0 \)
(iv) \( 0x + 5y - 2 = 0 \)
In simple words: If a variable is missing in an equation, we can write it with a coefficient of zero, like writing \( 0y \) when \( y \) is missing.

Exam Tip: Always write the missing variable with a \( 0 \) coefficient (e.g., \( 0y \)) to show that it represents a 2D line rather than a 1D coordinate point.

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 \( x \) represent the total distance covered (in km) and \( y \) represent the total fare (in Rs.). The fare is structured as follows: - Fare for the first kilometer = Rs. 20 - Remaining distance after the first kilometer = \( (x - 1) \) km - Fare for the subsequent distance = Rs. \( 6(x - 1) \) The linear equation representing the total fare is: \( y = 20 + 6(x - 1) \)
\( \implies y = 20 + 6x - 6 \)
\( \implies y = 6x + 14 \) To draw the graph, we find at least three points that satisfy this equation: 1. For \( x = 1 \): \( y = 6(1) + 14 = 20 \). Point: \( (1, 20) \) 2. For \( x = 2 \): \( y = 6(2) + 14 = 26 \). Point: \( (2, 26) \) 3. For \( x = 4 \): \( y = 6(4) + 14 = 38 \). Point: \( (4, 38) \) We plot these coordinate points on the Cartesian plane and draw a line passing through them. In simple words: The fare is Rs. 20 for the first kilometer and Rs. 6 for each extra kilometer, giving the equation \( y = 6x + 14 \). Find and plot points like \( (1,20) \) and \( (2,26) \) on a graph and draw a line through them.

Exam Tip: When drawing real-life graphs, note that distance and fare cannot be negative, so the graph starts from \( x \ge 0 \) in the first quadrant.

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 total count of boxes be represented by \( x \), and the complete cost in rupees be represented by \( y \). Since one box is priced at Rs. 25, the cost of \( x \) boxes will be \( 25 \) times \( x \). This gives us the linear relation:
\( y = 25x \)

To plot the graph of this equation, we can find a few coordinates:
- 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: \( (2, 50) \)
- When \( x = 3 \), \( y = 25(3) = 75 \). Point: \( (3, 75) \)

We can represent this relationship visually on the grid below:

In simple words: Each box costs Rs. 25, so the total bill is 25 times the number of boxes. This creates a straight upward-sloping line starting from zero.
Exam Tip: Remember to clearly label both axes with the quantities they represent and mark the scale coordinates to obtain full 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 mass of hydrogen in water be represented by \( y \) grams, and the mass of oxygen be represented by \( x \) grams. Since the ratio of hydrogen to oxygen is \( 2:1 \), we can set up the proportion:
\( \frac{y}{x} = \frac{2}{1} \)

Cross-multiplying, we find the linear equation:
\( y = 2x \)

Let us determine a couple of points to plot this relationship on a coordinate system:
- If \( x = 0 \), then \( y = 2(0) = 0 \). Point: \( (0, 0) \)
- If \( x = 3 \), then \( y = 2(3) = 6 \). Point: \( (3, 6) \)
- If \( x = 6 \), then \( y = 2(6) = 12 \). Point: \( (6, 12) \)

Now we can draw this line on the graph:

From the graph, finding \( 6 \) grams on the horizontal oxygen axis and moving vertically up to the line reveals that the matching hydrogen value is \( 12 \) grams.
\( \implies \text{Hydrogen} = 12 \text{ grams} \)
In simple words: Since water always has twice as much hydrogen as oxygen, if you have 6 grams of oxygen, you must have 12 grams of hydrogen.
Exam Tip: Draw dashed projection lines from the axes to the point on your graph to show the examiner exactly how you read the value.

