Chapter 13: Two Variables, One Line

Class 9 Ganita Manjari Part 2 · notes, exercises and mini test.

Learning map

What this chapter is teaching

A linear equation in two variables models relationships between two changing quantities as a straight line in the coordinate plane. This chapter covers standard forms, geometric graphing, slope-intercept equations, algebraic solving methods (substitution and elimination), and consistency ratio tests.

RepresentModel real-world scenarios in standard form ax + by + c = 0 and slope-intercept form y = mx + d.
GraphPlot lines using intercepts or slope (m = Rise/Run) to visualize infinite solution sets.
SolveSolve systems of two equations algebraically using Substitution and Elimination methods.
ClassifyUse ratio tests (a_1/a_2, b_1/b_2, c_1/c_2) to determine whether lines intersect, coincide, or run parallel.
Study tip: Always check your solutions by substituting the values of x and y back into both original equations.
Coverage check: Sections 1–10 cover all textbook definitions, Examples 1–13, geometric and algebraic proofs, Think & Reflect discussions, and historical origins.

1Linear Equations in Two Variables & Standard Form

Radha stops at a fruit cart. The price tags say that mangoes cost ₹60 per kg and bananas cost ₹50 per kg. She buys x kg of mangoes and y kg of bananas, paying ₹280 in total. This situation is modeled by the linear equation:

60x + 50y = 280

Both sides represent the exact same quantity: the total cost in rupees. This is a linear equation in two variables.

Standard Form of a Linear Equation in Two Variables

Any equation that can be expressed in the form:

ax + by + c = 0

where a, b, c are real numbers, and a and b are not both zero (a² + b² ≠ 0), is called a linear equation in two variables in standard form. Here a is the coefficient of x, b is the coefficient of y, and c is the constant term.

  • Clearing Fractions: Coefficients can be fractions. For example, ¼x + (3 / 5)y - ½ = 0 can be multiplied by the LCM 20 to clear denominators, giving 5x + 12y - 10 = 0.
  • Single-Variable Equations as Special Cases: Equations like 4x + 7 = 0 are special cases of two variables with b = 0: 4x + 0y + 7 = 0. Similarly, 3y = 5 is 0x + 3y - 5 = 0.
Solved Example 1 (Page 86) — Expressing Equations in Standard Form:
  1. (i) 5x + 2y = 2.7: Rewrite as 5x + 2y - 2.7 = 0 &implies a = 5, b = 2, c = -2.7. (Alternatively, -5x - 2y + 2.7 = 0).
  2. (ii) (x / 2) - 4 = (3 / 2)y: Rewrite as ½x - (3 / 2)y - 4 = 0 &implies a = ½, b = -(3 / 2), c = -4. (Or multiply by 2: x - 3y - 8 = 0).
  3. (iii) 4 = 1.5x - 1.3y: Rewrite as 1.5x - 1.3y - 4 = 0 &implies a = 1.5, b = -1.3, c = -4.
  4. (iv) 3y = 5: Rewrite as 0x + 3y - 5 = 0 &implies a = 0, b = 3, c = -5.
PracticeExercise 13.1 Q1–Q3

2Algebraic Meaning of Solutions & The Infinite Solution Set

A solution of a linear equation in two variables is an ordered pair of real numbers (u, v) such that substituting x = u and y = v satisfies the equation (makes LHS = RHS).

Consider 3x + 2y = 12:

  • (2, 3) is a solution because 3(2) + 2(3) = 6 + 6 = 12.
  • (4, 0) is a solution because 3(4) + 2(0) = 12 + 0 = 12.
  • (1, 3) is not a solution because 3(1) + 2(3) = 9 ≠ 12.
Why Every Linear Equation in Two Variables Has Infinitely Many Solutions:
Express y explicitly in terms of x: y = -(3 / 2)x + 6. For every chosen real value of x, there is a corresponding real value of y. Hence, there are infinitely many solutions!
Think and Reflect (Page 88): Testing the pairs for 3x + 2y = 12:
  • (1, 9/2): 3(1) + 2(9/2) = 3 + 9 = 12 &implies Solution.
  • (0, 6): 3(0) + 2(6) = 12 &implies Solution (y-intercept).
  • (4, 1): 3(4) + 2(1) = 14 ≠ 12 &implies Not a solution.
  • (1/2, 21/4): 3(1/2) + 2(21/4) = 3/2 + 21/2 = 24/2 = 12 &implies Solution.
Solved Example 4 (Page 92) — Finding Unknown Parameters:

If (-3, 4) is a solution of both 3ax + 4y = -2 and 2x + by = 14, find a and b:

  1. Substitute x = -3, y = 4 into equation 1: 3a(-3) + 4(4) = -2 ⇒ -9a + 16 = -2 ⇒ -9a = -18 ⇒ a = 2.
  2. Substitute x = -3, y = 4 into equation 2: 2(-3) + b(4) = 14 ⇒ -6 + 4b = 14 ⇒ 4b = 20 ⇒ b = 5.
Think and Reflect (Page 92) — The Origin Condition (c = 0):

Ranju, Farhan, Toshi, and Meena's mathematical deduction:

In standard form ax + by + c = 0, substituting the origin (0, 0) gives a(0) + b(0) + c = 0 ⇒ c = 0.
Theorem: A straight line passes through the origin (0, 0) if and only if its constant term c = 0.
Special cases: x = 0 (a=1, b=0, c=0) is the equation of the y-axis; y = 0 (a=0, b=1, c=0) is the equation of the x-axis.

