NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 13 Two Variables One Line Examples

Examples

Example 1. Write each of the following equations in standard form, ax + by + c = 0, and indicate the values of a, b and c in each case:
(i) 5x + 2y = 2.7   (ii) \( \frac{x}{2} \) − 4 = \( \frac{3}{2} \)y   (iii) 4 = 1.5x − 1.3y   (iv) 3y = 5
Answer:
To reach standard form, shift every term to the left side so that the right side is 0.
(i) Take 2.7 to the left: 5x + 2y − 2.7 = 0. So a = 5, b = 2, c = −2.7.
(Multiplying by −1 gives −5x − 2y + 2.7 = 0, with a = −5, b = −2, c = 2.7, which is equally correct.)
(ii) Take \( \frac{3}{2} \)y to the left: \( \frac{x}{2} \) − \( \frac{3}{2} \)y − 4 = 0. So a = \( \frac{1}{2} \), b = \( -\frac{3}{2} \), c = −4.
(Doubling every term gives x − 3y − 8 = 0, with a = 1, b = −3, c = −8.)
(iii) Subtract 4 from both sides and write the x-term first: 1.5x − 1.3y − 4 = 0. So a = 1.5, b = −1.3, c = −4.
(Or −1.5x + 1.3y + 4 = 0, with a = −1.5, b = 1.3, c = 4.)
(iv) There is no x-term, so write it with a zero coefficient: 0 × x + 3y − 5 = 0. So a = 0, b = 3, c = −5.

Example 2. Find any two solutions of 2x + y = 7. Draw the graph of this equation. Use the graph to find 4 more solutions.
Answer:
Two solutions:
x = 0: 0 + y = 7, so y = 7. Solution (0, 7).
y = 0: 2x = 7, so x = \( \frac{7}{2} \). Solution (\( \frac{7}{2} \), 0).
Graph: plot A(0, 7) and B(3.5, 0) and draw the straight line through them.

Four more solutions read from the line: (1, 5), (2, 3), (3, 1) and (\( \frac{3}{2} \), 4).
Check: 2(1) + 5 = 7, 2(2) + 3 = 7, 2(3) + 1 = 7 and 2(\( \frac{3}{2} \)) + 4 = 3 + 4 = 7 ✓.

Example 3. Find any two solutions for each of the following equations and draw their graphs:
(i) 4x + 3y = 12   (ii) 2x + 5y = 0
Answer:
(i) x = 0 gives 3y = 12, y = 4, so (0, 4) is a solution. y = 0 gives 4x = 12, x = 3, so (3, 0) is a solution.
Plot B(0, 4) and A(3, 0) and join them.

(ii) x = 0 gives 5y = 0, so y = 0: the point (0, 0). Putting y = 0 gives the same point, so try other values of x.
x = 1: 2 + 5y = 0, so y = \( -\frac{2}{5} \). x = 5: 10 + 5y = 0, so y = −2.
Solutions: (0, 0), (1, \( -\frac{2}{5} \)) and (5, −2). Plotting A, B, C shows they lie on one line, which passes through the origin.

Example 4. In the equations 3ax + 4y = −2 and 2x + by = 14, a and b are unknown numbers. If (−3, 4) is the solution of both equations, find the values of a and b.
Answer:
Since (−3, 4) satisfies both equations, put x = −3 and y = 4 in each.
Step 1 (first equation): 3a(−3) + 4(4) = −2 → −9a + 16 = −2 → −9a = −18 → a = 2.
Step 2 (second equation): 2(−3) + b(4) = 14 → −6 + 4b = 14 → 4b = 20 → b = 5.
Check: 3(2)(−3) + 16 = −18 + 16 = −2 ✓ and 2(−3) + 5(4) = −6 + 20 = 14 ✓.
a = 2 and b = 5.

Example 5. Consider the straight line that passes through A(−1, 0), B(1, 2) and C(4, 5). Plot the points and find the slope of the line.
Answer:
Plot the three points and draw the line through them. From B go across horizontally, and from C go down vertically; the two meet at P(4, 2).

