Chapter 2: Introduction to Linear PolynomialsPage 37 – 42All 14 Questions on This Page
Class 9 Maths Chapter 2 End-of-Chapter Exercises Solutions
Step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 2 End-of-Chapter Exercises (Questions 1–14, Pages 37–39). Covers constructing polynomials with specified degrees and coefficients, polynomial evaluation and symbolic substitution, solving linear equations with fractional coefficients, algebraic modeling of numbers and digits, linear pattern savings formulas, graphing linear equations, identifying slopes and y-intercepts, determining parallel lines, linear temperature scale conversions, direct linear variation (work and distance), determining linear polynomials from points and identity conditions, matchstick pattern modeling, and pencils of linear functions.
💡 Quick navigation: Tap any question number in the bar below to jump straight to that question and its step-by-step solution.← Chapter Overview
The degree of a polynomial in one variable is the highest exponent of the variable with a non-zero coefficient. In any term axⁿ, a is the coefficient. When only the degree and one coefficient are specified, infinitely many polynomials can be written by choosing different coefficients for the remaining terms.
Degree = Highest exponent with non-zero coefficient
2
Value of a Polynomial by Substitution
To find the value of a polynomial P(x) at a given numerical value or symbolic parameter x = c, substitute c for every occurrence of the variable and simplify.
P(c) is obtained by substituting variable = c
3
Solving Linear Equations with Fractional Terms
When an equation contains rational numbers, transpose constant fractions by using equivalent fractions with a common denominator, then multiply by the reciprocal of the variable's coefficient.
(a/b)x + (c/d) = k ⇒ Isolate x using equivalent fractions
4
Algebraic Modeling of Number Relationships
When a positive number is a multiple of another, express them as x and kx. Adding a positive quantity to both preserves their order (kx + c > x + c), allowing unambiguous formulation of linear equations.
Larger + c = 2 · (Smaller + c)
5
Algebraic Modeling of Two-Digit Numbers
A two-digit number with tens digit x and units digit y has the value 10x + y. When the digits are interchanged, the value becomes 10y + x. The sum of the number and its reverse is always a multiple of 11: (10x + y) + (10y + x) = 11(x + y).
Original = 10x + y, Reversed = 10y + x
6
Slopes, Intercepts, and Parallel Lines
Expressing an equation in the standard linear form y = ax + b immediately identifies the slope a and the y-intercept (0, b). Two lines are parallel if and only if they possess identical slopes (a₁ = a₂) but different y-intercepts (b₁ ≠ b₂).
y = ax + b ⇒ Slope = a, y-intercept = (0, b); Parallel if a₁ = a₂
7
Linear Conversion Between Scales
A linear equation relating two units of measurement allows direct evaluation of one unit given the other. To reverse the conversion, use standard algebraic inversion by isolating the variable step by step.
y = (9/5)(x − 273) + 32 ⇒ Linear scale transformation
8
Direct Linear Variation (Work and Distance)
When a constant force F acts in the direction of motion, work done is directly proportional to distance travelled (w = Fd). Its graph is a straight ray emerging from the origin (0, 0) with slope equal to the force.
w = Fd ⇒ Origin-crossing linear ray with slope F
9
Linear Growing Patterns and Matchstick Models
In iterative geometric chains where each newly appended unit shares a boundary with the preceding one, the count increases by a constant common increment. The general rule is M = (initial) + (added per stage) · (n − 1).
M = 6 + 5(n − 1) = 5n + 1
10
Families of Linear Functions (Pencil of Lines)
Factoring f(x) = a(x + 1) for a > 0 reveals that regardless of the slope a, f(−1) = 0. All such lines intersect at the common point (−1, 0) and rise from left to right.
f(x) = a(x + 1) ⇒ Common x-intercept (−1, 0) for all a > 0
Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.
Solution
We need to construct a polynomial in the variable x satisfying two conditions:
•The degree of the polynomial must be 3 (the highest power of x must be 3 with a non-zero coefficient).
•The coefficient of the x² term must be −7.
One such polynomial is:
x³ − 7x² + 2x + 1
In this polynomial:
•The highest exponent of x is 3, so its degree is 3.
•The x² term is −7x², so the coefficient of x² is −7.
Therefore, x³ − 7x² + 2x + 1 is a valid polynomial.