PracticeExercise 13.2 Q1–Q7

3Geometric Visualisation & Graphing One Line

The geometric graph of a linear equation in two variables is always a straight line. The fundamental connection between algebra and geometry is:

Point on Line &implies SolutionEvery point (x, y) lying on the line is an algebraic solution of the equation.
Solution &implies Point on LineEvery algebraic solution (x, y) corresponds to a coordinate point lying on the straight line.
The Intercept Method for Plotting a Line:
  • Set x = 0 to find where the line crosses the y-axis (y-intercept).
  • Set y = 0 to find where the line crosses the x-axis (x-intercept).
  • Join the two points with a straightedge and extend in both directions.
Solved Example 2 (Page 89–90) — Graph of 2x + y = 7:
  1. Set x = 0 ⇒ y = 7 ⇒ A(0, 7).
  2. Set y = 0 ⇒ 2x = 7 ⇒ x = 3.5 ⇒ B(3.5, 0).
  3. Plotting the line reveals further integer solutions: (1, 5), (2, 3), (3, 1), (4, -1).
Solved Example 3 (Page 90–91) — Intercept vs. Origin Graphs:
  • (i) 4x + 3y = 12: x = 0 ⇒ y = 4 ((0, 4)); y = 0 ⇒ x = 3 ((3, 0)). Graph passes through (0, 4) and (3, 0).
  • (ii) 2x + 5y = 0: Since c = 0, it passes through (0, 0). A second point is found by setting x = 5 ⇒ 2(5) + 5y = 0 ⇒ y = -2 ⇒ (5, -2). Another point is (1, -0.4).
PracticeExercise 13.2 Q2–Q5

4Understanding Slope & The Slope-Intercept Form (y = mx + d)

In mathematics, the steepness of a line is called its slope or gradient (m). It is the ratio of vertical change (Rise) to horizontal change (Run):

m = (Rise / Run) = (y_2 - y_1 / x_2 - x_1)
Solved Examples 5 & 6 (Page 97–98) — Slope Calculations:
  • Example 5: Line through A(-1, 0), B(1, 2), C(4, 5). Using B and C: m = (5 - 2 / 4 - 1) = (3 / 3) = 1.
  • Example 6 (Order Invariance): Line through A(-4, 5) and B(-1, 2):
    Taking A to B: m = (2 - 5 / -1 - (-4)) = (-3 / 3) = -1.
    Taking B to A: m = (5 - 2 / -4 - (-1)) = (3 / -3) = -1.
    Key Fact: The slope is strictly independent of the order of the two chosen points!

Mathematical Proof 1: Why m in y = mx + d Represents the Slope

  1. Let A(x_1, y_1) and B(x_2, y_2) be any two distinct points on the line y = mx + d.
  2. Since both points satisfy the equation: y_1 = mx_1 + d and y_2 = mx_2 + d.
  3. Subtracting the two equations gives the Rise:
    y_2 - y_1 = (mx_2 + d) - (mx_1 + d) = mx_2 - mx_1 = m(x_2 - x_1)
  4. Dividing by the Run (x_2 - x_1) (with x_2 ≠ x_1):
    (y_2 - y_1 / x_2 - x_1) = (m(x_2 - x_1) / x_2 - x_1) = m
  5. Hence, the coefficient m in y = mx + d identically equals the geometric slope between any two points on the line! Q.E.D.
Converting Standard Form to Slope-Intercept Form:
For ax + by + c = 0 with b ≠ 0:
by = -ax - c ⇒ y = (-(a / b))x + (-(c / b))
Therefore, the slope is m = -a/b and the y-intercept is d = -c/b (crossing at (0, -c/b)).
Example: 5x + y = 3 ⇒ y = -5x + 3. Slope m = -5 (down 5 units for every 1 unit right), y-intercept (0, 3).

The Four Geometric Cases of Slope

Positive Slope (m > 0)Slants upwards from left to right. As x increases, y increases (e.g. y = 2x + 1).
Negative Slope (m < 0)Slants downwards from left to right. As x increases, y decreases (e.g. y = -3x + 4).
Zero Slope (m = 0)Horizontal line parallel to the x-axis. y is constant (e.g. y = 5).
Undefined SlopeVertical line parallel to the y-axis (x = 2). Run is 0, division by zero is undefined!
PracticeExercise 13.3 Q1–Q4

5Pairs of Linear Equations & Real-World Modeling

When two unknown quantities are subject to two simultaneous conditions, we obtain a pair of linear equations in two variables (a system of linear equations):

a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0
  • Rakesh's Puzzle (p. 104): Two numbers have sum 25 and difference 11:
    x + y = 25 and x - y = 11 ⇒ x = 18, y = 7.
  • Science Exhibition Tickets (p. 104): Group A pays ₹600 for 2 adult and 3 child tickets; Group B pays ₹700 for 3 adult and 2 child tickets:
    2x + 3y = 600 and 3x + 2y = 700.
PracticeExercise 13.4 Q1–Q3

6Algebraic Methods: Substitution & Elimination

To solve a pair of linear equations, we reduce the two equations in two variables to a single linear equation in one variable.

Method 1: Substitution Method

Solved Example 7 (Page 106–107) — Substitution: Solve 7x - 15y = 2 … (1) and x + 2y = 3 … (2):
  1. Step 1: From equation (2), express x in terms of y: x = 3 - 2y … (3).
  2. Step 2: Substitute (3) into equation (1):
    7(3 - 2y) - 15y = 2 ⇒ 21 - 14y - 15y = 2 ⇒ -29y = -19 ⇒ y = 19/29.
  3. Step 3: Substitute y back into (3):
    x = 3 - 2((19 / 29)) = (87 - 38 / 29) = 49/29.
Solved Example 8 (Page 107) — Triangle Angles: Two angles of a triangle are x^° and y^° with y = 4x, and the third angle is 50^°:
x + y + 50 = 180 ⇒ x + 4x = 130 ⇒ 5x = 130 ⇒ x = 26^°, y = 104^°.

