CBSE Class 9 Mathematics Linear Equations In Two Variables Worksheet Set 04

1. Linear Equations
2. Solution of a Linear Equation
3. Graph of a Linear Equation in Two Variables
4. Equations of Lines Parallel to x-axis and y-axis
• An equation of the form ax + by + c = 0 where a, b and c are real numbers such that a and b are not both zero is called a linear equation in two variables.
•  A pair of values of x and y which satisfy the equation ax + by + c = 0 is called a solution of the equation.
•  Graph: The graph of every linear equation in two variables is a straight line. Every point on the graph of a linear equation in two variables is two variables is a solution of the linear equation. Conversely, every solution of the linear equation is a point on the graph of the linear equation.
•  A linear equation in two variables has infinitely many solutions.
•  The graph of every linear equation in two variables is a straight line.
•  y = 0 is the equation of x-axis and x = 0 is equation of y-axis.
•  The graph of x = a is a straight line parallel to the y-axis.
•  The graph of y = a is a straight line parallel to the x-axis.
•  An equation of the type y = mx represent a line passing through the origin.

Question. If (2, 0) is a solution of the linear equation 2x+ 3y= k, then the value of k is
(a) 4
(b) 6
(c) 5
(d) 2

Answer: A

Question. The equation 2x+ 5y = 7 has a unique solution, if x, y are:
(a) Natural numbers
(b) Positive real numbers
(c) Real numbers
(d) Rational numbers

Answer: A

Question. The equation x= 7, in two variables, can be written as
(a) 1.x + 1.y = 7
(b) 1.x + 0.y = 7
(c) 0.x + 1.y = 7
(d) 0.x + 0.y = 7

Answer: B

Question. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form:
(a) (-9/2,m)
(b) (n,-9/2)
(c) (0,-9/2)
(d) (-9,0)

Answer: A

Question. The graph of the linear equation 2x + 3y= 6 cuts the y-axis at the point
(a) (2, 0)
(b) (0, 3)
(c) (3, 0)
(d) (0, 2)

Answer: D

Question. Any point on the x-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)

Answer: C

Question. The graph of the linear equation 2x+ 3 y= 6 is a line which meets the x-axis at the point
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3)

Answer: C

Question. Any point on the line y = x is of the form
(a) (a, a)
(b) (0, a)
(c) (a, 0)
(d) (a, – a)

Answer: A

Question. The linear equation 2x– 5 y= 7 has
(a) A unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution

Answer: C

Question. The equation of x-axis is of the form
(a) x = 0
(b) y = 0
(c) x + y = 0
(d) x = y

Answer: B

Question. If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form
(a) y – x = 0
(b) x + y = 0
(c) –2x + y = 0
(d) –x + 2y = 0

Answer: B

Question. The graph of y = 6 is a line
(a) parallel to x -axis at a distance 6 units from the origin
(b) parallel to y -axis at a distance 6 units from the origin
(c) making an intercept 6 on the x-axis.
(d) making an intercept 6 on both the axes.

Answer: A

Question. The positive solutions of the equation ax + by + c = 0 always lie in the
(a) 1st quadrant
(b) 2nd quadrant
(c) 3rd quadrant
(d) 4th quadrant

Answer: A

Question. x = 5, y= 2 is a solution of the linear equation
(a) x + 2y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5x + y = 7

Answer: C

Question. The point of the form (a, a) always lies on:
(a) x-axis
(b) y-axis
(c) On the line y= x
(d) On the line x+ y = 0

Answer: C

Question. How many linear equations in x and y can be satisfied by x = 1 and y= 2?
(a) Only one
(b) Two
(c) Infinitely many
(d) Three

Answer: C

Question. The point of the form (a, –a) always lies on the line
(a) x = a
(b) y = –a
(c) y = x
(d) x + y = 0

Answer: D

Question. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation:
(a) Changes
(b) Remains the same
(c) Changes in case of multiplication only
(d) Changes in case of division only

Answer: B

Question. Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.
Answer. As every point on the graph of linear equation in two variables represent a solution of the equation, so the given statement is false.