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/sec2 , (ii) 6 m/sec2.
Answer: Let the pulling force be \( y \) (in Newtons) and the acceleration produced in the cart be \( x \) (in \( \text{m/s}^2 \)). We know that force is directly proportional to acceleration, where the mass serves as the constant of proportionality:
\( y = mx \)

Substituting the given constant mass value \( m = 6 \text{ kg} \), the linear equation is:
\( y = 6x \)

Let's prepare coordinates to plot this linear relation:
- If \( x = 0 \), then \( y = 0 \). Point: \( (0, 0) \)
- If \( x = 5 \), then \( y = 6(5) = 30 \). Point: \( (5, 30) \)
- If \( x = 6 \), then \( y = 6(6) = 36 \). Point: \( (6, 36) \)

Below is the graphical representation of this equation:

By reading values from our graph:
(i) For an acceleration of \( 5 \text{ m/sec}^2 \), the corresponding force is \( 30 \text{ N} \).
(ii) For an acceleration of \( 6 \text{ m/sec}^2 \), the corresponding force is \( 36 \text{ N} \).
In simple words: Force equals mass times acceleration. Since the mass is 6, the force is just 6 multiplied by the acceleration value.
Exam Tip: Be sure to write down the units (Newtons for force, m/sec² for acceleration) in your final numerical answers to secure full marks.

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 \( x \) represent the total parking duration in hours, and let \( y \) represent the total accumulated parking charges in rupees.
For the first hour, the fee is Rs. 50.
For the remaining duration of \( (x - 1) \) hours, the fee is Rs. 10 per hour.
Consequently, we construct the total cost equation as:
\( y = 50 + 10(x - 1) \)
\( \implies y = 50 + 10x - 10 \)
\( \implies y = 10x + 40 \) (valid for \( x \ge 1 \))

Let's find key coordinates on this line:
- If \( x = 1 \), \( y = 10(1) + 40 = 50 \). Point: \( (1, 50) \)
- If \( x = 2 \), \( y = 10(2) + 40 = 60 \). Point: \( (2, 60) \)
- If \( x = 8 \), \( y = 10(8) + 40 = 120 \). Point: \( (8, 120) \)

Here is the graphical display of the parking rates:

By reading values from our graph:
(i) The cost for 2 hours of parking is Rs. 60.
(ii) The cost for 8 hours of parking is Rs. 120.
In simple words: The first hour costs Rs. 50, and each hour after that adds Rs. 10 to the bill.
Exam Tip: Since parking duration cannot realistically be less than zero, restrict your drawn graph line to positive values in the first quadrant only.

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 made by player A be \( x \) and the runs made by player B be \( y \). Because the sum of their individual runs equals the overall score of 40, we construct the equation:
\( x + y = 40 \)

Let's find coordinates to plot this linear relation:
- If \( x = 0 \), then \( y = 40 \). Point: \( (0, 40) \)
- If \( x = 40 \), then \( y = 0 \). Point: \( (40, 0) \)
- If \( x = 30 \), then \( y = 10 \). Point: \( (30, 10) \)

Here is the graph representing the partnership:

By referencing the graph, if player B's score is \( 10 \), moving horizontally from \( 10 \) to the line and looking down shows player A's score is \( 30 \).
\( \implies \text{Player A's score} = 30 \text{ runs} \)
In simple words: The two scores must add up to 40. So if player B made 10 runs, player A must have made 30 runs.
Exam Tip: Since cricket runs are whole numbers, you can mark the calculated integer coordinate points clearly on the graph line.

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 simplify the equation step-by-step. First, find a common denominator for the terms on the left-hand side. The lowest common multiple of 8, 24, and 3 is 24:
\( \frac{3(3x + 6) - (11x - 8) + 8(x)}{24} = \frac{3x}{4} - \frac{x+7}{24} \)

Now, simplify the left numerator:
\( \frac{9x + 18 - 11x + 8 + 8x}{24} = \frac{3x}{4} - \frac{x+7}{24} \)
\( \frac{6x + 26}{24} = \frac{3x}{4} - \frac{x+7}{24} \)