Method 2: Elimination Method

Solved Example 9 (Page 108) — Monthly Incomes & Expenditures: Incomes ratio is 9 : 7 (9x, 7x) and expenditures ratio is 4 : 3 (4y, 3y). Each saves ₹2000 per month:
  1. 9x - 4y = 2000 … (1) and 7x - 3y = 2000 … (2).
  2. Multiply (1) by 3 and (2) by 4 to equalize coefficients of y:
    27x - 12y = 6000 … (3)
    28x - 12y = 8000 … (4)
  3. Subtract (3) from (4): (28x - 27x) - (-12y - (-12y)) = 8000 - 6000 ⇒ x = 2000.
  4. Substitute x = 2000 into (1): 9(2000) - 4y = 2000 ⇒ 4y = 16000 ⇒ y = 4000.
  5. Answer: Monthly incomes are 9(2000) = ₹18,000 and 7(2000) = ₹14,000.
Solved Example 10 (Page 109) — Reversible Two-Digit Number: Sum of a two-digit number and its reverse is 66, and digits differ by 2:
  1. Let number be 10x + y. Reversed number is 10y + x.
  2. (10x + y) + (10y + x) = 66 ⇒ 11(x + y) = 66 ⇒ x + y = 6 … (1).
  3. Digits differ by 2: x - y = 2 … (2).
  4. Adding (1) and (2): 2x = 8 ⇒ x = 4. Then y = 2.
  5. Answer: The number is 42 (or 24).
PracticeExercise 13.5 Q1, Q7, Q8

7Consistency & The Ratio Rules

When applying the Elimination Method to a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0, three cases arise:

Coefficients RatioAlgebraic ResultGeometric LinesNumber of SolutionsConsistency
(a_1 / a_2) ≠ (b_1 / b_2)One variable remainsIntersecting linesUnique solutionConsistent & Independent
(a_1 / a_2) = (b_1 / b_2) = (c_1 / c_2)0 = 0 (identity)Coincident linesInfinitely manyConsistent & Dependent
(a_1 / a_2) = (b_1 / b_2) ≠ (c_1 / c_2)0 = k (k ≠ 0, contradiction)Parallel linesNo solutionInconsistent
PracticeExercise 13.5 Q3, Q4, Q6End Q8, Q9, Q10

8Graphical Method & Solved Geometry Examples

Graphing both equations displays their geometric relationship directly:

Solved Example 11 (Page 113–114) — Unique Solution: Graph x + 3y = 6 (points (0, 2), (6, 0)) and 2x - 3y = 12 (points (0, -4), (3, -2)).
The lines intersect at point B(6, 0).
Ratio check: a_1/a_2 = 1/2 ≠ b_1/b_2 = -1. Unique solution: x = 6, y = 0.
Solved Example 12 (Page 114–115) — Parallel Lines (No Solution): x + 2y - 4 = 0 and 2x + 4y - 12 = 0.
Ratios: a_1/a_2 = 1/2, b_1/b_2 = 2/4 = 1/2, c_1/c_2 = -4/-12 = 1/3.
Since a_1/a_2 = b_1/b_2 ≠ c_1/c_2, the lines are parallel. No solution.
Solved Example 13 (Page 115–116) — Coincident Lines (Infinitely Many Solutions): Romila buys 2 erasers and 3 sheets for ₹9 (2x + 3y = 9); Sonali buys 4 erasers and 6 sheets for ₹18 (4x + 6y = 18).
Ratios: a_1/a_2 = 2/4 = 1/2, b_1/b_2 = 3/6 = 1/2, c_1/c_2 = -9/-18 = 1/2.
The graphs coincide as the exact same line. Infinitely many solutions.

Mathematical Proof 2: Two Lines with Equal Slopes are Parallel

  1. Let two lines be L_1: y = mx + d_1 and L_2: y = mx + d_2 with d_1 ≠ d_2.
  2. Suppose for contradiction that L_1 and L_2 intersect at point (x_0, y_0).
  3. Then y_0 = mx_0 + d_1 and y_0 = mx_0 + d_2 ⇒ mx_0 + d_1 = mx_0 + d_2.
  4. Subtracting mx_0 from both sides gives d_1 = d_2, which directly contradicts d_1 ≠ d_2!
  5. Therefore, no intersection point can exist. The lines are strictly parallel. Q.E.D.
PracticeExercise 13.5 Q2, Q4End Q3

9A Pinch of History & Mahāvīrāchārya's Problem

Solving systems of linear equations is one of mathematics’ oldest disciplines:

  • Babylonia (c. 2000 BCE): Cuneiform tablets show problems involving areas and sums of two unknown lengths.
  • Ancient China (c. 200 BCE): The classic Nine Chapters on the Mathematical Art (Jiuzhang Suanshu) devotes chapter 8, fangcheng (“rectangular arrays”), to solving linear systems using columns of counting rods on a board — an algorithm identical to Gaussian elimination, developed 1,800 years before Gauss!
  • India (c. 500–850 CE): Āryabhaṭa (499 CE) and Brahmagupta (628 CE) formulated systematic methods for linear systems in multiple unknowns.
Mahāvīrāchārya's Problem (c. 850 CE, Gaṇita Sāra Saṅgraha) — Exercise 13.5 Q7:

“The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of each!”

  1. Equations: 9x + 7y = 107 … (1) and 7x + 9y = 101 … (2).
  2. Mahāvīrāchārya's Symmetric Shortcut: Notice that the coefficients of x and y are interchanged!
  3. Step 1 — Add both equations:
    16x + 16y = 208 ⇒ x + y = (208 / 16) = 13 … (3)
  4. Step 2 — Subtract equation (2) from (1):
    2x - 2y = 6 ⇒ x - y = 3 … (4)
  5. Step 3 — Add (3) and (4):
    2x = 16 ⇒ x = 8. Then y = 13 - 8 = 5.
  6. Answer: Each citron costs ₹8, and each fragrant wood-apple costs ₹5!

