CBSE Class 9 Maths Ganita Manjari Part 2 Ch 13 Two Variables One Line Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Draw the graph of 3x − 2y = 6 on squared paper. From your graph, read off the solution for which x = 4 and the solution for which y = −6, and state the coordinates of the two points where the line meets the axes. Confirm each of your four readings by substituting into the equation.
Answer:
Intercepts: y = 0 gives x = 2 → (2, 0); x = 0 gives y = −3 → (0, −3). Plot and join:

From the graph: x = 4 gives (4, 3); y = −6 gives (−2, −6).
Checks: 12 − 6 = 6 ✓; −6 + 12 = 6 ✓; 6 − 0 = 6 ✓; 0 + 6 = 6 ✓.
In simple words: Plot the intercepts, draw the line, and read other points from it.

Teacher's Note:
1. Use a neat scale on both axes.
2. Confirm every reading algebraically.

Question 2. Draw the graphs of 2x + y = 8 and x − y = 1 on one pair of axes and read off the solution of the pair from your drawing. Then solve the pair by elimination and confirm that the two answers agree.
Answer:
2x + y = 8: (0, 8), (4, 0). x − y = 1: (0, −1), (1, 0).

The lines cross at (3, 2).
Elimination: adding gives 3x = 9, x = 3; then y = 2. (3, 2) ✓.
They agree because a point on both lines satisfies both equations.
In simple words: The crossing point of the graphs is the solution found by algebra.

Teacher's Note:
1. Find two points per line.
2. Graph and algebra must agree.

Question 3. Draw the graphs of x + 2y = 6 and 2x + 4y = 20 on one pair of axes. Describe what your drawing shows about the solutions of this pair, and confirm your conclusion using the ratios of the coefficients.
Answer:
x + 2y = 6: (0, 3), (6, 0). 2x + 4y = 20: (0, 5), (10, 0).

The lines are parallel and never meet, so there is no solution.
Ratios: 1/2 = 2/4 = 1/2, but −6/−20 = 3/10 ≠ 1/2. So a₁/a₂ = b₁/b₂ ≠ c₁/c₂: no solution ✓.
Slopes are both −1/2; intercepts 3 and 5 differ.
In simple words: Same slope, different intercepts: parallel lines, no solution.

Teacher's Note:
1. Parallel lines = inconsistent pair.
2. Use both graph and ratios.

Question 4. Draw the graphs of 3x − y = 3 and 9x − 3y = 9 on one pair of axes. Explain what your drawing shows, and write down three different ordered pairs that satisfy both equations.
Answer:
Both equations pass through (1, 0) and (0, −3), so they give the same line.

The lines are coincident: infinitely many solutions. (9x − 3y = 9 is 3 times the first.)
Three pairs: (1, 0), (2, 3), (0, −3). E.g. (2, 3): 6 − 3 = 3 and 18 − 9 = 9 ✓.
In simple words: The second equation is the first in disguise, so the lines are the same.

Teacher's Note:
1. Same two points → same line.
2. All three ratios equal (1/3).

Question 5. Two lines have the same slope m but different y-intercepts d₁ and d₂. Prove that they have no point in common, and hence that the corresponding pair of linear equations has no solution.
Answer:
Lines: y = mx + d₁ and y = mx + d₂, with d₁ ≠ d₂.
Suppose (p, q) lies on both. Then q = mp + d₁ and q = mp + d₂.
So mp + d₁ = mp + d₂, giving d₁ = d₂, a contradiction.
Hence there is no common point, and the pair has no solution.
At every x the vertical gap is d₁ − d₂, a constant non-zero number. If d₁ = d₂ the lines would coincide.
In simple words: Equal slopes keep the lines a fixed distance apart, so they never meet.

Teacher's Note:
1. Proof by contradiction.
2. Note why d₁ ≠ d₂ is needed.