Run BP = 4 − 1 = 3 and rise CP = 5 − 2 = 3.
Slope = \( \frac{CP}{BP} \) = \( \frac{5-2}{4-1} \) = \( \frac{3}{3} \) = 1.
Using A and B: \( \frac{2-0}{1-(-1)} \) = \( \frac{2}{2} \) = 1. Using A and C: \( \frac{5-0}{4-(-1)} \) = \( \frac{5}{5} \) = 1.
Every pair of points gives the same value, so the slope of the line is 1.

Example 6. Consider the line which passes through A(−4, 5) and B(−1, 2). Find the slope of the line.
Answer:
Take A(−4, 5) as (x₁, y₁) and B(−1, 2) as (x₂, y₂).
Slope = \( \frac{y_2 - y_1}{x_2 - x_1} \) = \( \frac{2-5}{-1-(-4)} \) = \( \frac{-3}{3} \) = −1.
Reversing the order: \( \frac{5-2}{-4-(-1)} \) = \( \frac{3}{-3} \) = −1, the same value.

The slope is −1. The minus sign tells us the line falls as we move from left to right.

Example 7. Solve the following pair of equations by substitution method:
7x − 15y = 2 ... (1)
x + 2y = 3 ... (2)
Answer:
Step 1: From (2), x = 3 − 2y ... (3).
Step 2: Put this in (1): 7(3 − 2y) − 15y = 2 → 21 − 14y − 15y = 2 → −29y = −19 → y = \( \frac{19}{29} \).
Step 3: From (3): x = 3 − 2 × \( \frac{19}{29} \) = \( \frac{87 - 38}{29} \) = \( \frac{49}{29} \).
Check (1): 7 × \( \frac{49}{29} \) − 15 × \( \frac{19}{29} \) = \( \frac{343 - 285}{29} \) = \( \frac{58}{29} \) = 2 ✓.
Check (2): \( \frac{49}{29} \) + 2 × \( \frac{19}{29} \) = \( \frac{87}{29} \) = 3 ✓.
x = \( \frac{49}{29} \), y = \( \frac{19}{29} \).

Teacher's Note:
1. Choose the equation where a variable has coefficient 1 to begin the substitution.
2. Fractional answers are fine; always verify them in both equations.

Example 8. If two angles of a triangle are x° and y° with y = 4x, and the third angle is 50°, what are the angles?
Answer:
By the angle sum property: x + y + 50 = 180 ... (1). Also y = 4x ... (2).
Substituting (2) in (1): x + 4x + 50 = 180 → 5x = 130 → x = 26.
Then y = 4 × 26 = 104.
Check: 26° + 104° + 50° = 180° ✓.
The angles are 26°, 104° and 50°.

Example 9. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.
Answer:
Let the incomes be ₹9x and ₹7x, and the expenditures be ₹4y and ₹3y.
Saving = income − expenditure: 9x − 4y = 2000 ... (1) and 7x − 3y = 2000 ... (2).
Step 1: Make the y-coefficients equal: (1) × 3 gives 27x − 12y = 6000 ... (3); (2) × 4 gives 28x − 12y = 8000 ... (4).
Step 2: (4) − (3): x = 2000.
Step 3: In (1): 18000 − 4y = 2000 → 4y = 16000 → y = 4000.
Incomes: 9 × 2000 = 18000 and 7 × 2000 = 14000.
Check: expenditures are 16000 and 12000, in the ratio 4 : 3 ✓, and each saving is ₹2000 ✓.
The monthly incomes are ₹18,000 and ₹14,000.

Teacher's Note:
1. For a ratio 9 : 7, write the quantities as 9x and 7x.
2. Pick multipliers (3 and 4) that make one variable's coefficients equal, then subtract.

Example 10. The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Answer:
Let the tens digit be x and the units digit be y. Number = 10x + y; reversed number = 10y + x.
Sum: (10x + y) + (10y + x) = 66 → 11x + 11y = 66 → x + y = 6 ... (1).
The digits differ by 2, which can happen in two ways.
Case 1: x − y = 2. Adding to (1): 2x = 8, so x = 4, y = 2. Number = 42.
Case 2: y − x = 2. Adding to (1): 2y = 8, so y = 4, x = 2. Number = 24.
Check: 42 + 24 = 66 ✓ and 4 − 2 = 2 ✓.
So there are two such numbers, 42 and 24. If we assume the tens digit is the larger one, the number is 42.

