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NCERT Class 9 Mathematics Introduction to Linear Polynomials: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “Introduction to Linear Polynomials”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Write a degree-3 polynomial in x whose x² coefficient is −7.

   1. A polynomial of degree 3 has the highest power of x as 3.
   2. The general form of such a polynomial is ax³ + bx² + cx + d, where a is not zero.
   3. The question asks for the coefficient of the x² term to be −7, so we set b = −7.
   4. We can choose any non-zero value for a and any values for c and d to complete the polynomial.
   5. Choosing a = 1, c = 2, and d = 1 gives the polynomial x³ − 7x² + 2x + 1.

   Answer: x³ − 7x² + 2x + 1

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2. Exercises Q2

   Evaluate (i) 5x² − 3x + 7 at x = 1, (ii) 4t³ − t² + 6 at t = a.

   1. (i) Substitute x = 1 into the expression 5x² − 3x + 7.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (i) Calculate 5(1)² − 3(1) + 7.
   3. (i) Simplify to 5 − 3 + 7.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   4. (i) The result is 9.
   5. (ii) Substitute t = a into the expression 4t³ − t² + 6.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. (ii) The result is 4a³ − a² + 6.

   Answer: (i) 9; (ii) 4a³ − a² + 6

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3. Exercises Q3

   A number times 5/2, plus 2/3, gives −7/12. Find the number.

   1. Let the required number be x.
   2. According to the question, 5/2 times x plus 2/3 equals −7/12.
   3. Write the equation as (5/2)x + 2/3 = −7/12.
   4. Subtract 2/3 from both sides to isolate the term with x.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   5. (5/2)x = −7/12 − 2/3.
   6. Convert 2/3 to twelfths, which is 8/12.
   7. (5/2)x = −7/12 − 8/12.
   8. (5/2)x = −15/12.
   9. Simplify the fraction −15/12 to −5/4.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   10. (5/2)x = −5/4.
   11. Multiply both sides by 2/5 to solve for x.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   12. x = (−5/4) × (2/5).
   13. x = −1/2.

   Answer: −1/2

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4. Exercises Q4

   A positive number is 5 times another. Adding 21 to both makes one new number twice the other. Find the numbers.

   1. Let the smaller number be x, so the larger number is 5x.
   2. After adding 21 to both, the numbers become x + 21 and 5x + 21.
   3. If instead x + 21 were twice 5x + 21, then x + 21 = 10x + 42 gives x = −21/9, which is negative, so that case is not possible.
   4. According to the problem, the larger new number is twice the smaller new number: 5x + 21 = 2(x + 21).
   5. Expand the right side to get 5x + 21 = 2x + 42.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   6. Subtract 2x from both sides to get 3x + 21 = 42.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. Subtract 21 from both sides to get 3x = 21.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   8. Divide by 3 to find x = 7.
   9. The larger number is 5x = 5 × 7 = 35.

   Answer: The numbers are 7 and 35.

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5. Exercises Q5

   Start with ₹800 and save ₹250 a month: amount after (i) 6 months, (ii) 2 years, and the linear pattern.

   1. (i) The initial amount is ₹800 and the monthly saving is ₹250.
   2. (i) For 6 months, the total saved is 250 × 6 = ₹1500.
   3. (i) The total amount after 6 months is 800 + 1500 = ₹2300.
   4. (ii) Two years equal 24 months.
   5. (ii) The total saved in 24 months is 250 × 24 = ₹6000.
   6. (ii) The total amount after 2 years is 800 + 6000 = ₹6800.
   7. Let n be the number of months passed.
   8. The linear pattern for the total amount is 800 + 250n.

   Answer: (i) 2300; (ii) 6800; 800 + 250n

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6. Exercises Q6

   A two-digit number's digits differ by 3; the number plus its reversal is 143. Find both numbers.

   1. Let the units digit be x and the tens digit be x + 3.
   2. The number is 10(x + 3) + x, which simplifies to 11x + 30.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   3. The reversed number is 10x + (x + 3), which simplifies to 11x + 3.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   4. The sum of the number and its reversal is (11x + 30) + (11x + 3) = 22x + 33.
   5. Set the sum equal to 143: 22x + 33 = 143.
   6. Subtract 33 from both sides: 22x = 110.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. Divide by 22: x = 5.
   8. The units digit is 5 and the tens digit is 5 + 3 = 8.
   9. If the units digit is the larger one instead, the same two digits 5 and 8 come out, giving the same pair of numbers.
   10. The two numbers are 85 and 58.

   Answer: 85, 58

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7. Exercises Q7

   Graph each line and give its slope, y-intercept and the point where it cuts the y-axis: (i) y = −3x + 4, (ii) 2y = 4x + 7, (iii) 5y = 6x − 10, (iv) 3y = 6x − 11. Which lines are parallel?