Question 6. A line has the equation ax + by + c = 0, where b ≠ 0. Take any two points of this line and prove that \( \frac{y_2 - y_1}{x_2 - x_1} \) always works out to \( -\frac{a}{b} \), whichever two points are chosen.
Answer:
Let (x₁, y₁) and (x₂, y₂) be distinct points on the line: ax₁ + by₁ + c = 0 and ax₂ + by₂ + c = 0.
Subtracting: a(x₂ − x₁) + b(y₂ − y₁) = 0, so b(y₂ − y₁) = −a(x₂ − x₁).
x₂ ≠ x₁: otherwise b(y₂ − y₁) = 0, and as b ≠ 0, y₂ = y₁, making the points the same.
Dividing by b(x₂ − x₁): \( \frac{y_2 - y_1}{x_2 - x_1} \) = \( -\frac{a}{b} \).
The result contains only a and b, not the points, so it is the same for every choice. If b = 0 the line is vertical and the slope is undefined.
In simple words: The slope depends only on the coefficients, never on which points you pick.

Teacher's Note:
1. Justify division by x₂ − x₁.
2. Matches y = −(a/b)x − c/b.

Question 7. The incomes of two people are in the ratio 8:5 and their expenditures are in the ratio 5:3. Each of them saves ₹3000 a month. Find their monthly incomes. Solve the pair by elimination, then solve it again by substitution, and comment on which method you found shorter here.
Answer:
Incomes 8x, 5x; expenditures 5y, 3y. Savings: 8x − 5y = 3000 and 5x − 3y = 3000.
Elimination: ×3 and ×5: 24x − 15y = 9000, 25x − 15y = 15 000. Subtract: x = 6000. Then 5y = 45 000, y = 9000.
Substitution: y = (5x − 3000)/3. Then 8x − 5(5x − 3000)/3 = 3000 → 24x − 25x + 15 000 = 9000 → x = 6000, y = 9000.
Incomes: ₹48 000 and ₹30 000.
Check: savings 48 000 − 45 000 = 3000 and 30 000 − 27 000 = 3000 ✓.
Elimination was shorter, since no coefficient was 1 and substitution brought in fractions.
In simple words: Write incomes and expenses with ratio multipliers; savings give the equations.

Teacher's Note:
1. Use 8x, 5x for a ratio 8:5.
2. Substitution suits a coefficient of 1.

Question 8. The digits of a two-digit number add up to 12. Four times the number equals seven times the number obtained by reversing its digits. Find the number, and verify that it satisfies both conditions of the question.
Answer:
Digits x (tens), y (units): x + y = 12.
4(10x + y) = 7(10y + x) → 40x + 4y = 70y + 7x → 33x = 66y → x = 2y.
3y = 12, y = 4, x = 8. The number is 84.
Verify: 8 + 4 = 12 ✓; 4 × 84 = 336 and 7 × 48 = 336 ✓.
In simple words: The tens digit is twice the units digit, and they add to 12.

Teacher's Note:
1. Simplify 33x = 66y first.
2. Verify both conditions.

Question 9. A cyclist covers a certain distance at a uniform speed. Had the speed been 6 km/h more, the journey would have taken 2 hours less; had the speed been 4 km/h less, it would have taken 4 hours more. Find the distance covered and the original speed.
Answer:
Speed s, time t, distance st.
(s + 6)(t − 2) = st → −2s + 6t = 12 → 3t − s = 6.
(s − 4)(t + 4) = st → 4s − 4t = 16 → s − t = 4.
s = t + 4 → 3t − t − 4 = 6 → t = 5, s = 9.
Distance = 9 × 5 = 45 km, speed 9 km/h.
Check: 45/15 = 3 h (2 less) and 45/5 = 9 h (4 more) ✓.
In simple words: The st terms cancel, leaving a simple linear pair.

Teacher's Note:
1. Expand carefully; st cancels.
2. Check with the original story.