Next, find a common denominator for the right-hand side terms (LCM of 4 and 24 is 24):
\( \frac{6x + 26}{24} = \frac{6(3x) - (x + 7)}{24} \)
\( \frac{6x + 26}{24} = \frac{18x - x - 7}{24} \)
\( \frac{6x + 26}{24} = \frac{17x - 7}{24} \)

Since both sides have the same denominator, we can equate their numerators directly:
\( 6x + 26 = 17x - 7 \)
\( 26 + 7 = 17x - 6x \)
\( 33 = 11x \)
\( x = 3 \)

Thus, the value of \( x \) is 3.
In simple words: First we make all the denominators 24, combine the fractions, and then solve the basic equation to find that x is 3.

Exam Tip: Be careful with the negative sign in front of fractions like \( -\frac{11x-8}{24} \) - it applies to both terms inside, changing \( -8 \) to \( +8 \).

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 us represent the unknown number as \( x \).
First, "one-fourth of the sum of the number and seven" is written as:
\( \frac{1}{4}(x + 7) \)

Second, "four less than three times the number" is written as:
\( 3x - 4 \)

Setting these two expressions equal as described in the problem yields:
\( \frac{1}{4}(x + 7) = 3x - 4 \)

Multiply both sides of the equation by 4 to clear the fraction:
\( x + 7 = 4(3x - 4) \)
\( x + 7 = 12x - 16 \)

Now, rearrange terms to solve for \( x \):
\( 7 + 16 = 12x - x \)
\( 23 = 11x \)
\( x = \frac{23}{11} \)

Thus, the required number is \( \frac{23}{11} \).
In simple words: We turn the word puzzle into a math equation. Solving the equation gives us the final fractional answer of 23 divided by 11.

Exam Tip: Always double-check your algebraic formulation by plugging the final value back into the original word statement to see if it makes sense.

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 solve the given equation first:
\( 2x + 1 = x - 3 \)
\( 2x - x = -3 - 1 \)
\( x = -4 \)

(i) **Representation on a number line:**
In one variable, the solution is represented as a single point at \( -4 \) on the number line.

(ii) **Representation on the Cartesian plane:**
In two variables, the equation can be expressed as \( 1x + 0y = -4 \). This is represented as a vertical line parallel to the y-axis, crossing through the point \( (-4, 0) \).

In simple words: In one dimension, x = -4 is just a point. On a 2D plane, x = -4 is a vertical line that stays 4 steps to the left of the center.
Exam Tip: Remember that equations containing only x are vertical lines, whereas equations containing only y represent horizontal lines on the Cartesian grid.

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 solve each of the three linear equations to find two sets of values:

**(i) For the equation \( 4x + 3y = 12 \):**
- Let \( x = 0 \):
\( 4(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \)
So, \( (0, 4) \) is the first solution.
- Let \( y = 0 \):
\( 4x + 3(0) = 12 \implies 4x = 12 \implies x = 3 \)
So, \( (3, 0) \) is the second solution.

**(ii) For the equation \( 2x + 5y = 0 \):**
- Let \( x = 0 \):
\( 2(0) + 5y = 0 \implies 5y = 0 \implies y = 0 \)
So, \( (0, 0) \) is the first solution.
- Let \( x = 5 \):
\( 2(5) + 5y = 0 \implies 10 + 5y = 0 \implies 5y = -10 \implies y = -2 \)
So, \( (5, -2) \) is the second solution.

**(iii) For the equation \( 3y + 4 = 0 \):**
This equation does not contain \( x \), meaning \( y \) is fixed while \( x \) can be any real number:
\( 3y = -4 \implies y = -\frac{4}{3} \)
- Choosing \( x = 0 \) gives the coordinate: \( (0, -\frac{4}{3}) \)
- Choosing \( x = 1 \) gives the coordinate: \( (1, -\frac{4}{3}) \)
In simple words: To find coordinates, pick any value for x and solve for y, or pick a value for y and solve for x.

Exam Tip: Substituting 0 for one variable is the fastest and easiest way to calculate a valid coordinate pair.