Note
The answer is not unique. Any polynomial of the form ax³ − 7x² + cx + d where a ≠ 0 and c, d are any real numbers satisfies the given conditions. For example, 2x³ − 7x² + x and 5x³ − 7x² − 3 are also valid polynomials.
Answer:A valid polynomial is: x³ − 7x² + 2x + 1 | Note: The answer is not unique (any non-zero x³ term with −7x² satisfies the condition)
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution
Let the smaller positive number be x.
Then the other positive number is:
5x
If 21 is added to both numbers, the new numbers are:
x + 21 and 5x + 21
Since both original numbers are positive and 5x > x, adding 21 to both preserves the order:
5x + 21 > x + 21
Therefore, the larger new number becomes twice the smaller new number:
5x + 21 = 2(x + 21)
Expand the right side:
5x + 21 = 2x + 42
Subtract 2x from both sides:
3x + 21 = 42
Subtract 21 from both sides:
3x = 21
x = 7
Therefore, the smaller number is:
7
The larger number is:
5x = 5(7) = 35
Verification
After adding 21 to both numbers:
7 + 21 = 28
35 + 21 = 56
Notice that:
56 = 2 × 28
One new number is indeed twice the other new number. Thus, the solution is verified.
*6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Solution
Let the tens digit be x and the units digit be y.
Then the original two-digit number is:
10x + y
After interchanging the digits, the number becomes:
10y + x
According to the question, the sum of the original number and the reversed number is 143:
(10x + y) + (10y + x) = 143
11x + 11y = 143
Dividing throughout by 11 gives:
x + y = 13...(1)
We are also given that the digits differ by 3:
|x − y| = 3
This yields two possible cases for the two digits x and y:
Case 1: When x − y = 3 (Tens digit is greater)
We have the simultaneous equations:
x + y = 13
x − y = 3
Adding the two equations:
2x = 16
x = 8
Substitute x = 8 into x + y = 13:
8 + y = 13
y = 5
Thus, the original two-digit number is:
10(8) + 5 = 85
Case 2: When y − x = 3 (Units digit is greater)
We have the simultaneous equations:
x + y = 13
y − x = 3
Adding the two equations:
2y = 16
y = 8
Substitute y = 8 into x + y = 13:
x + 8 = 13
x = 5
Thus, the original two-digit number is:
10(5) + 8 = 58
Hence, the two possible numbers are:
85 and 58
Verification
•For 85:
Digits difference = 8 − 5 = 3
Reversed number = 58
Sum = 85 + 58 = 143
•For 58:
Digits difference = 8 − 5 = 3
Reversed number = 85
Sum = 58 + 85 = 143
Therefore, both 85 and 58 satisfy all conditions given in the problem.
*7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = −3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x − 10
(iv) 3y = 6x − 11
Are any of the lines parallel?
Solution
We write each equation in the standard form:
y = ax + b
where:
•a represents the slope (inclination / steepness);
•b represents the y-intercept;
•The point where the line cuts the y-axis has coordinates (0, b).
For drawing each straight line on the coordinate plane, two distinct points are sufficient.
(i) y = −3x + 4
This equation is already written in the form y = ax + b.
Therefore:
Slope = −3
y-intercept = 4
When x = 0:
y = 4
So the line cuts the y-axis at (0, 4).
For another point, take x = 1:
y = −3(1) + 4
= 1
So another point is (1, 1).
Plot (0, 4) and (1, 1), then join them with a straight line.
(ii) 2y = 4x + 7
Divide throughout by 2 to obtain the standard form:
y = 2x + 72
Therefore:
Slope = 2
y-intercept = 72
When x = 0:
y = 72
So the line cuts the y-axis at (0, 7/2).
For another point, take x = 1:
y = 2(1) + 72
= 112
So another point is (1, 11/2).
Plot (0, 7/2) and (1, 11/2), then join them with a straight line.
(iii) 5y = 6x − 10
Divide throughout by 5 to obtain the standard form:
y = 65 x − 2
Therefore:
Slope = 65
y-intercept = −2
When x = 0:
y = −2
So the line cuts the y-axis at (0, −2).
For an easy second point, take x = 5:
y = 65(5) − 2
= 6 − 2
= 4
So another point is (5, 4).
Plot (0, −2) and (5, 4), then join them with a straight line.