Question 10. In a cyclic quadrilateral each pair of opposite angles adds up to 180°. In the cyclic quadrilateral ABCD, ∠A = (4x + 10)°, ∠B = (3x − 5)°, ∠C = (2y + 40)°, ∠D = (5y)°. Form a pair of linear equations in x and y, solve it, and find all four angles. Check that the four angles you obtain add up to 360°.
Answer:
∠A + ∠C = 180: 4x + 2y + 50 = 180 → 2x + y = 65.
∠B + ∠D = 180: 3x − 5 + 5y = 180 → 3x + 5y = 185.
y = 65 − 2x → 3x + 325 − 10x = 185 → x = 20, y = 25.
∠A = 90°, ∠B = 55°, ∠C = 90°, ∠D = 125°.
Total: 90 + 55 + 90 + 125 = 360° ✓.
In simple words: Opposite angles of a cyclic quadrilateral add to 180°.

Teacher's Note:
1. Pair A with C and B with D.
2. Check both pairs and the total.

Question 11. A woman's present age is equal to the sum of the present ages of her five children. Fifteen years from now, the sum of her children's ages will be twice her age at that time. Taking her present age as x years and the present sum of her children's ages as y years, frame a pair of linear equations, solve it, and find her present age.
Answer:
Now: x − y = 0.
In 15 years the children's sum rises by 5 × 15 = 75: y + 75 = 2(x + 15) → 2x − y = 45.
Subtracting: x = 45, y = 45.
Her present age is 45 years. Check: 120 = 2 × 60 ✓.
In simple words: Five children each gain 15 years, so their total gains 75.

Teacher's Note:
1. The 75 is the key step.
2. Check with the words.

Question 12. At a certain moment Jacob's digital watch reads a minutes after two o'clock. Twenty minutes later it reads b minutes after three o'clock. He notices that a is five times b. Frame a pair of linear equations in a and b, solve it, and state the time shown by the watch when he looked at it the second time.
Answer:
Counting from 2:00: a + 20 = 60 + b → a − b = 40.
a = 5b.
4b = 40 → b = 10, a = 50.
Second reading: 3:10 (first reading 2:50).
In simple words: Twenty minutes after 2:50 is 3:10, and 50 = 5 × 10.

Teacher's Note:
1. Measure both times from the same starting point.
2. Answers must be between 0 and 59.

Question 13. Solve the pair 4x − y = 4 and 3x + 2y = 14 graphically. The two lines, together with the y-axis, enclose a triangle; find the area of that triangle.
Answer:
4x − y = 4: (0, −4), (1, 0). 3x + 2y = 14: (0, 7), (14/3, 0).

The lines cross at (2, 4). Check: 8 − 4 = 4 and 6 + 8 = 14 ✓.
Triangle vertices: (0, 7), (0, −4), (2, 4). Base on the y-axis = 7 − (−4) = 11; height = 2.
Area = ½ × 11 × 2 = 11 square units.
In simple words: Use the y-axis side as base; the height is the x-coordinate of the crossing point.

Teacher's Note:
1. Shade the triangle on the graph.
2. Base along an axis makes area easy.

Question 14. The equation 5x − 3y + 7 = 0 is given. Write a second linear equation in two variables so that the resulting pair (i) has no solution, (ii) has infinitely many solutions, (iii) has exactly one solution. Justify each of your three answers using the ratios of the coefficients.
Answer:
(i) 10x − 6y + 3 = 0: 5/10 = −3/−6 = 1/2 but 7/3 ≠ 1/2 → no solution.
(ii) 10x − 6y + 14 = 0: 5/10 = −3/−6 = 7/14 = 1/2 → infinitely many.
(iii) 2x + y − 4 = 0: 5/2 ≠ −3/1 → exactly one (x = 5/11, y = 34/11).
Many other answers are possible.
In simple words: Scale a and b only for parallel; scale everything for coincident; change the ratio for crossing.

Teacher's Note:
1. Accept any correct examples.
2. Justify each with ratios.

