Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 13: Two Variables, One Line (0904CH13)

Ganita Manjari Part II · 54 questions · 7 exercises

Chapter code: 0904ch13 · Book: Ganita Manjari Part II · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 13. Every exercise has a clear, step-by-step answer for chapter code 0904CH13.

Jump to:

Exercise Set 13.1(3 questions)

Write a linear equation in two variables in standard form ax+by+c=0 in which a=3, b=0 and c=−1/5.

School Level Solution

Substitute the given coefficients into ax+by+c=0:3x+0y−1/5=0.Multiplying every term by 5 gives 15x−1=0. The coefficient of y is zero, which is allowed: this is a linear equation in two variables whose graph is a vertical line.

Complete the table by expressing each equation in standard form ax + by + c = 0 and stating the coefficients of x and y and the constant term: (i) y − 15 = √2x; (ii) 3y − 2x = 0; (iii) 5x = 3y; (iv) x = 8; (v) 3y = 1.

School Level Solution

Standard form: collect all terms on the left so the equation is ax + by + c = 0. The coefficient triple is (a,b,c).(i) y − 15 = √2x ⇒ √2x − y + 15 = 0; (a,b,c) = (√2, −1, 15).(ii) 3y − 2x = 0 ⇒ −2x + 3y + 0 = 0; (a,b,c) = (−2, 3, 0).(iii) 5x = 3y ⇒ 5x − 3y + 0 = 0; (a,b,c) = (5, −3, 0).(iv) x = 8 ⇒ x + 0y − 8 = 0; (a,b,c) = (1, 0, −8).(v) 3y = 1 ⇒ 0x + 3y − 1 = 0; (a,b,c) = (0, 3, −1).Multiplying an equation by −1 gives an equivalent standard form, provided all three coefficients change sign together.

(i) A notebook costs twice as much as a pen. If notebook cost is ₹t and pen cost is ₹p, Charlie writes t=2p while Meera writes p=2t. Who is correct? (ii) Two Indian batsmen together score 176 runs. Is x+y=176 correct if their scores are x and y?

School Level Solution

(i) “The notebook costs twice the pen” means t=2×p. Therefore t=2p is correct. The equation p=2t would say the pen costs twice as much as the notebook, so Meera’s representation reverses the relationship.(ii) If the two scores are x and y, their combined score is x+y. Since it is 176, x+y=176 is a correct mathematical representation. A single equation does not determine the two scores uniquely; it represents every pair whose sum is 176.

Exercise Set 13.2(7 questions)

Verify whether (4,3) is a solution of 5x−6y=2.

School Level Solution

Substitute x=4 and y=3:5x−6y=5(4)−6(3)=20−18=2.This equals the right-hand side, 2. Hence (4,3) is a solution.

Find any two solutions for each equation: (i) 7x−3y=21; (ii) 2x+3y=5.

School Level Solution

(i) For 7x−3y=21: set y=0 to obtain x=3, giving (3,0). Set x=0: −3y=21, so y=−7, giving (0,−7).(ii) For 2x+3y=5: set x=1, then 2+3y=5, so y=1 and (1,1). Set y=−1, then 2x−3=5, so x=4 and (4,−1). Substitution verifies all four ordered pairs.

For 2mx+3y=7 and 4x+ny=−10, if (2,−1) solves both equations, find m and n.

School Level Solution

Substitute x=2, y=−1 into 2mx+3y=7:2m(2)+3(−1)=7 ⇒ 4m−3=7 ⇒ 4m=10 ⇒ m=5/2.Now substitute in 4x+ny=−10:4(2)+n(−1)=−10 ⇒ 8−n=−10 ⇒ n=18.

Find two solutions in different quadrants for each equation and identify the quadrants: (i) 5x+3y=7; (ii) 5x−3y=7; (iii) −5x+3y=7; (iv) −5x−3y=7.

School Level Solution

Substitute each selected x-value and solve for y.(i) 5x+3y=7: x=1 gives 5+3y=7, y=2/3 (QI); x=2 gives 10+3y=7, y=−1 (QIV).(ii) 5x−3y=7: x=1 gives 5−3y=7, y=−2/3 (QIV); x=2 gives 10−3y=7, y=1 (QI).(iii) −5x+3y=7: x=−1 gives 5+3y=7, y=2/3 (QII); x=2 gives −10+3y=7, y=17/3 (QI).(iv) −5x−3y=7: x=−1 gives 5−3y=7, y=−2/3 (QIII); x=−2 gives 10−3y=7, y=1 (QII).Each pair satisfies its equation and lies in two different quadrants. Plotting either pair for an equation and joining the points gives its straight-line graph.