(iv) 3y = 6x − 11
Divide throughout by 3 to obtain the standard form:
y = 2x − 113
Therefore:
Slope = 2
y-intercept = −113
When x = 0:
y = −113
So the line cuts the y-axis at (0, −11/3).
For another point, take x = 1:
y = 2(1) − 113
= −53
So another point is (1, −5/3).
Plot (0, −11/3) and (1, −5/3), then join them with a straight line.
Points-to-Plot Summary Table
Equation
Standard Form (y = ax + b)
First point (x = 0)
Second point
(i) y = −3x + 4
y = −3x + 4
(0, 4)
(1, 1)
(ii) 2y = 4x + 7
y = 2x + 7/2
(0, 7/2)
(1, 11/2)
(iii) 5y = 6x − 10
y = 6/5 x − 2
(0, −2)
(5, 4)
(iv) 3y = 6x − 11
y = 2x − 11/3
(0, −11/3)
(1, −5/3)
Combined Coordinate Graph of All Four Lines
Parallel Lines Analysis
From the equations in standard form y = ax + b:
•Slope of Line (i) = −3
•Slope of Line (ii) = 2
•Slope of Line (iii) = 6/5
•Slope of Line (iv) = 2
Notice that Line (ii) and Line (iv) have identical slopes:
Slope of (ii) = Slope of (iv) = 2
Their y-intercepts are different:
72 ≠ −113
Since lines (ii) and (iv) have equal slopes and distinct y-intercepts, they are distinct parallel lines that will never intersect.
Therefore, Lines (ii) and (iv) are parallel.
Final Summary of Slopes, Intercepts, and y-Axis Points
Part
Equation
Slope
y-intercept
Point cutting y-axis
(i)
y = −3x + 4
−3
4
(0, 4)
(ii)
2y = 4x + 7
2
7/2
(0, 7/2)
(iii)
5y = 6x − 10
6/5
−2
(0, −2)
(iv)
3y = 6x − 11
2
−11/3
(0, −11/3)
Answer:(i) Slope = −3, y-intercept = 4, cuts y-axis at (0, 4) | (ii) Slope = 2, y-intercept = 72, cuts y-axis at (0, 72) | (iii) Slope = 65, y-intercept = −2, cuts y-axis at (0, −2) | (iv) Slope = 2, y-intercept = −113, cuts y-axis at (0, −113) | Lines (ii) and (iv) are parallel
*8. If the temperature of a liquid can be measured in Kelvin units as x K and 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.
Solution
The relation between temperature in Kelvin (x K) and temperature in Fahrenheit (y °F) is given by the linear equation:
y = 9/5 (x − 273) + 32
(i) When x = 313 K
Substitute x = 313 into the given equation:
y = 9/5 (313 − 273) + 32
= 95 (40) + 32
= 72 + 32
= 104
Therefore, the temperature of the liquid in Fahrenheit is:
104 °F
(ii) When y = 158 °F
Substitute y = 158 into the given equation:
158 = 9/5 (x − 273) + 32
Subtract 32 from both sides:
158 − 32 = 9/5 (x − 273)
126 = 95 (x − 273)
Multiply both sides by 5/9 (the reciprocal of 9/5):
126 × 59 = x − 273
14 × 5 = x − 273
70 = x − 273
Add 273 to both sides:
x = 70 + 273
x = 343
Therefore, the temperature of the liquid in Kelvin is:
343 K
Verification
Substitute x = 343 into the original linear equation:
y = 9/5 (343 − 273) + 32
= 95 (70) + 32
= 9 × 14 + 32
= 126 + 32
= 158 °F
The result matches the given value 158 °F, verifying that the solution is correct.
*9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Solution
Work done by a body is defined as the product of the constant force and the distance travelled in the direction of the force:
Work = Force × Distance
Let:
•w be the work done;
•d be the distance travelled.
We are given that the constant force is 3 units.
Therefore, the linear equation in two variables is:
w = 3d
This represents a direct linear variation between w and d.
Points for Drawing the Graph
We choose non-negative values of distance d (since physical distance d ≥ 0) to find the corresponding work w:
Distance d (units)
0
1
2
3
Work w = 3d (units)
0
3
6
9
Plotted Point (d, w)
(0, 0)
(1, 3)
(2, 6)
(3, 9)
Graph of Work against Distance
Work Done when Distance is 2 Units
Substitute d = 2 into the linear equation w = 3d:
w = 3(2)
= 6
Therefore, the work done is 6 units.