Teacher's Note:
1. 'Differ by 2' does not say which digit is bigger, so both cases must be tried.
2. The two answers are reverses of each other, which is a good check.

Example 11. Solve the pair of linear equations x + 3y = 6 and 2x − 3y = 12 graphically. Comment on the nature of the solution.
Answer:
For x + 3y = 6: x = 0 gives y = 2; x = 6 gives y = 0. Points A(0, 2), B(6, 0).
For 2x − 3y = 12: x = 0 gives y = −4; x = 3 gives y = −2. Points P(0, −4), Q(3, −2).
Draw lines AB and PQ.

The lines meet at B(6, 0). Check: 6 + 0 = 6 ✓ and 12 − 0 = 12 ✓.
Solution: x = 6, y = 0. The lines cross at exactly one point, so the pair has a unique solution.
Ratio check: \( \frac{a_1}{a_2} \) = \( \frac{1}{2} \) and \( \frac{b_1}{b_2} \) = \( \frac{3}{-3} \) = −1, which are unequal ✓.

Example 12. Determine if the lines x + 2y − 4 = 0 and 2x + 4y − 12 = 0 intersect, coincide or are parallel to each other. Then verify your answer by graphing the equations.
Answer:
a₁ = 1, b₁ = 2, c₁ = −4; a₂ = 2, b₂ = 4, c₂ = −12.
\( \frac{a_1}{a_2} \) = \( \frac{1}{2} \), \( \frac{b_1}{b_2} \) = \( \frac{2}{4} \) = \( \frac{1}{2} \), \( \frac{c_1}{c_2} \) = \( \frac{-4}{-12} \) = \( \frac{1}{3} \).
Since \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \), the lines are parallel and the pair has no solution.
Graph: x + 2y − 4 = 0 passes through R(0, 2) and S(4, 0); 2x + 4y − 12 = 0 passes through P(0, 3) and Q(6, 0).

The two lines never meet; both have slope \( -\frac{1}{2} \). This confirms they are parallel.

Example 13. Romila went to a stationery shop and purchased 2 erasers and 3 A4 sheets for ₹9. Her friend Sonali saw the new variety of erasers and A4 sheets Romila bought, and she also bought 4 erasers and 6 A4 sheets of the same kind for ₹18. Represent this situation graphically.
Answer:
Let one eraser cost ₹x and one A4 sheet cost ₹y. Then 2x + 3y = 9 and 4x + 6y = 18.
Table for both equations: x = 0, y = 3; x = 3, y = 1; x = 4.5, y = 0. The two equations give the same points.

The two graphs are the same line (coincident lines).
Ratios: \( \frac{2}{4} \) = \( \frac{3}{6} \) = \( \frac{-9}{-18} \) = \( \frac{1}{2} \), all equal, so there are infinitely many solutions.
For instance, ₹3 per eraser with ₹1 per sheet, or ₹1.50 per eraser with ₹2 per sheet, both fit. The data is not enough to fix the prices.

Think and Reflect and In-text Questions

Think and Reflect (p. 88). Are (1, \( \frac{9}{2} \)), (0, 6), (4, 1) and (\( \frac{1}{2} \), \( \frac{21}{4} \)) solutions of the equation 3x + 2y = 12? Can you find any other solution? How many solutions can you find?
Answer:
Substitute each pair in 3x + 2y and compare with 12.
(1, \( \frac{9}{2} \)): 3 + 9 = 12 ✓, a solution.
(0, 6): 0 + 12 = 12 ✓, a solution.
(4, 1): 12 + 2 = 14 ≠ 12, not a solution.
(\( \frac{1}{2} \), \( \frac{21}{4} \)): \( \frac{3}{2} \) + \( \frac{21}{2} \) = \( \frac{24}{2} \) = 12 ✓, a solution.
Other solutions: use y = \( \frac{12 - 3x}{2} \). x = −2 gives y = 9; x = 2 gives 3; x = 4 gives 0; x = 6 gives −3. So (−2, 9), (2, 3), (4, 0), (6, −3) are also solutions.

How many? Each value of x gives exactly one y, and x can take infinitely many values, so there are infinitely many solutions. All of them lie on one line; (4, 1) is off the line.