For the graph of 3x−7y=21, does C=(2,3) lie on the line and satisfy the equation? Can a point not on the line satisfy the equation?

School Level Solution

Substitute C=(2,3): 3(2)−7(3)=6−21=−15, which is not 21. Thus C does not satisfy the equation and does not lie on the graph. By definition, the graph consists exactly of the points whose coordinates satisfy the equation. Therefore a point not on the line cannot satisfy it.

State 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 a valid linear equation in two variables; (v) 2x+3y=7 has infinitely many solutions; (vi) (1,2) solves 2x+3y=7.

School Level Solution

(i) False: a non-vertical linear equation has infinitely many ordered-pair solutions; for example, (0,7/3) and (1,5/3) solve 2x+3y=7.(ii) False: only equations with zero constant term (in standard form ax+by+c=0, c=0) necessarily pass through (0,0); 2x+3y=7 does not.(iii) False: for 2x+3y=7, (2,1) is an integer solution and hence rational.(iv) True: x=3 can be written x+0y−3=0, a linear equation in two variables.(v) True: choose any x and solve y=(7−2x)/3, giving infinitely many solutions.(vi) False: 2(1)+3(2)=8≠7.

(i) Compare 3x+4y=7 and 6x+8y=14. (ii) Show that ax+by=c and kax+kby=kc have the same solutions whenever k≠0.

School Level Solution

(i) Divide 6x+8y=14 by 2 to get 3x+4y=7. Thus the equations are equivalent, so every solution of either satisfies the other.(ii) If ax+by=c, multiplying both sides by non-zero k gives kax+kby=kc. Conversely, divide the second equation by k (permitted because k≠0) to recover ax+by=c. Hence the two equations have the same solution set.

Exercise Set 13.3(4 questions)

Three ski hills have the following rises and horizontal runs: A rises 60 m over 70 m; B rises 60 m over 110 m; C rises 80 m over 100 m. Rank their steepness from least to greatest.

School Level Solution

Steepness is represented by slope = rise/run.A: 60/70=6/7≈0.857. B: 60/110=6/11≈0.545. C: 80/100=0.8.Since 0.545<0.8<0.857, the order from least to greatest steepness is B, C, A.

A loading-dock ramp rises 2.5 m over a horizontal run of 4 m. Find its slope.

School Level Solution

Slope = vertical rise / horizontal run = 2.5/4 = 25/40 = 5/8 = 0.625. The slope is dimensionless when both measurements use metres.

Find the slope of line l in each of the four diagrams in the textbook.

School Level Solution

(i) The line is horizontal, so its rise is 0 for every non-zero run. Its slope is 0.(ii) Two marked grid points are (−4, 4) and (4, −4). Therefore, $$m=\frac{−4−4}{4−(−4)}=\frac{−8}{8}=\mathbf{−1}.$$(iii) Two marked grid points are (−4, −2) and (2, 2). Therefore, $$m=\frac{2−(−2)}{2−(−4)}=\frac{4}{6}=\mathbf{\frac{2}{3}}.$$(iv) The line is vertical, so its horizontal run is 0. Division by zero is not defined; hence the slope is undefined.

A wheelchair ramp must have a slope of 1 cm vertical rise for each 12 cm of horizontal length. What horizontal length is required for stairs with total height 18 cm?

School Level Solution

The slope guideline means rise/run=1/12. With a rise of 18 cm,18/run=1/12 ⇒ run=18×12=216 cm = 2.16 m.

Exercise Set 13.4(3 questions)

A family buys movie tickets and snack boxes on two days. If a ticket costs ₹x and a snack box costs ₹y, write equations for (i) 2 tickets and 3 snack boxes cost ₹850; (ii) 4 tickets and 1 snack box cost ₹1100.

School Level Solution

(i) Two tickets cost 2x and three snack boxes cost 3y, so 2x+3y=850.(ii) Four tickets cost 4x and one snack box costs y, so 4x+y=1100. These equations model the two purchases.

A taxi fare consists of fixed charge ₹x plus ₹y per kilometre. Sahil travels 6 km for ₹122 and Meena 8 km for ₹160. Form a pair of equations.