Question 15. Find the values of a and b for which the pair (a − 1)x + 3y = 2 and 6x + (1 − 2b)y = 6 has infinitely many solutions. Write out the pair for those values and show that one equation is a multiple of the other.
Answer:
Need \( \frac{a-1}{6} \) = \( \frac{3}{1-2b} \) = \( \frac{2}{6} \) = \( \frac{1}{3} \).
(a − 1)/6 = 1/3 → a = 3. 3/(1 − 2b) = 1/3 → 1 − 2b = 9 → b = −4.
Pair: 2x + 3y = 2 and 6x + 9y = 6.
3 × (2x + 3y = 2) gives 6x + 9y = 6, the second equation exactly ✓.
In simple words: The constant ratio 1/3 fixes both unknowns.

Teacher's Note:
1. Start with the ratio that has no unknown.
2. Verify by multiplying.

Question 16. Consider the pair kx + 2y = 5 and 3x + y = 1.
(i) Find the values of k for which the pair has a unique solution.
(ii) Find the value of k for which it has no solution.
(iii) Show that there is no value of k for which the pair has infinitely many solutions.
Answer:
(i) k/3 ≠ 2/1 → k ≠ 6.
(ii) k/3 = 2 → k = 6. Then c₁/c₂ = −5/−1 = 5 ≠ 2, so k = 6 gives no solution.
(iii) Infinitely many needs 2/1 = 5/1, i.e. 2 = 5, which is false for every k. So no such k.
k only affects the first ratio, not the clash between the second and third.
In simple words: k can make the lines parallel but never make them the same line.

Teacher's Note:
1. Check ratio 3 at k = 6.
2. Explain why k cannot fix 2 ≠ 5.

Question 17. Robot 1 starts at the origin and repeatedly moves 3 units to the right and then 4 units upward. Robot 2 starts at (12, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Find the equations of the two straight lines along which the corners of the two paths lie, and find where those lines cross. Then check whether the crossing point is actually a corner reached by each robot, and say how many repeats each robot needs to arrive there.
Answer:
Robot 1 corners (3k, 4k) lie on 4x − 3y = 0.
Robot 2 corners (12 + k, 2k) lie on 2x − y = 24.
Solving: y = 2x − 24 → 4x − 6x + 72 = 0 → x = 36, y = 48. Crossing at (36, 48).
Robot 1: 3k = 36 → k = 12, 4k = 48 ✓ → 12 repeats.
Robot 2: 2k = 48 → k = 24, 12 + 24 = 36 ✓ → 24 repeats.
So both robots do reach (36, 48).
In simple words: The corner points lie on lines; the crossing point must also be a whole-number repeat for both.

Teacher's Note:
1. Robots move in steps, not along the line.
2. Whole-number check is essential.

Question 18. The relation between a temperature C in degrees Celsius and the same temperature F in degrees Fahrenheit is F = \( \frac{9}{5} \)C + 32.
(i) State the slope of this line and explain in words what it means.
(ii) Convert 35 °C to degrees Fahrenheit.
(iii) Convert 77 °F to degrees Celsius.
(iv) Find the temperature at which the two scales give the same reading.
Answer:
(i) Slope = \( \frac{9}{5} \) = 1.8: each rise of 1 °C is a rise of 1.8 °F. (Intercept 32: 0 °C = 32 °F.)
(ii) F = 1.8 × 35 + 32 = 63 + 32 = 95 °F.
(iii) 77 = 1.8C + 32 → 1.8C = 45 → C = 25 °C.
(iv) Put F = C: C = 1.8C + 32 → −0.8C = 32 → C = −40. So −40 °C = −40 °F.
In simple words: The slope 9/5 is the Fahrenheit change for one Celsius degree.

Teacher's Note:
1. (iv) is a pair of equations: F = 1.8C + 32 and F = C.
2. Slopes differ, so exactly one crossing.