Question. The graph of the linear equation x+ 2y= 7 passes through the point (0, 7).
Answer. Substituting x = 0 and y = 7 in the given equation x + y 2y = 7, we get 0 + 2(7) = 7 ⇒ 14 = 7, which is false.
The point (0, 7) does not satisfy the equation.
Hence, the given statement is false.

Question. Find the solution of the linear equation x+ 2y = 8 which represents a point on
(i) x-axis (ii) y-axis
Answer. We know that the point which lies on x-axis has its ordinate 0.
Putting y = 0 in the equation x + 2y = 8, we get x + 2(0) = 8 ⇒ x = 8
A point which lies on y-axis has its abscissa 0.
Putting x = 0 in the equation x + 2y = 8, we get 0 + 2y = 8 ⇒ y = 4

Question. Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.
Answer. A linear equation such that each point on it graph has an ordinate 3 times its abscissa is y = 3x. 6. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.
Answer. The point (3, 4) lies on the graph of 3y = ax + 7.
Substituting x = 3 and y = 4 in the given equation 3y = ax + 7,we get
∴ 3× 4 = a × 3 + 7
⇒ 12 = 3a + 7 ⇒ 3a = 5 ⇒ a = 5/3

Question. How many solution(s) of the equation 2x+ 1 = x– 3 are there on the:
(i) Number line
(ii) Cartesian plane?
Answer. (i) The number of solution(s) of the equation 2x + 1 = x – 3 which are on the number line is one.
2x +1 = x − 3 ⇒ 2x − x = −3 − 1 ⇒ x = −4
∴ x = −4 is the solution of the given equation.
(ii) The number of solution(s) of the equation 2x + 1 = x – 3 which are on the Cartesian plane are infinitely many solutions.

Question. For what value of c, the linear equation 2x+ cy = 8 has equal values of x and y for its solution.
Answer. The value of c for which the linear equation 2x + cy = 8 has equal values of x and y i.e., x = y for its solution is
2x + cy = 8 ⇒ 2x + cx = 8 [∵ y = x]
⇒ cx = 8 − 2x
∵ c = 8 - 2x/x, x ≠ 0

Question. Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?
Answer. y varies of directly as x.
⇒ y ∝ x,
∴ y = kx
Substituting y = 12 when x = 4, we get
12 = k × 4 ⇒ k = 12 ÷ 4 = 3
Hence, the required equation is y = 3x.
The value of y when x = 5 is y = 3 × 5 =15.

Question. Show that the points A (1, 2), B (– 1, – 16) and C (0, – 7) lie on the graph of the linear equation y= 9 x – 7.
Answer. For A (1, 2), we have 2 = 9 (1) – 7 = 9 – 7 = 2
For B (–1, –16), we have –16 = 9(–1) – 7 = – 9 – 7 = – 16
For C (0, –7), we have – 7 = 9 (0) – 7 = 0 – 7 = – 7
We see that the line y = 9x – 7 is satisfied by the points A (1, 2), B (–1, –16) and C (0, –7).
Therefore, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7 and therefore, lie on the graph of the linear equation y = 9x – 7.

Question. The linear equation that converts Fahrenheit (F) to Celsius (^oC) is given by the relation: C = 5F - 160 / 9
(i) If the temperature is 86°F, what is the temperature in Celsius?
(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?
(iii) 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?
(iv) What is the numerical value of the temperature which is same in both the scales?
Answer. C = 5F - 160 / 9
(i) Putting F = 86^o, we get C = 5 (86) - 160 / 9 = 430 - 160 / 9 = 270 / 9 = 30^o
Hence, the temperature in Celsius is 30^o C.
(ii) Putting C = 35^o, we get 35^o 5 (F) - 160 / 9 ⇒315^o = 5F - 160
⇒ 5F = 315 +160 = 475
∴ F = 475 / 5 = 95^o
Hence, the temperature in Fahrenheit is 95 F.
(iii) Putting C = 0^o, we get
0 = 5F - 160 / 9 ⇒ 0 = 5 F - 160
⇒ 5F = 160
∴ F = 160 / 5 = 32^o
Now, putting F = 0^o, we get
C = 5F - 160 / 9 ⇒ C = 5(0) - 160 / 9 = - {160/9}^o
If the temperature is 0^o C, the temperature in Fahrenheit is 32^o and if the temperature is ⁰F, then the temperature in Celsius is - {160/9}^oC
(iv) Putting C = F, in the given relation, we get
F = 5F - 160 / 9 ⇒ 9F = 5F - 160
⇒ 4F = - 160
∴ F = - 160 / 4 = - 40^o
Hence, the numerical value of the temperature which is same in both the scales is – 40.
The linear equation that converts Kelvin (x) to Fahrenheit (y) is given by the relation:
y = 9/5 (x - 273) + 32