School Level Solution

Fixed charge is x, and the distance charge is y per kilometre. For 6 km, the total is x+6y=122. For 8 km, it is x+8y=160. Hence the system is x+6y=122, x+8y=160.

Adult tickets cost ₹150 and children’s tickets cost ₹100. There are 200 people and ₹25,000 is collected. If x adult and y child tickets are sold, form the equations.

School Level Solution

Total attendance gives x+y=200. Ticket revenue gives 150x+100y=25,000, or 3x+2y=500 after dividing by 50. From x+y=200, 2x+2y=400. Subtracting this from 3x+2y=500 gives x=100, so y=100. Thus 100 adult tickets and 100 children’s tickets fit the data.

Exercise Set 13.5(9 questions)

Form and solve a pair of linear equations for each: (i) sum of two integers is +5 and their difference is −21; (ii) the difference between two numbers is 26 and one is three times the other; (iii) 7 bats and 6 balls cost ₹8880, while 3 bats and 5 balls cost ₹4000; (iv) taxi fares are ₹155 for 10 km and ₹220 for 15 km; (v) adding 2 to numerator and denominator gives 9/11, adding 3 gives 5/6; (vi) adding 1 to the numerator and subtracting 1 from denominator gives 1, adding 1 only to denominator gives 1/2; (vii) five years ago Nuri was three times Sonu’s age; ten years later she will be twice his age; (viii) digits sum to 9 and nine times the number equals twice the reversed number; (ix) ₹2000 is paid in ₹50 and ₹100 notes, 25 notes total; (x) library fee is fixed for first 3 days plus extra daily charge; 7 days cost ₹27 and 5 days cost ₹21.

School Level Solution

(i) Let the integers be x,y. x+y=5 and x−y=−21. Adding gives 2x=−16 ⇒ x=−8; hence y=13.(ii) Let the larger number x and smaller y. x−y=26 and x=3y. Thus 3y−y=26 ⇒ y=13, x=39.(iii) Let bat cost b and ball cost c. 7b+6c=8880 ...(1), 3b+5c=4000 ...(2). Multiply (1) by 5 and (2) by 6: 35b+30c=44400; 18b+30c=24000. Subtract: 17b=20400, so b=₹1200. Substitute into (2): 3600+5c=4000 ⇒ c=₹80.(iv) Let fixed charge f and per-km fare r. f+10r=155, f+15r=220. Subtracting gives 5r=65 ⇒ r=₹13/km. Then f=155−130=₹25. For 25 km, fare=25+13×25=₹350.(v) Let the fraction be x/y. (x+2)/(y+2)=9/11 ⇒ 11x−9y=−4 ...(1). (x+3)/(y+3)=5/6 ⇒ 6x−5y=−3 ...(2). Multiplying (1) by 5 and (2) by 9: 55x−45y=−20, 54x−45y=−27. Thus x=7; substituting in (2), 42−5y=−3 ⇒ y=9. Fraction=7/9.(vi) (x+1)/(y−1)=1 gives y=x+2. Also x/(y+1)=1/2 gives 2x=y+1. Substitute: 2x=x+3 ⇒ x=3 and y=5. The fraction is 3/5.(vii) Let current ages be N and S. N−5=3(S−5) ⇒ N=3S−10. N+10=2(S+10) ⇒ N=2S+10. Equating, 3S−10=2S+10 ⇒ S=20, N=50.(viii) Let tens digit a and units digit b. a+b=9. The number is 10a+b, reversed is 10b+a, and 9(10a+b)=2(10b+a). Simplify: 88a=11b, hence b=8a. Therefore 9a=9 ⇒ a=1,b=8, number 18.(ix) Let x be the number of ₹50 notes and y the number of ₹100 notes. x+y=25 and 50x+100y=2000, i.e. x+2y=40. Subtract x+y=25: y=15, so x=10.(x) Let F be the fee for the first three days and r the charge for each additional day. F+4r=27 and F+2r=21. Subtract: 2r=6 ⇒ r=₹3; F=21−6=₹15.

Form and solve graphically a pair of equations for 10 quiz students if there are 4 more girls than boys.