Question 19. A lending library charges a fixed amount for the first four days that a book is kept, and a further fixed amount for each day after that. A book kept for 9 days cost ₹38, and a book kept for 13 days cost ₹54. Find the fixed charge and the charge for each extra day, and then find the cost of keeping a book for 20 days.
Answer:
9 days = 5 extra days: x + 5y = 38. 13 days = 9 extra days: x + 9y = 54.
Subtracting: 4y = 16, y = 4; x = 18.
Fixed charge ₹18, extra day ₹4.
20 days = 16 extra days: 18 + 64 = ₹82.
In simple words: Only days beyond the first four are charged extra.

Teacher's Note:
1. Count extra days correctly (9 − 4 = 5).
2. A common mistake is using 9 and 13 directly.

Question 20. A pair of linear equations is given by a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where a₂, b₂ and c₂ are all non-zero. By carrying out the elimination step in general, prove each of the following.
(i) If \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \), the pair has exactly one solution, and it is given by x = \( \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \), y = \( \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \).
(ii) If \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) = \( \frac{c_1}{c_2} \), the pair has infinitely many solutions.
(iii) If \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \), the pair has no solution.
Answer:
Multiply (1) by b₂ and (2) by b₁ and subtract: (a₁b₂ − a₂b₁)x + (b₂c₁ − b₁c₂) = 0.
Similarly removing x: (a₂b₁ − a₁b₂)y + (a₂c₁ − a₁c₂) = 0.
(i) a₁/a₂ ≠ b₁/b₂ means a₁b₂ − a₂b₁ ≠ 0. Dividing: x = \( \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and y = \( \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \).
Substituting back, the terms cancel in pairs, so these values satisfy both equations: exactly one solution.
(ii) Let the common ratio be k (k ≠ 0). Then (1) is k × (2), so both equations have the same solutions: infinitely many.
(iii) a₁ = ka₂, b₁ = kb₂, c₁ ≠ kc₂. (1) − k × (2) gives c₁ − kc₂ = 0, a false statement. So no solution.
In simple words: Everything depends on whether a₁b₂ − a₂b₁ is zero.

Teacher's Note:
1. Show the general elimination once.
2. k ≠ 0 because a₁, b₁ are not both zero.

Case Study Based Questions

Question 21. At the Fruit Cart: A fruit seller charges ₹40 for a kilogram of guavas and ₹90 for a kilogram of apples. A customer buys some guavas and some apples and pays ₹350 altogether. Let x be the number of kilograms of guavas and y the number of kilograms of apples.
(a) Write the equation connecting x and y in standard form with the smallest possible integer coefficients.
(b) Find the value of y when x = 5.
(c) The answer to part (b) is an acceptable weight of fruit, but it would not be acceptable if the cart sold guavas and apples only in sealed one-kilogram packets. Explain why.
(d) Find the only solution in which x and y are both whole numbers.
Answer:
(a) 40x + 90y = 350 → 4x + 9y − 35 = 0.
(b) 20 + 9y = 35 → y = \( \frac{15}{9} \) = \( \frac{5}{3} \) kg.
(c) 5/3 kg can be weighed loose, but packets cannot be split; the number of packets must be a whole number.
(d) Try y = 0, 1, 2, 3: only y = 3 makes 35 − 9y = 8 a multiple of 4, giving x = 2. So (2, 3).
In simple words: Of infinitely many solutions, only (2, 3) uses whole kilograms.

Teacher's Note:
1. Divide by 10 for smallest coefficients.
2. Check: 80 + 270 = 350.

Question 22. Three Ramps: Three ramps are built at the entrance of a building. Their rises and their horizontal runs are marked in the figure below.
(a) Find the slope of ramp A.
(b) Find the slope of ramp B.
(c) Find the slope of ramp C.
(d) Arrange the three ramps in order of steepness, from the gentlest to the steepest.
Answer:
(a) 3/15 = \( \frac{1}{5} \).
(b) 4/16 = \( \frac{1}{4} \).
(c) 5/30 = \( \frac{1}{6} \).
(d) 12/60, 15/60, 10/60 → C, A, B. C has the largest rise but the longest run.
In simple words: Steepness is rise compared with run, not rise alone.