10Chapter Summary — Key Principles

  • ax + by + c = 0 (a, b not both zero) represents a straight line with infinitely many solutions (x, y).
  • The line cuts the axes at (0, -c/b) (y-intercept) and (-c/a, 0) (x-intercept).
  • Slope m = (y_2 - y_1 / x_2 - x_1) = -a/b. In y = mx + d, m is the slope and d is the y-intercept.
  • A system of two equations has:
    • A unique solution if a_1/a_2 ≠ b_1/b_2 (intersecting lines).
    • Infinitely many solutions if a_1/a_2 = b_1/b_2 = c_1/c_2 (coincident lines).
    • No solution if a_1/a_2 = b_1/b_2 ≠ c_1/c_2 (parallel lines).
  • Two algebraic methods — Substitution and Elimination — solve any solvable system directly.

Exercise Set 13.1 — Solutions

Standard form, coefficients and mathematical representation.

1Write a linear equation in two variables in which a = 3, b = 0 and c = −1/5.
Answer / method
Step 1: Substitute a = 3, b = 0 and c = −1/5 in ax + by + c = 0. Step 2: 3x + 0y − 1/5 = 0. Conclusion: 3x − 1/5 = 0 is the required linear equation in two variables.
2Complete the table after expressing y − 15 = √2x, 3y − 2x = 0, 5x = 3y, x = 8 and 3y = 1 in standard form. Give the coefficient of x, coefficient of y and constant term.
Answer / method
Step 1: Put all terms on the left side. Step 2: y − 15 = √2x becomes √2x − y + 15 = 0, so a = √2, b = −1, c = 15. Step 3: 3y − 2x = 0 becomes −2x + 3y + 0 = 0, so a = −2, b = 3, c = 0. Step 4: 5x = 3y becomes 5x − 3y + 0 = 0, so a = 5, b = −3, c = 0. Step 5: x = 8 becomes x + 0y − 8 = 0, so a = 1, b = 0, c = −8. Step 6: 3y = 1 becomes 0x + 3y − 1 = 0, so a = 0, b = 3, c = −1.
3(i) The cost of a notebook is twice the cost of a pen. If a pen costs ₹p and a notebook costs ₹t, Charlie wrote t = 2p, whereas Meera wrote p = 2t. Which representation is correct? (ii) Two Indian batsmen together scored 176 runs. If their scores are x and y, is x + y = 176 correct?
Answer / method
Step 1: Let pen cost = ₹p and notebook cost = ₹t. Step 2: Notebook cost is twice pen cost, so t = 2p. Conclusion for (i): Charlie is correct; p = 2t reverses the meaning. Step 3: If the two scores are x and y, their total is x + y. Step 4: The total is 176, so x + y = 176. Conclusion for (ii): Manisha's representation is correct.

Exercise Set 13.2 — Solutions

Solutions, graphs, quadrants and equivalent equations.

1Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.
Answer / method
Step 1: Substitute x = 4 and y = 3 into 5x − 6y. Step 2: 5(4) − 6(3) = 20 − 18 = 2. Step 3: The result equals the right side, 2. Conclusion: (4, 3) is a solution.
2Find any two solutions for each equation: (i) 7x − 3y = 21. (ii) 2x + 3y = 5.
Answer / method
Step 1: For 7x − 3y = 21, choose x = 3. Then 21 − 3y = 21, so y = 0. One solution is (3, 0). Step 2: Choose x = 0. Then −3y = 21, so y = −7. Another solution is (0, −7). Step 3: For 2x + 3y = 5, choose x = 1. Then 2 + 3y = 5, so y = 1. One solution is (1, 1). Step 4: Choose x = 4. Then 8 + 3y = 5, so y = −1. Another solution is (4, −1). Step 5: Substitution verifies all four ordered pairs.
3In 2mx + 3y = 7 and 4x + ny = −10, m and n are unknown. If (2, −1) is the solution of both equations, find m and n.
Answer / method
Step 1: Substitute (2, −1) in 2mx + 3y = 7. Step 2: 2m(2) + 3(−1) = 7, so 4m − 3 = 7. Step 3: 4m = 10, hence m = 5/2. Step 4: Substitute (2, −1) in 4x + ny = −10. Step 5: 4(2) + n(−1) = −10, so 8 − n = −10. Step 6: n = 18.
4Find two solutions in different quadrants for each: (i) 5x + 3y = 7, (ii) 5x − 3y = 7, (iii) −5x + 3y = 7, (iv) −5x − 3y = 7. Identify the quadrants and verify by graphing.
Answer / method
Step 1: Choose convenient x-values and calculate y. Step 2: (i) 5x + 3y = 7. With x = 1, y = 2/3, which is in Quadrant I. With x = 2, y = −1, which is in Quadrant IV. Step 3: (ii) 5x − 3y = 7. With x = 1, y = −2/3, which is in Quadrant IV. With x = 2, y = 1, which is in Quadrant I. Step 4: (iii) −5x + 3y = 7. With x = −1, y = 2/3, which is in Quadrant II. With x = 1, y = 4, which is in Quadrant I. Step 5: (iv) −5x − 3y = 7. With x = −1, y = −2/3, which is in Quadrant III. With x = 1, y = −4, which is in Quadrant IV. Step 6: Plot each pair and join the two points. The resulting straight line verifies the equation.
5For the graph of 3x − 7y = 21, does C(2, 3) lie on the line? Does it satisfy the equation? Can a point not on the line satisfy the equation?
Answer / method
Step 1: Substitute C(2, 3) in 3x − 7y = 21. Step 2: 3(2) − 7(3) = 6 − 21 = −15. Step 3: Since −15 ≠ 21, C does not satisfy the equation and does not lie on the line. Conclusion: a point not on the graph cannot satisfy the equation, because the graph contains exactly all the solutions.
6State True or False and justify: (i) a linear equation in two variables has only one solution; (ii) its graph always passes through the origin; (iii) it can never have rational solutions; (iv) x = 3 is valid; (v) 2x + 3y = 7 has infinitely many solutions; (vi) (1, 2) solves 2x + 3y = 7.
Answer / method
Step 1: (i) False. Choose x freely and calculate y; normally infinitely many ordered pairs result. Step 2: (ii) False. The graph passes through (0, 0) only when the constant term is zero. Step 3: (iii) False. For example, 2x + 3y = 7 has the rational solution (2, 1). Step 4: (iv) True. x = 3 can be written as x + 0y − 3 = 0. Step 5: (v) True. One equation in two variables has infinitely many solutions when it is a valid linear equation. Step 6: (vi) False. 2(1) + 3(2) = 8, not 7.
7(i) Compare the solutions of 3x + 4y = 7 and 6x + 8y = 14. (ii) Show that ax + by = c and kax + kby = kc, where k ≠ 0, have the same solutions.
Answer / method
Step 1: 6x + 8y = 14 is exactly 2(3x + 4y = 7). Step 2: Multiplying an equation by a non-zero number does not change its solutions. Conclusion for (i): both equations have the same set of solutions. Step 3: If ax + by = c, multiply every term by k, where k ≠ 0. Step 4: The result is kax + kby = kc. Step 5: Dividing the second equation by k returns the first equation. Conclusion for (ii): the two equations have exactly the same solutions.