Think and Reflect (p. 92). Four friends, Ranju, Meena, Farhan and Toshi, are solving problems.
Ranju: If c = 0 in the standard form of a line ax + by + c = 0, then the line must pass through the origin. If I substitute x = 0 and y = 0 in ax + by = 0, the equation is satisfied. So the origin lies on the line.
Farhan: Let us try 2x + 3y = 0. Here a = 2, b = 3 and c = 0. If we substitute x = 0, we get 3y = 0 or y = 0. So (0, 0) lies on the line.
Toshi: Suppose b = c = 0, say 5x = 0. This becomes x = 0, which is the equation of the y-axis, and the y-axis passes through the origin.
Meena: If a = c = 0, say 7y = 0, that means y = 0, which is the x-axis, and that also passes through the origin.
They conclude: whenever c = 0, the line ax + by + c = 0 always passes through the origin, whatever the values of a and b. Do you agree with them?
Answer:
Yes, they are right.
With c = 0 the equation is ax + by = 0. Putting (0, 0): a × 0 + b × 0 = 0 = RHS, true for every a and b. So the origin is always on the line.
Their examples cover every possible case (a and b are never both zero):
1. a ≠ 0 and b ≠ 0 (Farhan, 2x + 3y = 0): a slanting line through the origin.
2. b = 0, a ≠ 0 (Toshi, 5x = 0, i.e. x = 0): the y-axis.
3. a = 0, b ≠ 0 (Meena, 7y = 0, i.e. y = 0): the x-axis.
The converse holds too: if (0, 0) lies on ax + by + c = 0, then 0 + 0 + c = 0, so c = 0. Hence a line passes through the origin if and only if c = 0.

Teacher's Note:
1. Substituting (0, 0) is the one-line proof; the friends' examples only illustrate it.
2. Mention the converse so students see the condition works both ways.

Think and Reflect (p. 97). The line AB passes through A(2, 0) and B(4, 3).
1. Consider the point on the line AB whose x-coordinate is 8. What is the y-coordinate of this point? Express this as an ordered pair.
2. What is the x-coordinate of the point on the line AB whose y-coordinate is 6? Express this as an ordered pair.
Answer:
Slope of AB = \( \frac{3 - 0}{4 - 2} \) = \( \frac{3}{2} \): every 2 units to the right, the line goes up 3 units.
1. From x = 2 to x = 8 is a run of 6, i.e. three steps of 2, so the rise is 3 × 3 = 9. y = 0 + 9 = 9.
Check with the slope formula: \( \frac{y - 0}{8 - 2} \) = \( \frac{3}{2} \) → y = \( \frac{3}{2} \) × 6 = 9. The point is (8, 9).
2. Let the point be (x, 6). Using A: \( \frac{6 - 0}{x - 2} \) = \( \frac{3}{2} \) → 3(x − 2) = 12 → x − 2 = 4 → x = 6. The point is (6, 6).

Think and Reflect (p. 99). The line y = −5x + 3 passes through the points A(0, 3), B(1, −2) and C(−1, 8). Can you use the points A, B and C to verify that the slope of this line is indeed −5?
Answer:
Using A and B: \( \frac{-2 - 3}{1 - 0} \) = \( \frac{-5}{1} \) = −5.
Using B and C: \( \frac{8 - (-2)}{-1 - 1} \) = \( \frac{10}{-2} \) = −5.
Using A and C: \( \frac{8 - 3}{-1 - 0} \) = \( \frac{5}{-1} \) = −5.

Every pair gives −5, so yes, the slope is −5, which is the value of m in y = −5x + 3.
(Special cases: y = 5 is horizontal, slope 0; x = 2 is vertical, slope undefined.)