Verification by Plotting on the Graph
From the graph:
•Locate d = 2 on the horizontal distance axis.
•Move vertically up to meet the line at the point (2, 6).
•Move horizontally from this point to the vertical axis, which indicates w = 6 units.
Thus, the graph verifies that when the distance travelled is 2 units, the work done is indeed 6 units.
Answer:Linear equation: w = 3d | Work done when distance is 2 units: 6 units (verified by the point (2, 6) on the graph)
*10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Solution
Let the linear polynomial be:
p(x) = ax + b
(i) Finding the polynomial p(x)
Since the graph passes through the point (1, 5):
p(1) = 5
a(1) + b = 5
a + b = 5...(1)
Since the graph passes through the point (3, 11):
p(3) = 11
a(3) + b = 11
3a + b = 11...(2)
Subtract equation (1) from equation (2):
(3a + b) − (a + b) = 11 − 5
2a = 6
a = 3
Substitute a = 3 into equation (1):
3 + b = 5
b = 2
Therefore, the linear polynomial is:
p(x) = 3x + 2
(ii) Coordinates where the graph cuts the axes
Intersection with the y-axis
On the y-axis, the x-coordinate is 0:
p(0) = 3(0) + 2 = 2
So the graph cuts the y-axis at:
(0, 2)
Intersection with the x-axis
On the x-axis, the value of the polynomial is 0:
p(x) = 0
3x + 2 = 0
3x = −2
x = −23
So the graph cuts the x-axis at:
(−2/3, 0)
(iii) Graph and Verification
To plot the line p(x) = 3x + 2 on the Cartesian coordinate plane, we use the points found:
Point
Coordinates (x, y)
Significance / Source
Point A
(1, 5)
Given point
Point B
(3, 11)
Given point
y-intercept
(0, 2)
Cuts y-axis at x = 0
x-intercept
(−2/3, 0)
Cuts x-axis at y = 0
Verification
•Both given points (1, 5) and (3, 11) lie precisely on the straight line drawn.
•The line crosses the y-axis at (0, 2), exactly matching the calculated y-intercept.
•The line crosses the x-axis at (−2/3, 0), exactly matching the calculated x-intercept.
Thus, the graph completely verifies the polynomial and both axis intercepts.
Answer:(i) p(x) = 3x + 2 | (ii) y-axis: (0, 2), x-axis: (−23, 0) | (iii) Verified by the collinear plot of all four points on the graph
*11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5.
(ii) The polynomial p(x) − q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.
Find the polynomials p(x) and q(x).
*12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Growing Pattern of Hexagons (Stages 1–3)
Figure: The first three stages of the matchstick hexagon pattern as given in the textbook.
Solution
Let us analyze the pattern of matchsticks across the stages:
•Stage 1 consists of 1 isolated regular hexagon. A hexagon has 6 sides, so it requires 6 matchsticks.
•At each subsequent stage, one new hexagon is attached to the last hexagon of the preceding stage. Because the new hexagon shares exactly 1 side with the preceding hexagon, it requires only 6 − 1 = 5 additional matchsticks.
Thus, every new stage adds exactly 5 matchsticks.
(i) Next Two Stages (Stage 4 and Stage 5)
Continuing the addition of 5 matchsticks per stage:
•Stage 4: 16 + 5 = 21 matchsticks
•Stage 5: 21 + 5 = 26 matchsticks
The drawings of Stage 4 and Stage 5 continue the chain pattern:
Stage 4 Diagram
Stage 5 Diagram
(ii) Complete the Table
Filling in the table with the values calculated for Stages 1 to 5, …, and the general nth stage:
Stage Number
1
2
3
4
5
…
n
Number of matchsticks
6
11
16
21
26
…
5n + 1
(iii) Rule to determine matchsticks for the nth stage
At Stage 1, there are 6 matchsticks.
Each additional stage (from 2 to n) adds 5 matchsticks for each of the (n − 1) additional hexagons.
Therefore, the number of matchsticks M for the nth stage is:
M = 6 + 5(n − 1)
M = 6 + 5n − 5
M = 5n + 1
(iv) Matchsticks required for the 15th stage
Substitute n = 15 into the rule M = 5n + 1:
M = 5(15) + 1
M = 75 + 1
M = 76
Therefore, 76 matchsticks are required for the 15th stage.