Exercise Set 13.3 — Solutions

Slope, steepness, ramps and graphical measurement.

1Three ski hills have rises and runs: A: 60 m and 70 m; B: 60 m and 110 m; C: 80 m and 100 m. Rank them from least to greatest steepness.
Answer / method
Step 1: Slope = rise/run. Step 2: A: slope = 60/70 = 6/7 ≈ 0.857. Step 3: B: slope = 60/110 = 6/11 ≈ 0.545. Step 4: C: slope = 80/100 = 4/5 = 0.8. Conclusion: least to greatest steepness is B, C, A.
2A loading-dock ramp rises 2.5 metres over a run of 4 metres. Find its slope.
Answer / method
Step 1: Rise = 2.5 m and run = 4 m. Step 2: Slope = rise/run = 2.5/4 = 0.625. Step 3: As a fraction, 0.625 = 5/8. Conclusion: the ramp slope is 5/8, or 0.625.
3Find the slope of line l in each of the four textbook graph diagrams (i), (ii), (iii) and (iv).
The four graph diagrams were not included in the supplied text, so the numerical slopes cannot be read reliably here.
Answer / method
Step 1: Read any two clear points (x₁, y₁) and (x₂, y₂) from each diagram. Step 2: Calculate slope = (y₂ − y₁)/(x₂ − x₁). Step 3: Keep the sign: a rising line has positive slope and a falling line has negative slope. Step 4: The four numerical answers cannot be identified because the four textbook diagrams were not included in the supplied material.
4An accessibility ramp has a guideline of 1 cm vertical rise to 12 cm horizontal length. What horizontal length is needed for stairs of height 18 cm?
Answer / method
Step 1: The guideline says 1 cm vertical rise needs 12 cm horizontal run. Step 2: For an 18 cm rise, horizontal length = 18 × 12. Step 3: 18 × 12 = 216 cm. Conclusion: the ramp should have a horizontal length of 216 cm, or 2.16 m.

Exercise Set 13.4 — Solutions

Making pairs of linear equations from everyday situations.

1A movie ticket costs ₹x and a snack box costs ₹y. On one day, 2 tickets and 3 snack boxes cost ₹850; on another day, 4 tickets and 1 snack box cost ₹1100. Frame the pair.
Answer / method
Step 1: Let one ticket cost ₹x and one snack box cost ₹y. Step 2: First day: 2 tickets and 3 boxes cost 850, so 2x + 3y = 850. Step 3: Second day: 4 tickets and 1 box cost 1100, so 4x + y = 1100. Conclusion: the pair is 2x + 3y = 850 and 4x + y = 1100.
2A taxi has fixed charge ₹x and additional charge ₹y per kilometre. Sahil travels 6 km for ₹122 and Meena travels 8 km for ₹160. Frame the pair.
Answer / method
Step 1: Let fixed taxi charge be ₹x and charge per kilometre be ₹y. Step 2: Sahil's fare gives x + 6y = 122. Step 3: Meena's fare gives x + 8y = 160. Conclusion: the pair is x + 6y = 122 and x + 8y = 160.
3Adult tickets cost ₹150 and child tickets cost ₹100. There are 200 people and ₹25,000 is collected. If adult tickets are x and child tickets are y, frame the pair.
Answer / method
Step 1: Let adult tickets be x and child tickets be y. Step 2: Total people: x + y = 200. Step 3: Total amount: 150x + 100y = 25000. Conclusion: the pair is x + y = 200 and 150x + 100y = 25000.

Exercise Set 13.5 — Solutions

Applications of simultaneous linear equations.