Think and Reflect (p. 104). The cost of one paratha and 2 bowls of dahi is ₹90. Also, the cost of 2 parathas and 7 bowls of dahi is ₹260. Can you figure out the cost of one paratha and one bowl of dahi and explain your reasoning?
Answer:
Let a paratha cost ₹x and a bowl of dahi ₹y: x + 2y = 90 ... (1), 2x + 7y = 260 ... (2).
Reasoning: doubling the first order gives 2 parathas + 4 bowls for ₹180. The second order has the same 2 parathas but 3 more bowls and costs ₹80 more, so 3 bowls cost ₹80.
Algebraically: (2) − 2 × (1): 3y = 80, y = \( \frac{80}{3} \) = 26\( \frac{2}{3} \).
From (1): x = 90 − \( \frac{160}{3} \) = \( \frac{110}{3} \) = 36\( \frac{2}{3} \).
Check: \( \frac{110}{3} \) + \( \frac{160}{3} \) = \( \frac{270}{3} \) = 90 ✓; \( \frac{220}{3} \) + \( \frac{560}{3} \) = \( \frac{780}{3} \) = 260 ✓.
One paratha costs ₹\( \frac{110}{3} \) ≈ ₹36.67 and one bowl of dahi ₹\( \frac{80}{3} \) ≈ ₹26.67. (With the prices in the book the answers are not whole rupees; the method is what matters.)

Teacher's Note:
1. The 'double the first order' argument is elimination in words; show students the link.
2. Non-integer prices are fine here; do not let students think they have made a mistake.

In-text Rakesh's Puzzle (p. 104). Do you remember Rakesh's puzzle from Grade 7? "I have thought of two numbers", he said. "Their sum is 25 and their difference is 11." We solved the problem by making a table and listing possible combinations (e.g. 10 and 15: sum 25, difference −5; 20 and 5: sum 25, difference 15; 18 and 7: sum 25, difference 11). We can also solve this using a pair of linear equations.
Answer:
Let the numbers be x and y: x + y = 25 ... (1), x − y = 11 ... (2).
Adding: 2x = 36, x = 18. From (1): y = 7.
Check: 18 + 7 = 25 ✓; 18 − 7 = 11 ✓.
The numbers are 18 and 7, the same as the last row of the table.

In-text Question (p. 104). A science exhibition charges an entry fee that is different for adults and children. Group A paid ₹600 for 2 adult tickets and 3 child tickets; Group B paid ₹700 for 3 adult tickets and 2 child tickets. If one adult ticket costs ₹x and one child ticket ₹y, then 2x + 3y = 600 and 3x + 2y = 700. A solution to these equations will give us the answer that we seek. Can you see why? How much do individual adult and child tickets cost?
Answer:
Why: x and y are exactly the unknown prices. The true prices must make the statements about both groups true at the same time, so they must satisfy both equations together. The pair (x, y) that does this is the answer.
Prices: 2x + 3y = 600 ... (1), 3x + 2y = 700 ... (2).
Adding: 5x + 5y = 1300 → x + y = 260 ... (3). (2) − (1): x − y = 100 ... (4).
(3) + (4): 2x = 360, x = 180; y = 80.
Check: 360 + 240 = 600 ✓; 540 + 160 = 700 ✓.
An adult ticket costs ₹180 and a child ticket ₹80.

Think and Reflect (p. 106). Solve the puzzle and discuss your strategy with your friends. Can you determine the value represented by each shape? (Row totals are on the right, column totals below.)

Answer:
Let triangle = T, rectangle = R, hexagon = H, circle = C.
Row 1: 2T + 2R = 28 → T + R = 14 ... (1).
Row 2: 2H + 2R = 30 → H + R = 15 ... (2).
Row 3: 3C + T = 18 ... (3). Row 4: 3C + R = 20 ... (4).
Strategy: column 4 (T, R, C, C) gives T + R + 2C = 22. By (1), 14 + 2C = 22, so C = 4.
From (3): 12 + T = 18, T = 6. From (4): 12 + R = 20, R = 8 (check (1): 6 + 8 = 14 ✓). From (2): H = 7.
Column checks: column 2 (R, R, T, R) = 8 + 8 + 6 + 8 = 30 ✓; column 3 (R, H, C, C) = 8 + 7 + 4 + 4 = 23 ✓; column 4 = 6 + 8 + 4 + 4 = 22 ✓.
Missing total, column 1 (T, H, C, C) = 6 + 7 + 4 + 4 = 21.
Triangle = 6, Rectangle = 8, Hexagon = 7, Circle = 4, and ? = 21.

Teacher's Note:
1. Look for a row or column that combines an equation you already have; column 4 contains T + R.
2. Use the remaining column totals as checks.