(v) Can 200 matchsticks form a stage in this pattern?
Set the rule equal to 200:
5n + 1 = 200
Subtract 1 from both sides:
5n = 199
Divide by 5:
n = 1995
n = 39.8
Justification: The stage number n must be a positive integer (whole number of hexagons). Since 199 is not divisible by 5, n is not a whole number.
Therefore, 200 matchsticks cannot form a stage in this pattern (Stage 39 requires 196 matchsticks and Stage 40 requires 201 matchsticks).
Answer:(i) Stage 4 = 21 matchsticks, Stage 5 = 26 matchsticks | (ii) Completed table: 6, 11, 16, 21, 26, …, 5n + 1 | (iii) Rule: M = 5n + 1 | (iv) 76 matchsticks | (v) No, 200 matchsticks cannot form a stage because n = 1995 = 39.8 is not an integer
*13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, −1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
Solution
We are given two linear polynomials:
p(x) = ax + b
q(x) = cx + d
Finding the polynomial p(x)
Since the graph of p(x) passes through (2, 3):
p(2) = 3
2a + b = 3...(1)
Since the graph of p(x) passes through (6, 11):
p(6) = 11
6a + b = 11...(2)
Subtract equation (1) from equation (2):
(6a + b) − (2a + b) = 11 − 3
4a = 8
a = 2
Substitute a = 2 into equation (1):
2(2) + b = 3
4 + b = 3
b = −1
Therefore, the first polynomial is:
p(x) = 2x − 1
Finding the polynomial q(x)
We are given that the graph of q(x) is parallel to the graph of p(x).
Two straight lines are parallel if and only if their slopes are equal.
The slope of p(x) is a = 2, so the slope of q(x) must also be 2:
c = 2
So q(x) has the form:
q(x) = 2x + d
Since the graph of q(x) passes through the point (4, −1):
q(4) = −1
2(4) + d = −1
8 + d = −1
d = −9
Therefore, the second polynomial is:
q(x) = 2x − 9
Coordinates where these lines meet the x-axis
A line meets the x-axis where its y-value (or polynomial value) is zero.
Note
Although the textbook asks for 'the point where these lines meet the x-axis', the two lines are distinct parallel lines with different y-intercepts (−1 ≠ −9). Therefore, each line intersects the x-axis at its own unique point:
For the line p(x):
Set p(x) = 0:
2x − 1 = 0
2x = 1
x = 12
Thus, the line p(x) meets the x-axis at:
(1/2, 0)
For the line q(x):
Set q(x) = 0:
2x − 9 = 0
2x = 9
x = 92
Thus, the line q(x) meets the x-axis at:
(9/2, 0)
Answer:p(x) = 2x − 1 and q(x) = 2x − 9 | Coordinates where the lines meet the x-axis: (12, 0) for p(x) and (92, 0) for q(x)
*14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution
We are given the general linear function:
f(x) = ax + a, where a > 0
Factoring out the common parameter a:
f(x) = a(x + 1)
1. Common x-intercept (Fixed Point)
To find the x-intercept of any line in this family, set f(x) = 0:
a(x + 1) = 0
We are given that a > 0, which means a ≠ 0. Dividing both sides by a:
x + 1 = 0
x = −1
This means that regardless of what positive number a is chosen, the graph always passes through the fixed point:
(−1, 0)
Therefore, all linear functions in this family share the exact same x-intercept at (−1, 0).
2. Common Slope Characteristic
In the equation f(x) = ax + a, the coefficient of x is the slope of the line:
Slope = a
Since we are given a > 0:
•All functions have a strictly positive slope.
•Every line rises from left to right.
3. Note on the y-intercept
At x = 0:
f(0) = a(0) + a = a
So the y-intercept is (0, a). As a varies, the y-intercept shifts along the positive y-axis, but it is always positive.
Conclusion
All linear functions of the form f(x) = ax + a with a > 0 have in common that:
1. They all have the same x-intercept at (−1, 0) (they all pass through the point (−1, 0)).
2. They all have a strictly positive slope (a > 0), so they all rise from left to right.
Answer:All linear functions of the form f(x) = ax + a (a > 0) have the same x-intercept (−1, 0) and all have a positive slope (rising from left to right).