1Form a pair and solve: (i) sum of two integers is 5 and difference is −21; (ii) difference is 26 and one number is three times the other; (iii) 7 bats and 6 balls cost ₹8880, while 3 bats and 5 balls cost ₹4000; (iv) taxi totals ₹155 for 10 km and ₹220 for 15 km; (v) a fraction becomes 9/11 after adding 2 to numerator and denominator and 5/6 after adding 3; (vi) adding 1 to numerator and subtracting 1 from denominator gives 1, while adding 1 only to denominator gives 1/2; (vii) Nuri was thrice Sonu's age five years ago and twice ten years later; (viii) digit sum is 9 and nine times the number is twice its reverse; (ix) ₹2000 is withdrawn using ₹50 and ₹100 notes, 25 notes in all; (x) a library charges a fixed amount for three days and an extra daily charge thereafter, with ₹27 for seven days and ₹21 for five days.
Answer / method
Step 1: (i) Let the integers be x and y. x + y = 5 and x − y = −21. Adding gives 2x = −16, so x = −8. Then y = 13. Step 2: (ii) Let the numbers be x and y, with x − y = 26 and x = 3y. Then 3y − y = 26, so y = 13 and x = 39. Step 3: (iii) Let bat cost be b and ball cost be l. 7b + 6l = 8880 and 3b + 5l = 4000. Elimination gives l = 80 and b = 1200. Step 4: (iv) Let fixed charge be f and per-kilometre charge be r. f + 10r = 155 and f + 15r = 220. Subtracting gives 5r = 65, so r = 13 and f = 25. For 25 km, fare = 25 + 25(13) = 350. Step 5: (v) Let fraction be x/y. (x+2)/(y+2)=9/11 gives 11x−9y=−4. (x+3)/(y+3)=5/6 gives 6x−5y=−3. Solving gives x = 7 and y = 9. Fraction = 7/9. Step 6: (vi) Let fraction be x/y. (x+1)/(y−1)=1 gives y−x=2. x/(y+1)=1/2 gives 2x−y=1. Solving gives x = 3, y = 5. Fraction = 3/5. Step 7: (vii) Let present ages be N and S. N−5=3(S−5) and N+10=2(S+10). Solving gives S = 20 and N = 50. Step 8: (viii) Let the tens and units digits be a and b. a+b=9 and 9(10a+b)=2(10b+a). This gives b=8a, so a=1 and b=8. Number = 18. Step 9: (ix) Let ₹50 notes be x and ₹100 notes be y. x+y=25 and 50x+100y=2000. Dividing the second by 50 gives x+2y=40. Subtracting gives y=15 and x=10. Step 10: (x) Let fixed charge be f and extra daily charge be r. f+4r=27 and f+2r=21. Subtracting gives r=3 and f=15.
2Ten Grade 9 students took part in a quiz. The number of girls is 4 more than the number of boys. Find the number of boys and girls.
Answer / method
Step 1: Let boys = b and girls = g. Step 2: Total students gives b + g = 10. Step 3: Girls are four more, so g = b + 4. Step 4: b + b + 4 = 10, so 2b = 6 and b = 3. Step 5: g = 3 + 4 = 7. Conclusion: 3 boys and 7 girls.
3Use a₁/a₂, b₁/b₂ and c₁/c₂ to decide whether the pairs represent intersecting, parallel or coincident lines: (i) 5x − 4y + 8 = 0 and 7x + 6y − 9 = 0; (ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0; (iii) 6x − 3y + 10 = 0 and 2x − y + 9 = 0.
Answer / method
Step 1: (i) Ratios are 5/7 and (−4)/6. They are unequal, so the lines intersect at one point and the pair has a unique solution. Step 2: (ii) 9/18 = 3/6 = 12/24 = 1/2. All three ratios are equal, so the lines are coincident and there are infinitely many solutions. Step 3: (iii) 6/2 = (−3)/(−1) = 3, but 10/9 is not 3. Since the first two ratios are equal and the third is different, the lines are parallel and there is no solution.
4Which pairs have solutions? If they do, find them graphically: (i) x + y = 5, 2x + 2y = 10; (ii) x − y = 8, 3x − 3y = 16; (iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0; (iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0.
Answer / method
Step 1: (i) The second equation is twice the first. The lines coincide, so there are infinitely many solutions: every point on x + y = 5. Step 2: (ii) The first gives x − y = 8, while the second gives x − y = 16/3. The lines are parallel, so there is no solution. Step 3: (iii) 2x + y = 6 and 4x − 2y = 4. From the first, y = 6 − 2x. Substitute: 4x − 2(6 − 2x) = 4, so 8x = 16 and x = 2. Then y = 2. Solution: (2, 2). Step 4: (iv) The first gives x − y = 1, while the second gives x − y = 5/4. The lines are parallel, so there is no solution.
5Half the perimeter of a rectangular garden is 36 m, and its length is 4 m more than its width. Find its dimensions.
Answer / method
Step 1: Let width = w m. Then length = w + 4 m. Step 2: Half the perimeter equals length + width, so (w + 4) + w = 36. Step 3: 2w + 4 = 36, so 2w = 32 and w = 16. Step 4: Length = 16 + 4 = 20. Conclusion: dimensions are 20 m by 16 m.
6For 2x + 3y − 8 = 0, write another equation so that the pair represents (i) intersecting lines, (ii) parallel lines, (iii) coincident lines.
Answer / method
Step 1: Given equation is 2x + 3y − 8 = 0. Step 2: For intersecting lines, choose an equation not proportional to it, for example x + y − 1 = 0. Step 3: For parallel lines, keep the same x and y coefficients but change the constant: 2x + 3y − 5 = 0. Step 4: For coincident lines, multiply every term by a non-zero number: 4x + 6y − 16 = 0.
7Mahāvīrāchārya's problem: 9 citrons and 7 fragrant wood-apples cost 107, while 7 citrons and 9 fragrant wood-apples cost 101. Find each price without directly using elimination or substitution. Explain what is special about the equations.
Answer / method
Step 1: Let citron price be x and fragrant wood-apple price be y. Step 2: The equations are 9x + 7y = 107 and 7x + 9y = 101. Step 3: The coefficients of x and y are interchanged, so adding gives 16x + 16y = 208, hence x + y = 13. Step 4: Subtracting gives 2x − 2y = 6, hence x − y = 3. Step 5: Add the last two equations: 2x = 16, so x = 8. Step 6: Then y = 13 − 8 = 5. Conclusion: one citron costs 8 units and one fragrant wood-apple costs 5 units. The special interchange makes addition and subtraction immediately give x+y and x−y.
8Five pencils and seven pens cost ₹50, while seven pencils and five pens cost ₹46. Find the cost of each pencil and pen.
Answer / method
Step 1: Let pencil cost be p and pen cost be q. Step 2: 5p + 7q = 50 and 7p + 5q = 46. Step 3: Adding gives 12p + 12q = 96, so p + q = 8. Step 4: Subtracting the second equation from the first gives −2p + 2q = 4, so q − p = 2. Step 5: Adding p + q = 8 and q − p = 2 gives 2q = 10, so q = 5. Step 6: p = 8 − 5 = 3. Conclusion: pencil ₹3 and pen ₹5.
9Find the height of the stool shown in the textbook figure.
The supplied question image is cut off before the measurements are visible.
Answer / method
Step 1: The supplied figure is cut off before the numerical measurements are visible. Step 2: A stool-height question needs the labelled heights or lengths from the complete diagram. Conclusion: the height cannot be calculated reliably from the supplied material. The complete textbook figure is required.