Think and Reflect (p. 107). In Example 7 (7x − 15y = 2 and x + 2y = 3), what if we had expressed y in terms of x, i.e. taken Equation (2) and written y = \( \frac{1}{2} \)(3 − x)? Would we still get the same solution?
Answer:
Yes. Substitute y = \( \frac{1}{2} \)(3 − x) in 7x − 15y = 2:
7x − \( \frac{15}{2} \)(3 − x) = 2 → (× 2) 14x − 15(3 − x) = 4 → 14x − 45 + 15x = 4 → 29x = 49 → x = \( \frac{49}{29} \).
y = \( \frac{1}{2} \)(3 − \( \frac{49}{29} \)) = \( \frac{1}{2} \) × \( \frac{38}{29} \) = \( \frac{19}{29} \).
Same solution: x = \( \frac{49}{29} \), y = \( \frac{19}{29} \). The pair has only one solution, so the order of substitution does not matter; x = 3 − 2y was just more convenient because it avoided fractions at the start.

Think and Reflect (p. 109). The method used in Example 9 is called the Elimination Method because we eliminate one of the variables to obtain a linear equation in the other variable. In the example we eliminated y. Try the same problem (9x − 4y = 2000 and 7x − 3y = 2000) by eliminating x instead of y. Also try to solve the two equations using the Substitution Method. Assess the pros and cons of each method.
Answer:
(a) Eliminating x: (1) × 7: 63x − 28y = 14000 ... (3); (2) × 9: 63x − 27y = 18000 ... (4).
(4) − (3): y = 4000. Then 9x − 16000 = 2000, 9x = 18000, x = 2000.
(b) Substitution: from (1), x = \( \frac{2000 + 4y}{9} \). In (2): 7 × \( \frac{2000 + 4y}{9} \) − 3y = 2000 → (× 9) 14000 + 28y − 27y = 18000 → y = 4000; x = \( \frac{18000}{9} \) = 2000.
Every method gives x = 2000, y = 4000: incomes ₹18,000 and ₹14,000.
Elimination, pros: neat with whole-number coefficients, avoids fractions, fast when a variable already has equal or opposite coefficients. Cons: needs suitable multipliers, and numbers can grow large (63, 14000).
Substitution, pros: natural when a variable is already alone or has coefficient 1 (like y = 4x in Example 8). Cons: often brings in fractions such as \( \frac{2000 + 4y}{9} \), making the work longer and error-prone.
Here, eliminating y (multipliers 3 and 4) was the easiest.

Teacher's Note:
1. Choose the variable whose coefficients need the smallest multipliers to match.
2. Substitution is best when some coefficient is already 1.

Think and Reflect (p. 109). Does a pair of linear equations always have a unique solution? Can you analyse this using the Elimination Method? While subtracting the equations to eliminate a particular variable, if the other variable remains, then clearly there is a unique solution. But what if the second variable gets eliminated too while eliminating the first variable? Can you think of a situation where this happens?
Answer:
No, the solution is not always unique.
When we match the x-coefficients and subtract, y also disappears if its coefficients have become equal at the same time. This happens when the left side of one equation is a multiple of the other's. Only a statement about numbers is left:
1. A true statement such as 0 = 0: the equations are really the same, so there are infinitely many solutions. Example: x − y = 10 and 10x − 10y = 100 (10 × first, then subtract: 0 = 0).
2. A false statement such as 0 = 1: the equations contradict each other, so there is no solution. Example: x − y = 10 and 10x − 10y = 101 (subtract: 0 = 1).

Think and Reflect (p. 110). How many solutions are there to the equations 10x − 5y = 20 and 4x − 2y = 8? Give 3 more examples of pairs of equations that lead to 0 = 0 after subtraction, and therefore have infinitely many solutions. Can you find a simple rule to check when this happens?
Answer:
Divide the first by 5 and the second by 2: both become 2x − y = 4. They are the same equation.
By elimination: (first) × 2 and (second) × 5 both give 20x − 10y = 40; subtracting leaves 0 = 0.
So there are infinitely many solutions: every solution of 2x − y = 4, such as (0, −4), (2, 0), (3, 2), (5, 6), ….
Three more examples:
1. x + y = 3 and 2x + 2y = 6 (second = 2 × first).
2. 3x − y = 1 and 9x − 3y = 3 (second = 3 × first).
3. x − 2y = 5 and −2x + 4y = −10 (second = −2 × first).
Rule: for a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 this happens when one equation is a constant multiple of the other, i.e. \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = \( \frac{c_1}{c_2} \).
For the given pair (10x − 5y − 20 = 0, 4x − 2y − 8 = 0): \( \frac{10}{4} \) = \( \frac{-5}{-2} \) = \( \frac{-20}{-8} \) = \( \frac{5}{2} \) ✓.