Question. If the temperature of a liquid can be measured in Kelvin units as x° K or in Fahrenheit units as y° F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5 (x - 273) + 32
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313°K.
(ii) If the temperature is 158° F, then find the temperature in Kelvin.
Answer. y = 9/5 (x - 273) + 32
(i) When the temperature of the liquid is x = 313^o K y = 9/5 (313 - 273) + 32 = 9/5 X 40 + 32 = 72^o + 32^o = 104^oF
(ii) When the temperature of the liquid is y = 158^o F 
158 = 9/5 (x - 273) + 32 ⇒ 9/5 (x - 273) = 158 - 32
⇒ x − 273 = 126 × 5/6 = 70
⇒ x − 273 = 70 = 273 + 70 = 343^o K

Worksheet - I: Linear Equation in Two Variables

Question 1. Draw the graph of equation \( 3x+y=6 \). Also find the points when the line intersect x-axis and y-axis.
Answer: Given equation:
\( 3x + y = 6 \)
Expressing \( y \) in terms of \( x \):
\( y = 6 - 3x \)
Let us determine three points that satisfy this linear relation:
1. When \( x = 0 \):
\( y = 6 - 3(0) = 6 \implies (0, 6) \)
2. When \( x = 1 \):
\( y = 6 - 3(1) = 3 \implies (1, 3) \)
3. When \( x = 2 \):
\( y = 6 - 3(2) = 0 \implies (2, 0) \)
Table of solutions:
\( (x, y) \in \{ (0, 6), (1, 3), (2, 0) \} \)
To find the intersection points with the coordinate axes:
- The line intersects the x-axis where \( y = 0 \): Setting \( y = 0 \) gives \( 3x = 6 \implies x = 2 \). Hence, the point of intersection with the x-axis is \( (2, 0) \).
- The line intersects the y-axis where \( x = 0 \): Setting \( x = 0 \) gives \( y = 6 \). Hence, the point of intersection with the y-axis is \( (0, 6) \).
In simple words: Substitute x = 0 into the equation to find where it touches the vertical axis at (0, 6), and set y = 0 to find where it crosses the horizontal axis at (2, 0).

Exam Tip: Whenever a problem asks for axis intersections, state both coordinates in the explicit ordered pair format (x, y) rather than writing only single numbers.

Question 2. Two years later a father will be eight years more than three times the age of the son. Taking the present age of father and son as x and y resp:
a, Write a linear equ: for the above and draw its graph.
b, From the graph find the age of father when son's age is 10 yrs.
Answer: Let the present age of the father be \( x \) years and the present age of the son be \( y \) years.
Two years from now:
Father's age \( = x + 2 \)
Son's age \( = y + 2 \)
According to the given condition:
\( x + 2 = 3(y + 2) + 8 \)
\( \implies x + 2 = 3y + 6 + 8 \)
\( \implies x + 2 = 3y + 14 \)
\( \implies x - 3y = 12 \)
(a) The required linear equation is \( x - 3y = 12 \), which can be written as \( x = 3y + 12 \).
To plot the graph, calculate points by selecting values for \( y \) (son's age):
- For \( y = 0 \implies x = 3(0) + 12 = 12 \implies (12, 0) \)
- For \( y = 4 \implies x = 3(4) + 12 = 24 \implies (24, 4) \)
- For \( y = 10 \implies x = 3(10) + 12 = 42 \implies (42, 10) \)
(b) From the graph, locate \( y = 10 \) on the vertical axis (son's age). Move horizontally to intersect the line and then vertically down to the horizontal axis to read the father's age \( x \).
The corresponding value is \( x = 42 \).
Thus, the father's age is 42 years when the son is 10 years old.
In simple words: Add 2 years to both ages, set up the relationship to get x - 3y = 12, and then look across from 10 on the son's axis to find the father is 42 years old.