Teacher's Note:
1. Use a common denominator of 60.
2. Point out the trap in ramp C.

Question 23. Filling a Tank: Water is run into an empty tank. The level y, in centimetres, after x minutes is given by the graph below.
(a) State the slope of the line and explain what it means for the filling of the tank.
(b) Find the level of the water after 7 minutes.
(c) Find how long the tank takes to reach a level of 50 cm.
(d) Determine, without extending the graph, whether the point (9, 36) lies on this line.
Answer:
(a) Through (0, 0) and (5, 20): slope = 20/5 = 4. The level rises 4 cm every minute (y = 4x).
(b) y = 4 × 7 = 28 cm.
(c) 50 = 4x → x = 12.5 minutes (12 min 30 s).
(d) 4 × 9 = 36 ✓, so (9, 36) lies on the line.
In simple words: The slope is the filling rate: 4 cm per minute.

Teacher's Note:
1. Slope = rate of change.
2. An equation avoids extending the graph.

Question 24. Choosing a Mobile Plan: Plan A charges a fixed ₹60 a month together with ₹0.40 for each minute of call time. Plan B charges a fixed ₹20 a month together with ₹0.60 for each minute. The two lines are drawn below, with t the number of minutes of call time in a month.
(a) Write an expression for the monthly cost under each plan in terms of t.
(b) Find the cost of 150 minutes of calls under each plan.
(c) Find the number of minutes for which the two plans cost exactly the same.
(d) State which plan is cheaper for someone who talks for more minutes than your answer to part (c).
Answer:
(a) Plan A: 60 + 0.40t; Plan B: 20 + 0.60t.
(b) A: 60 + 60 = ₹120; B: 20 + 90 = ₹110.
(c) 60 + 0.40t = 20 + 0.60t → 0.20t = 40 → t = 200 minutes (₹140 each).
(d) Plan A (lower per-minute rate).
In simple words: B starts cheaper but rises faster; after 200 minutes A is cheaper.

Teacher's Note:
1. The crossing point is the break-even.
2. Test a value on each side.

Question 25. Tickets for the Science Exhibition: A science exhibition charges one price for an adult ticket and another for a child ticket. One family paid ₹770 for 3 adult tickets and 4 child tickets. Another family paid ₹910 for 5 adult tickets and 2 child tickets. Let the adult ticket cost ₹x and the child ticket cost ₹y.
(a) Frame a pair of linear equations in x and y for the two payments.
(b) Eliminate y from your pair and find the cost of an adult ticket.
(c) Find the cost of a child ticket.
(d) Find what a group of 2 adults and 3 children would pay.
Answer:
(a) 3x + 4y = 770 and 5x + 2y = 910.
(b) 2 × second: 10x + 4y = 1820. Subtract the first: 7x = 1050 → x = ₹150.
(c) 450 + 4y = 770 → y = ₹80.
(d) 300 + 240 = ₹540.
In simple words: Double the second equation to match the child-ticket terms.

Teacher's Note:
1. Check both payments.
2. Multiply only one equation here.

Question 26. The Price Board at a Food Stall: A food stall displays two offers. One paratha with two bowls of dahi costs ₹110. Three parathas with four bowls of dahi cost ₹270. Let one paratha cost ₹x and one bowl of dahi cost ₹y.
(a) Frame a pair of linear equations in x and y.
(b) Find the cost of one bowl of dahi.
(c) Find the cost of one paratha.
(d) Find what one paratha together with one bowl of dahi costs.
Answer:
(a) x + 2y = 110 and 3x + 4y = 270.
(b) 3 × first: 3x + 6y = 330. Subtract: 2y = 60 → y = ₹30.
(c) x = 110 − 60 = ₹50.
(d) 50 + 30 = ₹80.
In simple words: Match the paratha terms, subtract, and the dahi price appears.

Teacher's Note:
1. Check: 150 + 120 = 270.
2. Elimination suits this pair.