End-of-Chapter Exercises — Solutions

Graphs, applications, parameters, robots and challenging word problems.

1For the graph y = 3x through (0, 0) and (2, 6): (i) identify the slope; (ii) identify the y-intercept; (iii) for a water tank following this graph, find height after 5 minutes and time for height 21 cm; (iv) test (4, 10); (v) find the intersection with x = 3.
Answer / method
Step 1: For y = 3x, use the two points (0, 0) and (2, 6). Step 2: Slope = (6 − 0)/(2 − 0) = 6/2 = 3. Step 3: Comparing y = mx + c with y = 3x gives m = 3 and c = 0. The y-intercept is (0, 0), showing the line passes through the origin. Step 4: Water after 5 minutes: y = 3(5) = 15 cm. Step 5: For height 21 cm, 21 = 3x, so x = 7 minutes. Step 6: At (4, 10), 3x = 12, not 10. Therefore the point is not on the line. Step 7: On x = 3, y = 3(3) = 9. The intersection is (3, 9).
2For F = (9/5)C + 32: (i) draw the Celsius–Fahrenheit graph; (ii) find F at 30°C; (iii) find C at 95°F; (iv) find F at 0°C and C at 0°F; (v) find a temperature numerically equal in both scales.
Answer / method
Step 1: Use F = (9/5)C + 32 and choose Celsius values for a table: (0,32), (10,50), (20,68), (30,86). Plot these points and join them. Step 2: At C = 30, F = (9/5)(30) + 32 = 54 + 32 = 86°F. Step 3: At F = 95, 95 = (9/5)C + 32, so 63 = (9/5)C and C = 35°C. Step 4: At C = 0, F = 32°F. At F = 0, 0 = (9/5)C + 32, so C = −160/9°C, approximately −17.78°C. Step 5: For the same numerical value, let F = C. Then C = (9/5)C + 32, so −4C/5 = 32 and C = −40. Therefore F = C = −40°.
3Solve graphically: 2x + y = 6 and 2x − y − 2 = 0.
Answer / method
Step 1: Rewrite the equations as 2x + y = 6 and 2x − y = 2. Step 2: Adding the equations gives 4x = 8, so x = 2. Step 3: Substitute x = 2 in 2x + y = 6: 4 + y = 6, so y = 2. Step 4: On a graph, the two lines meet at (2, 2). Conclusion: the graphical solution is (2, 2).
4Find the point of intersection of the lines shown on the cover page.
The cover-page graph was not included in the supplied text, so its intersection cannot be determined from the available material.
Answer / method
Step 1: The supplied text refers to a cover-page graph, but that graph is not included. Step 2: Read the coordinates of the crossing point from the complete cover image, or solve the two line equations if they are provided. Conclusion: the numerical intersection cannot be determined from the available source.
5Give a formula for the x-intercept of y = mx + c.
Answer / method
Step 1: At the x-intercept, y = 0. Step 2: Put y = 0 in y = mx + c: 0 = mx + c. Step 3: mx = −c, so x = −c/m, provided m ≠ 0. Conclusion: x-intercept = (−c/m, 0).
6Plan A costs ₹50 monthly plus ₹0.20 per minute. Plan B costs ₹30 monthly plus ₹0.30 per minute. When is each plan cheaper, and when are they equal?
Answer / method
Step 1: Let monthly minutes be t. Plan A costs 50 + 0.20t. Plan B costs 30 + 0.30t. Step 2: Plan A is cheaper when 50 + 0.20t < 30 + 0.30t. Thus 20 < 0.10t, so t > 200. Step 3: Plan B is cheaper when t < 200. Step 4: At equality, 50 + 0.20t = 30 + 0.30t, so t = 200. Conclusion: Plan B is cheaper below 200 minutes, both cost the same at 200 minutes, and Plan A is cheaper above 200 minutes.
7How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?
Answer / method
Step 1: Fix a slope m. Changing the y-intercept gives y = mx + c for different values of c. Conclusion for (i): infinitely many lines have a given slope. Step 2: If the slope and one point are fixed, the y-intercept is fixed by substituting the point in y = mx + c. Conclusion for (ii): exactly one line has a given slope and passes through a given point.
8For what values of p does 4x + py + 8 = 0 and 2x + 2y + 2 = 0 have a unique solution?
Answer / method
Step 1: The equations are 4x + py + 8 = 0 and 2x + 2y + 2 = 0. Step 2: The coefficient ratios are a₁/a₂ = 4/2 = 2 and b₁/b₂ = p/2. Step 3: A unique solution requires unequal coefficient ratios. Step 4: 2 ≠ p/2, so p ≠ 4. Conclusion: the pair has a unique solution for every p except p = 4.
9Find a and b so that (a + b)x − 2by = 5a + 2b + 1 and 3x − y = 14 have infinitely many solutions.
Answer / method
Step 1: For infinitely many solutions, all coefficient ratios must be equal. Step 2: (a+b)/3 = (−2b)/(−1) = 2b. Hence a + b = 6b, so a = 5b. Step 3: Also (5a + 2b + 1)/14 = 2b, so 5a + 2b + 1 = 28b. Step 4: Substitute a = 5b: 25b + 2b + 1 = 28b, so b = 1. Step 5: a = 5b = 5. Conclusion: a = 5 and b = 1.
10Find k so that x + 2y = 3 and (k − 1)x + (k + 1)y = k + 3 represent coincident lines.
Answer / method