Exam Tip: Remember to add 2 years to both persons' ages before writing the equation; adding the time jump to only one side is a frequent student error.

Question 3. Represent the solutions of equation \( 2x+1 = x-3 \)
a) on number line
b) in cartesian plane
Answer: First, simplify the given linear equation:
\( 2x + 1 = x - 3 \)
\( \implies 2x - x = -3 - 1 \)
\( \implies x = -4 \)
(a) Representation on the number line:
In one variable, the equation \( x = -4 \) is represented as a single unique point at coordinate \( -4 \) on the real number line. (b) Representation in the Cartesian plane:
In two variables, the equation can be written as \( 1 \cdot x + 0 \cdot y = -4 \). This represents a straight line parallel to the y-axis passing through the point \( (-4, 0) \), where every point on the line has an x-coordinate equal to \( -4 \).
In simple words: On a regular number line, x = -4 is just a single point. In a two-dimensional grid, it becomes a complete vertical line running parallel to the vertical axis.

Exam Tip: For the Cartesian plane part, write the equation as \( x + 0y = -4 \) to explicitly explain why the graph is a line rather than a single point.

Question 4. Solve the L.E for 'x'
\( \frac{2x - 3}{5} + \frac{x + 3}{4} = \frac{2x + 3}{4} \)
Answer: Given linear equation:
\( \frac{2x - 3}{5} + \frac{x + 3}{4} = \frac{2x + 3}{4} \)
Collect the terms with the common denominator 4 on the right-hand side:
\( \implies \frac{2x - 3}{5} = \frac{2x + 3}{4} - \frac{x + 3}{4} \)
Combine the fractions with the same denominator:
\( \implies \frac{2x - 3}{5} = \frac{(2x + 3) - (x + 3)}{4} \)
\( \implies \frac{2x - 3}{5} = \frac{2x + 3 - x - 3}{4} \)
\( \implies \frac{2x - 3}{5} = \frac{x}{4} \)
Cross-multiply to eliminate denominators:
\( \implies 4(2x - 3) = 5x \)
\( \implies 8x - 12 = 5x \)
Subtract \( 5x \) from both sides:
\( \implies 8x - 5x = 12 \)
\( \implies 3x = 12 \)
\( \implies x = \frac{12}{3} = 4 \)
Hence, the solution is \( x = 4 \).
In simple words: Group the fractions with denominator 4 together so they simplify to x/4, and then cross-multiply to easily find that x equals 4.

Exam Tip: Moving like-denominator terms together right away avoids calculating an overall LCM of 20 and prevents arithmetic mistakes.

Question 5. Ramesh is driving his car with a uniform speed of 90 km/hr. Draw the time distance graph on the graph paper. Find the distance travelled by him in
i) \( \frac{1}{2} \) hr
ii) 2 hrs.
Answer: Let the time taken be \( x \) hours and the distance travelled be \( y \) kilometers.
Since the speed is uniform at 90 km/hr:
\( \text{Distance} = \text{Speed} \times \text{Time} \)
\( \implies y = 90x \)
Points to plot the linear graph:
- When \( x = 0 \implies y = 0 \implies (0, 0) \)
- When \( x = 1 \implies y = 90(1) = 90 \implies (1, 90) \)
- When \( x = 2 \implies y = 90(2) = 180 \implies (2, 180) \)
Reading values directly from the graph:
(i) For time \( x = \frac{1}{2}\text{ hr} \):
Locate \( x = 0.5 \) on the time axis, trace vertically up to meet the graph line at point \( P(0.5, 45) \), and read the corresponding value on the distance axis:
\( \text{Distance} = 45\text{ km} \)
(ii) For time \( x = 2\text{ hrs} \):
Locate \( x = 2 \) on the time axis, trace vertically up to meet the graph line at point \( Q(2, 180) \), and read the distance axis:
\( \text{Distance} = 180\text{ km} \)
In simple words: The distance covered is 90 times the time spent driving. Plotting this straight line shows that in half an hour he covers 45 km, and in 2 hours he covers 180 km.