   1. (i) Rewrite y = −3x + 4 in slope-intercept form y = mx + c, so slope m = −3 and y-intercept c = 4.
   2. (i) The line cuts the y-axis at the point (0, 4).
   3. (ii) Divide 2y = 4x + 7 by 2 to get y = 2x + 7/2, so slope m = 2 and y-intercept c = 7/2.
   4. (ii) The line cuts the y-axis at the point (0, 7/2).
   5. (iii) Divide 5y = 6x − 10 by 5 to get y = (6/5)x − 2, so slope m = 6/5 and y-intercept c = −2.
   6. (iii) The line cuts the y-axis at the point (0, −2).
   7. (iv) Divide 3y = 6x − 11 by 3 to get y = 2x − 11/3, so slope m = 2 and y-intercept c = −11/3.
   8. (iv) The line cuts the y-axis at the point (0, −11/3).
   9. Lines (ii) and (iv) both have a slope of 2, so they are parallel.

   Answer: (i) slope −3, y-intercept (0, 4); (ii) slope 2, y-intercept (0, 7/2); (iii) slope 6/5, y-intercept (0, −2); (iv) slope 2, y-intercept (0, −11/3); (ii) and (iv) are parallel

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8. Exercises Q8

   y °F = (9/5)(x − 273) + 32 for x K. (i) 313 K in °F; (ii) 158 °F in K.

   1. (i) Substitute x = 313 into the formula y = (9/5)(x − 273) + 32.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (i) Calculate the difference inside the brackets: 313 − 273 = 40.
   3. (i) Multiply by 9/5: (9/5) × 40 = 72.
   4. (i) Add 32 to the result: 72 + 32 = 104.
   5. (ii) Substitute y = 158 into the formula 158 = (9/5)(x − 273) + 32.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. (ii) Subtract 32 from both sides: 158 − 32 = 126.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. (ii) Multiply both sides by 5/9 to isolate the term: 126 × (5/9) = 70.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   8. (ii) Add 273 to both sides to find x: 70 + 273 = 343.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   Answer: (i) 104 °F; (ii) 343 K

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9. Exercises Q9

   Work = force × distance: write it as a linear equation in w and d for a force of 3 units, graph it, and find the work for a distance of 2 units.

   1. The formula for work is Work = Force × Distance.
   2. Here, force is 3 units, so the equation is w = 3 × d.
   3. This gives the linear equation w = 3d.
   4. To graph it, find points: when d = 0, w = 0; when d = 1, w = 3; when d = 2, w = 6.
   5. Plot the points (0, 0), (1, 3), and (2, 6) on a graph and draw a straight line through them.
   6. To find work for a distance of 2 units, substitute d = 2 in the equation w = 3d.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   7. w = 3 × 2 = 6.

   Answer: w = 3d; 6 units

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10. Exercises Q10

    A linear p(x) passes through (1, 5) and (3, 11). (i) Find p(x); (ii) where its graph cuts the axes; (iii) graph and verify.

    1. (i) Let p(x) = ax + b.
    2. Substitute (1, 5) to get a + b = 5.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    3. Substitute (3, 11) to get 3a + b = 11.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    4. Subtract the first equation from the second to get 2a = 6, so a = 3.
    5. Substitute a = 3 into a + b = 5 to get b = 2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    6. Thus, p(x) = 3x + 2.
    7. (ii) To find the y-intercept, set x = 0 in p(x) = 3x + 2.
    8. p(0) = 3(0) + 2 = 2, so the graph cuts the y-axis at (0, 2).
    9. To find the x-intercept, set p(x) = 0.
    10. 3x + 2 = 0 gives 3x = −2, so x = −2/3.
    11. The graph cuts the x-axis at (−2/3, 0).
    12. (iii) Draw the line passing through (0, 2) and (−2/3, 0).
    13. Verify by checking that the line also passes through (1, 5) and (3, 11).

    Answer: (i) p(x) = 3x + 2; (ii) (0, 2), (−2/3, 0); (iii) Graph is a straight line passing through these points.

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11. Exercises Q11

    p(x) = ax + b, q(x) = cx + d with p(0) = 5, p(x) − q(x) zero at x = 3, and p(x) + q(x) = 6x + 4. Find p(x) and q(x).

    1. Given p(x) = ax + b and q(x) = cx + d.
    2. Since p(0) = 5, substituting x = 0 gives a(0) + b = 5, so b = 5.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    3. Thus p(x) = ax + 5.
    4. Given p(x) + q(x) = 6x + 4, we have (ax + 5) + (cx + d) = 6x + 4.
    5. Comparing coefficients of x: a + c = 6.
    6. Comparing constant terms: 5 + d = 4, so d = −1.
    7. Thus q(x) = cx − 1.
    8. Given p(x) − q(x) is zero at x = 3, we have (ax + 5) − (cx − 1) = 0 at x = 3.
    9. Simplifying: (a − c)x + 6 = 0. At x = 3: 3(a − c) + 6 = 0.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    10. So 3(a − c) = −6, which gives a − c = −2.
    11. We now have a + c = 6 and a − c = −2.
    12. Adding the two equations: 2a = 4, so a = 2.
    13. Substituting a = 2 into a + c = 6 gives 2 + c = 6, so c = 4.
        why?

        Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    14. Therefore p(x) = 2x + 5 and q(x) = 4x − 1.

    Answer: p(x) = 2x + 5, q(x) = 4x − 1

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12. Exercises Q12

    Matchstick hexagons in a row (each new hexagon shares a side with the last): (i) matchsticks for Stages 4 and 5; (ii) table for Stages 1 to 5 and n; (iii) rule for Stage n; (iv) Stage 15; (v) can 200 matchsticks make a stage?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. (i) Stage 1 uses 6 matchsticks. Each new hexagon added to the row shares one side with the previous one, so it requires 5 additional matchsticks.
    2. (i) For Stage 4, we add 3 more hexagons to the first one: 6 + 3 × 5 = 6 + 15 = 21 matchsticks.
    3. (i) For Stage 5, we add 4 more hexagons to the first one: 6 + 4 × 5 = 6 + 20 = 26 matchsticks.
    4. (ii) Stage 1: 6 matchsticks. Stage 2: 6 + 5 = 11. Stage 3: 11 + 5 = 16. Stage 4: 16 + 5 = 21. Stage 5: 21 + 5 = 26.
    5. (ii) The table of matchsticks for Stages 1 to 5 is: 6, 11, 16, 21, 26.
    6. (iii) Stage n has 6 matchsticks for the first hexagon and 5 for each of the other n − 1 hexagons: 6 + 5(n − 1) = 5n + 1.
    7. (iv) For Stage 15, substitute n = 15 into the rule: 5 × 15 + 1 = 75 + 1 = 76 matchsticks.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    8. (v) To check if 200 matchsticks can form a stage, set 5n + 1 = 200.
    9. (v) Solving for n: 5n = 199, so n = 199 ÷ 5 = 39.8.
    10. (v) Since n must be a whole number (you cannot have a fraction of a stage), 200 matchsticks cannot make a complete stage.

    Answer: (i) Stage 4: 21, Stage 5: 26; (ii) 6, 11, 16, 21, 26, …, 5n + 1; (iii) 5n + 1; (iv) 76; (v) no, since 5n + 1 = 200 gives n = 39.8, not a whole number

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13. Exercises Q13

    p(x) = ax + b passes through (2, 3) and (6, 11); q(x) = cx + d passes through (4, −1) and is parallel to p. Find p(x), q(x) and where each line meets the x-axis.

    1. p(x) = ax + b passes through (2, 3), so 2a + b = 3.
    2. p(x) = ax + b passes through (6, 11), so 6a + b = 11.
    3. Subtracting the first equation from the second gives 4a = 8, so a = 2.
    4. Substituting a = 2 into 2a + b = 3 gives 4 + b = 3, so b = −1.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    5. Thus, p(x) = 2x − 1.
    6. q(x) = cx + d is parallel to p(x), so its slope c is equal to a, which means c = 2.
    7. q(x) = 2x + d passes through (4, −1), so 2(4) + d = −1.
    8. Solving 8 + d = −1 gives d = −9.
    9. Thus, q(x) = 2x − 9.
    10. To find where p(x) meets the x-axis, set p(x) = 0, so 2x − 1 = 0.
    11. Solving 2x = 1 gives x = 1/2, so the point is (1/2, 0).
    12. To find where q(x) meets the x-axis, set q(x) = 0, so 2x − 9 = 0.
    13. Solving 2x = 9 gives x = 9/2, so the point is (9/2, 0).

    Answer: p(x) = 2x − 1; q(x) = 2x − 9; (1/2, 0); (9/2, 0)

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14. Exercises Q14

    What do all linear functions f(x) = ax + a with a > 0 have in common?

    1. The given linear function is f(x) = ax + a where a is greater than 0.
    2. To find the common point, set f(x) equal to 0 to find the x-intercept.
    3. Solving ax + a = 0 gives ax = −a.
    4. Dividing both sides by a (since a is not 0) results in x = −1.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    5. Substituting x = −1 into the function gives f(−1) = a(−1) + a = 0.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    6. This shows that for any value of a, the graph passes through the point (−1, 0).
    7. Also f(x) = a(x + 1): the slope a and the y-intercept a are equal and positive, so every such line rises to the right and cuts the y-axis above the origin at (0, a).

    Answer: All such linear functions pass through the point (−1, 0); each has slope equal to its y-intercept, a > 0, so each line rises to the right.

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