Think and Reflect (p. 110). What if one of a₂, b₂, c₂ is zero?
Answer:
Then the matching ratio (for example \( \frac{c_1}{c_2} \)) cannot be formed, since division by zero is not defined, so the ratio test cannot be used as it stands.
Use the multiplied form instead: a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ for some k ≠ 0.
So wherever the second equation has a zero coefficient, the first must also have a zero there, and the other coefficients must be in the same ratio.
Examples:
1. x + 2y = 0 and 3x + 6y = 0: c₁ = c₂ = 0 and \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = \( \frac{1}{3} \). Same line → infinitely many solutions.
2. x + 2y − 1 = 0 and 3x + 6y = 0: c₂ = 0 but c₁ ≠ 0. a and b are in the same ratio, the constants are not → parallel lines → no solution.
3. 2y − 4 = 0 and 3y − 6 = 0 (both a = 0): \( \frac{b_1}{b_2} \) = \( \frac{c_1}{c_2} \) = \( \frac{2}{3} \). Both are the line y = 2 → infinitely many solutions.

Teacher's Note:
1. Writing the condition as a₁ = ka₂ etc. avoids dividing by zero.
2. A zero in one equation must be matched by a zero in the other for the lines to coincide.

Think and Reflect (p. 110). Does a pair of linear equations always have either a unique solution or infinitely many solutions?
Answer:
No. A pair can also have no solution.
Example: x − y = 10 ... (1) and 10x − 10y = 101 ... (2). Multiplying (1) by 10 gives 10x − 10y = 100.
The left sides of this and (2) are identical but the right sides (100 and 101) differ. No (x, y) can make 10x − 10y equal to both, so there is no solution. On a graph, the lines are parallel.
So there are three possibilities: a unique solution, infinitely many solutions, or no solution.

Think and Reflect (p. 111). Give 3 more examples of pairs of equations that have no solution. Can you find a simple rule to check when this will happen?
Answer:
Examples:
1. x + y = 3 and x + y = 5 (the same x + y cannot be both 3 and 5).
2. 2x − y = 1 and 4x − 2y = 5 (doubling the first gives 4x − 2y = 2 ≠ 5).
3. x + 3y = 4 and 3x + 9y = 10 (tripling the first gives 3x + 9y = 12 ≠ 10).
In each case, making the x-coefficients equal also makes the y-coefficients equal, and subtracting leaves a false statement such as 0 = 3.
Rule: there is no solution when \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \).
For example 2 (2x − y − 1 = 0, 4x − 2y − 5 = 0): \( \frac{2}{4} \) = \( \frac{-1}{-2} \) = \( \frac{1}{2} \), but \( \frac{-1}{-5} \) = \( \frac{1}{5} \) ≠ \( \frac{1}{2} \) ✓.

Think and Reflect (p. 112). The ratio rules are: (1) if \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \), the pair has a unique solution; (2) if \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = \( \frac{c_1}{c_2} \), it has infinitely many solutions; (3) if \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \), it has no solution. Are the converses of these statements true?
Answer:
Yes, all three converses are true.
The three ratio conditions cover every possibility and no two can hold together. The three kinds of outcome (unique, infinitely many, none) also cannot happen together.
Converse of (1): if the solution is unique, conditions (2) and (3) cannot hold (they give infinitely many or none), so \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \).
Converse of (2): if there are infinitely many solutions, it is neither case (1) nor case (3), so \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = \( \frac{c_1}{c_2} \).
Converse of (3): if there is no solution, it is neither case (1) nor case (2), so \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \).
So the rules work in both directions and can be used to decide the nature of the solutions.