Exam Tip: Draw clear dashed projection lines from both axes to the points P and Q to show the examiner that your answers were derived directly from the graph.

Question 6. If (2, 3) and (4, 0) lie on the graph of equation ax+by = 1. Find value of a and b. Plot the graph of equation obtained.
Answer: Given equation of the line:
\( ax + by = 1 \)
Because \( (4, 0) \) lies on the graph, its coordinates must satisfy the equation:
\( a(4) + b(0) = 1 \)
\( \implies 4a = 1 \implies a = \frac{1}{4} \)
Since \( (2, 3) \) also lies on the graph, substitute \( x = 2 \), \( y = 3 \), and \( a = \frac{1}{4} \):
\( \left(\frac{1}{4}\right)(2) + b(3) = 1 \)
\( \implies \frac{1}{2} + 3b = 1 \)
\( \implies 3b = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \implies b = \frac{1}{6} \)
Thus, the values are \( a = \frac{1}{4} \) and \( b = \frac{1}{6} \).
Substituting these values into \( ax + by = 1 \):
\( \frac{1}{4}x + \frac{1}{6}y = 1 \)
Multiplying the entire equation by the LCM of denominators (12):
\( 3x + 2y = 12 \)
Points for plotting the line:
- For \( x = 0 \implies 2y = 12 \implies y = 6 \implies (0, 6) \)
- For \( y = 0 \implies 3x = 12 \implies x = 4 \implies (4, 0) \)
- For \( x = 2 \implies 3(2) + 2y = 12 \implies 2y = 6 \implies y = 3 \implies (2, 3) \)
In simple words: Plugging in (4, 0) cancels out b and instantly gives a = 1/4. Then substituting (2, 3) yields b = 1/6, resulting in the equation 3x + 2y = 12.

Exam Tip: Always insert the point with zero coordinate first so one variable cancels out completely, giving you the first unknown without solving simultaneous equations.

Question 7. The taxi fare in a city as follows: For the first 2 km the fare is Rs 100, and for the subsequent distance the fare is Rs 40/km. Draw the graph of the given information
Answer: Let the total distance travelled be \( x \) km and the total taxi fare be Rs. \( y \).
Fare breakdown:
- For the initial \( 2\text{ km} \), the charge is fixed at Rs. \( 100 \).
- The remaining distance is \( (x - 2)\text{ km} \).
- The charge for this remaining distance is Rs. \( 40 \) per km, amounting to \( 40(x - 2) \).
The total fare equation is:
\( y = 100 + 40(x - 2) \)
\( \implies y = 100 + 40x - 80 \)
\( \implies y = 40x + 20 \)
Points to plot the linear graph (valid for \( x \ge 2 \)):
- When \( x = 2 \implies y = 40(2) + 20 = 100 \implies (2, 100) \)
- When \( x = 3 \implies y = 40(3) + 20 = 140 \implies (3, 140) \)
- When \( x = 4 \implies y = 40(4) + 20 = 180 \implies (4, 180) \)
In simple words: The trip costs Rs. 100 for the first 2 km, plus Rs. 40 for every kilometer after that, producing the straight-line equation y = 40x + 20.

Exam Tip: Be careful not to write 40x; the Rs. 40 rate only applies to the distance beyond the first 2 km, which must be written as 40(x - 2).

Question 8. Equation of line parallel to x-axis and 2 units above the origin
(a) \( x = 2 \)
(b) \( x = -2 \)
(c) \( y = 2 \)
(d) \( y = -2 \)
Answer: (c) \( y = 2 \)
Any straight line parallel to the x-axis has a fixed y-coordinate and is expressed in the general form \( y = c \), where \( c \) denotes the directed distance from the x-axis. Since the given line lies 2 units above the origin (in the positive vertical direction), we have \( c = 2 \). Therefore, the equation of the line is \( y = 2 \).
In simple words: Lines parallel to the horizontal axis always keep the same vertical height, so 2 units above the origin means y equals 2.

Exam Tip: Remember that a line parallel to the x-axis has equation y = constant, whereas a line parallel to the y-axis has equation x = constant.