Question 27. Three Pictures, Three Answers: Each panel below shows the graphs of the two equations of a pair of linear equations in two variables.
(a) State the number of solutions of the pair shown in Panel 1.
(b) State the number of solutions of the pair shown in Panel 2.
(c) State the number of solutions of the pair shown in Panel 3.
(d) State which panel corresponds to the condition \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \).
Answer:
(a) Lines cross once: exactly one solution.
(b) Parallel lines: no solution.
(c) Dashed line lies on the solid one (coincident): infinitely many.
(d) Panel 2: same slope, but not the same line.
In simple words: Crossing = one, parallel = none, same line = infinitely many.

Teacher's Note:
1. Link each picture to its ratio condition.
2. Learn the three cases together.

Question 28. A Number Puzzle: A teacher says to the class: 'I have thought of two numbers. Their sum is 34 and their difference is 8.' Let the numbers be x and y, with x the larger of the two.
(a) Frame a pair of linear equations in x and y.
(b) Find the two numbers.
(c) Name the method you used in part (b) and say why it suits this particular pair.
(d) Describe what the graph of this pair of equations looks like.
Answer:
(a) x + y = 34 and x − y = 8.
(b) Adding: 2x = 42, x = 21; subtracting: 2y = 26, y = 13. The numbers are 21 and 13.
(c) Elimination: y has coefficients +1 and −1, so simply adding removes it.
(d) Two straight lines crossing at (21, 13); slopes −1 and 1, so they are perpendicular.
In simple words: Sum-and-difference puzzles are solved by adding and subtracting.

Teacher's Note:
1. Check: 21 + 13 = 34, 21 − 13 = 8.
2. One crossing point = unique solution.

Question 29. Notes from the Bank: A customer withdraws ₹4500 and asks for the amount in ₹200 notes and ₹500 notes only. She receives 15 notes altogether. Let x be the number of ₹200 notes and y the number of ₹500 notes.
(a) Frame a pair of linear equations in x and y.
(b) Find the number of ₹500 notes she receives.
(c) Find the number of ₹200 notes she receives.
(d) Determine whether ₹4500 could instead have been paid out in exactly 20 notes of these two kinds, and justify your answer.
Answer:
(a) x + y = 15 and 2x + 5y = 45 (from 200x + 500y = 4500).
(b) 2 × first: 2x + 2y = 30. Subtract: 3y = 15 → 5 notes.
(c) x = 15 − 5 = 10 notes.
(d) No. With x + y = 20: 3y = 5, y = 5/3, not a whole number. So 20 notes is impossible.
In simple words: A solution exists on paper, but notes must come in whole numbers.

Teacher's Note:
1. Check: 2000 + 2500 = 4500.
2. Whole-number check matters in real problems.

Question 30. The Long History of Simultaneous Equations: Pairs and larger systems of linear equations are among the oldest problems in mathematics, and several civilisations developed methods for them long before the modern notation existed.
(a) Name the civilisation whose cuneiform tablets of about 2000 BCE carry the earliest known traces of such problems.
(b) Name the ancient Chinese text that devotes an entire chapter, called fangcheng, to systems of linear equations, and describe briefly how the coefficients were handled in that method.
(c) Name the two Indian mathematicians of 499 CE and 628 CE who studied general systems of equations in several unknowns.
(d) Name the mathematician whose Gaṇita sāra saṅgraha of about 850 CE poses the problem of the prices of citrons and fragrant wood-apples.
Answer:
(a) The Babylonians.
(b) The Nine Chapters on the Mathematical Art. Coefficients were laid out with counting rods in columns (one per equation), and columns were combined to cancel unknowns one by one: the elimination method.
(c) Āryabhaṭa (499 CE) and Brahmagupta (628 CE).
(d) Mahāvīrācārya.
In simple words: Elimination is thousands of years old and was developed in many cultures.

Teacher's Note:
1. Link (d) to the citron problem in the short answers.
2. History helps students remember methods.