Think and Reflect (p. 112). Can there be methods other than elimination and substitution to reduce a pair of linear equations in two variables to a linear equation in one variable? If so, describe them.
Answer:
Yes. Some other methods:
1. Comparison (equating) method: write the same variable in terms of the other from both equations and set the two expressions equal. For x + y = 25 and x − y = 11: y = 25 − x and y = x − 11, so 25 − x = x − 11, giving x = 18, y = 7.
2. Adding and subtracting when the coefficients are interchanged, e.g. 9x + 7y = 107 and 7x + 9y = 101: adding gives 16(x + y) = 208, so x + y = 13; subtracting gives 2(x − y) = 6, so x − y = 3 (see Exercise Set 13.5, Question 7).
3. Graphical method: draw both lines and read off their point of intersection (Section 13.6).
4. Cross-multiplication method (in higher classes), which gives formulas for x and y directly in terms of the coefficients.

In-text Activity (p. 113). Consider the following pairs of linear equations in two variables.
Pair 1: 2x + y = 6 and x − y = 2
Pair 2: x + y = 4 and 2x + 2y = 8
Pair 3: 3x − 2y = 6 and 6x − 4y = 12
Pair 4: x − 2y = 4 and 2x − 4y = 6
For each pair, prepare a table of values (with at least two ordered pairs). Plot the points on the Cartesian plane. Draw the straight lines.
Answer:
Tables of values (three points each):
2x + y = 6: (0, 6), (3, 0), (1, 4).   x − y = 2: (0, −2), (2, 0), (4, 2).
x + y = 4: (0, 4), (4, 0), (1, 3).   2x + 2y = 8: (0, 4), (4, 0), (2, 2).
3x − 2y = 6: (0, −3), (2, 0), (4, 3).   6x − 4y = 12: (0, −3), (2, 0), (−2, −6).
x − 2y = 4: (0, −2), (4, 0), (2, −1).   2x − 4y = 6: (0, −1.5), (3, 0), (1, −1).




Observations:
Pair 1: the lines cross at one point, (\( \frac{8}{3} \), \( \frac{2}{3} \)) (adding the equations gives 3x = 8). Unique solution.
Pair 2: the second equation is 2 × the first, so both give the same line. Coincident; infinitely many solutions.
Pair 3: again the second is 2 × the first. Coincident; infinitely many solutions.
Pair 4: both lines have slope \( \frac{1}{2} \) but cut the y-axis at −2 and −1.5. Parallel; no solution.

Teacher's Note:
1. Plotting a third point for each line catches arithmetic slips.
2. Ask students to link each picture to the ratio test they already know.

Think and Reflect (p. 113). How do we find solutions from the graph? What can we say about the number of solutions when the two lines (i) intersect at a point, (ii) are parallel, (iii) are coincident?
Answer:
A solution of the pair must satisfy both equations, so it is a point lying on both lines. We read the coordinates of the common point(s) from the graph.
(i) Intersecting lines: exactly one common point, so a unique solution (the point of intersection).
(ii) Parallel lines: they never meet, so no solution.
(iii) Coincident lines: every point of one line is on the other, so infinitely many solutions.

Think and Reflect (p. 115). Can you prove that two lines of equal slope are parallel? (Hint: Consider the equations of the two lines to be y = mx + d₁ and y = mx + d₂.)
Answer:
Given: two different lines y = mx + d₁ and y = mx + d₂ with the same slope m and d₁ ≠ d₂.
To prove: they have no common point, i.e. they are parallel.

Proof: suppose (p, q) lies on both. Then q = mp + d₁ and q = mp + d₂.
Subtracting: 0 = d₁ − d₂, i.e. d₁ = d₂, which contradicts d₁ ≠ d₂.
So the lines have no common point. Two lines in a plane that never meet are parallel; hence two different lines with equal slope are parallel. (If d₁ = d₂ the equations are identical and the lines coincide.)
Another view: at any x, the vertical gap between the lines is (mx + d₁) − (mx + d₂) = d₁ − d₂, the same non-zero number everywhere, so the lines never meet.

Teacher's Note:
1. This is a proof by contradiction: assume a common point and reach d₁ = d₂.
2. The constant vertical gap gives a picture students remember easily.