Step 1: For coincident lines, (k−1)/1 = (k+1)/2 = (k+3)/3. Step 2: Equate the first two ratios: 2(k−1) = k + 1, so k = 3. Step 3: Check the third ratio: when k = 3, (k−1) = 2, (k+1)/2 = 2 and (k+3)/3 = 2. Conclusion: k = 3.
11(i) Robot 1 starts at the origin and moves 3 right, 4 up repeatedly. Robot 2 starts at (10, 0) and moves 1 right, 2 up repeatedly. Do their paths intersect? (ii) Robot 1 starts at (3, 0) and moves 5 right, 5 up. Robot 2 starts at (7, 0) and moves 5 right, 3 down. Do their paths intersect?
Answer / method
Step 1: Robot 1 moves 3 right and 4 up, so its path has slope 4/3 and passes through (0,0): y = 4x/3. Step 2: Robot 2 moves 1 right and 2 up, so its path has slope 2 and passes through (10,0): y = 2x − 20. Step 3: Equate the paths: 4x/3 = 2x − 20. Thus 4x = 6x − 60, so x = 30 and y = 40. Conclusion for (i): the paths intersect at (30, 40). Step 4: For (ii), Robot 1 has y = x − 3. Robot 2 has slope −3/5 through (7,0), so y = −3x/5 + 21/5. Step 5: Equating gives x − 3 = −3x/5 + 21/5, so x = 9/2 and y = 3/2. Step 6: This point has x = 4.5, which is before Robot 2's starting x-coordinate 7. Therefore the infinite lines meet there, but the forward paths traced by the robots do not intersect.
*12Jacob's watch reads a minutes after two o'clock. Fifteen minutes later it reads b minutes after three o'clock, and a is six times b. What time was the second reading?
Answer / method
Step 1: The first reading is 2:a, so the second reading 15 minutes later must cross 3 o'clock. Therefore a = b + 45. Step 2: Given a = 6b. Step 3: 6b = b + 45, so 5b = 45 and b = 9. Step 4: a = 6(9) = 54. Conclusion: the second reading was 3:09.
*13The digits of a two-digit number add to 15. Interchanging the digits increases the number by 9. Find the number.
Answer / method
Step 1: Let the tens digit be a and units digit be b. Step 2: a + b = 15. Step 3: Reversed number exceeds the original by 9: 10b + a = 10a + b + 9. Step 4: 9b − 9a = 9, so b − a = 1. Step 5: Adding a + b = 15 and b − a = 1 gives 2b = 16, so b = 8 and a = 7. Conclusion: the number is 78.
*14In cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles.
Answer / method
Step 1: Opposite angles of a cyclic quadrilateral are supplementary. Step 2: (x + 7) + (3y + 23) = 180 gives x + 3y = 150. Step 3: (y + 8) + (4x + 12) = 180 gives 4x + y = 160. Step 4: From 4x + y = 160, y = 160 − 4x. Substitute: x + 3(160 − 4x) = 150. Step 5: −11x = −330, so x = 30 and y = 40. Step 6: ∠A = 37°, ∠B = 48°, ∠C = 143° and ∠D = 132°.
*15A train takes 6 hours less for a distance when its speed is increased by 6 km/h, and 6 hours more when speed is decreased by 4 km/h. Find distance and original speed.
Answer / method
Step 1: Let original speed be v km/h and original time be t hours. Distance d = vt. Step 2: Increased speed gives d = (v + 6)(t − 6). Expanding and comparing with vt gives t − v = 6. Step 3: Decreased speed gives d = (v − 4)(t + 6). Comparing with vt gives 3v − 2t = 12. Step 4: Substitute t = v + 6 into 3v − 2t = 12: 3v − 2(v + 6) = 12. Step 5: v = 24 km/h and t = 30 hours. Step 6: Distance d = vt = 24 × 30 = 720 km. Conclusion: original speed is 24 km/h and distance is 720 km.
*16A father's age equals the sum of the ages of his four children. After 20 years, the children's age sum will be twice the father's age. Find the father's age.
Answer / method
Step 1: Let father's present age be F and the sum of the children's present ages be S. Step 2: Given F = S. Step 3: After 20 years, father is F + 20 and the four children together are S + 80. Step 4: Given S + 80 = 2(F + 20). Step 5: Substitute S = F: F + 80 = 2F + 40, so F = 40. Conclusion: the father is 40 years old. The four children's present ages together total 40 years.
Source note: The supplied text did not include the textbook diagrams for Exercise 13.3 Q3, the stool question, the cover-page intersection or some graph-based prompts. Their solutions explain the exact method and identify where the missing measurement is essential.

Mini Test - 10 Questions

Select one option for each question.

Q1. The standard form of a linear equation in two variables is:

Q2. The graph of a linear equation in two variables is a:

Q3. The slope of a line is:

Q4. A pair of coincident lines has:

Q5. The y-intercept of y = 3x + 5 is:

Q6. The slope of a ramp rising 2 m over a run of 8 m is:

Q7. If two coefficient ratios are unequal, the pair of lines has:

Q8. The solution of 2x + y = 6 and 2x − y = 2 is:

Q9. At what Celsius temperature are Celsius and Fahrenheit numerically equal?

Q10. A line with a fixed slope and a fixed point is:

Score: 0 / 10