School Level Solution

Let b be boys and g be girls. Total students: b+g=10. “Four more girls than boys” gives g−b=4. Add the equations: 2g=14 ⇒ g=7; then b=3.For a graph, write g=10−b and g=b+4. Plot the first through (0,10),(10,0), and the second through (0,4),(3,7). They intersect at (b,g)=(3,7), giving 3 boys and 7 girls.

Use coefficient ratios to classify: (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.

School Level Solution

(i) a₁/a₂=5/7, b₁/b₂=−4/6=−2/3; these differ. So the lines intersect at one point. Solving 5x−4y=−8 and 7x+6y=9 gives 29x=−6, so x=−6/29 and y=101/58.(ii) 18/9=6/3=24/12=2. All coefficient and constant ratios agree, so the equations describe coincident lines and have infinitely many common solutions.(iii) 6/2=3 and (−3)/(−1)=3, but 10/9≠3. Coefficients are proportional while constants are not, so the lines are parallel and distinct; there is no common solution.

Which pairs have solutions? For those that do, find the common solution graphically: (i) x+y=5 and 2x+2y=10; (ii) x−y=8 and 3x−3y=16; (iii) 2x+y−6=0 and 4x−2y−4=0; (iv) 2x−2y−2=0 and 4x−4y−5=0.

School Level Solution

(i) Dividing the second equation by 2 gives x+y=5, the first equation itself. The graphs coincide, so every point on x+y=5 is a common solution.(ii) The equations reduce to x−y=8 and x−y=16/3. These are distinct parallel lines; there is no solution.(iii) 2x+y=6 and 4x−2y=4. Doubling the first gives 4x+2y=12. Add the second: 8x=16 ⇒ x=2; then y=2. The graphs intersect at (2,2).(iv) The first reduces to x−y=1 and the second to x−y=5/4. The lines are parallel and distinct, so there is no solution.

Half the perimeter of a rectangular garden is 36 m, and its length is 4 m more than its width. Find its dimensions.

School Level Solution

Let width=w m, length=l m. Half the perimeter is l+w=36, and l=w+4. Substitute: (w+4)+w=36 ⇒ 2w=32 ⇒ w=16 m. Then l=20 m. Check: l+w=36 and l−w=4.

For 2x+3y−8=0, write another equation so that the pair represents (i) intersecting lines, (ii) parallel lines, and (iii) coincident lines.

School Level Solution

(i) Intersecting: choose x+y−1=0. Its x:y coefficient ratio 1:1 differs from 2:3, so there is one intersection.(ii) Parallel, distinct: choose 4x+6y−10=0. The x and y coefficients are twice those of the original, but the constant −10 is not twice −8 (which would be −16). So the lines are parallel but not coincident.(iii) Coincident: multiply the whole original equation by 2, obtaining 4x+6y−16=0. Both equations represent the same line.

Using the ancient problem, find the price of one citron x and one fragrant wood-apple y if 9x+7y=107 and 7x+9y=101. Solve without directly applying the usual elimination/substitution setup.

School Level Solution

Add the equations: 16x+16y=208 ⇒ x+y=13. Subtract the second equation from the first: 2x−2y=6 ⇒ x−y=3. Add these simpler equations: 2x=16, so x=8. Then y=13−8=5. Therefore one citron costs ₹8 and one fragrant wood-apple ₹5.

Five pencils and seven pens cost ₹50; seven pencils and five pens cost ₹46. Find the cost of each.

School Level Solution

Let pencil cost be p and pen cost be q. 5p+7q=50 and 7p+5q=46. Adding: 12p+12q=96 ⇒ p+q=8. Subtracting the first from the second: 2p−2q=−4 ⇒ p−q=−2. Add the last two relations: 2p=6 ⇒ p=3; hence q=5. Check: 5(3)+7(5)=50 and 7(3)+5(5)=46.

Find the height of the stool shown in the figure, using the measurements 25 cm and 85 cm.

School Level Solution

Let s be stool height and c be cat height. In the first picture, the cat is on the stool, so s+c=85. In the second, the cat is on the floor and the stool seat is 25 cm above its head, so s−c=25. Add the equations: 2s=110 ⇒ s=55 cm. Then c=85−55=30 cm. The stool is 55 cm high.

End-of-Chapter Exercises(16 questions)

For y=3x: (i) find its slope from (0,0) and (2,6); (ii) state the y-intercept and its meaning; (iii) for a water tank whose level y cm after x minutes follows the line, find the level after 5 minutes and time to reach 21 cm; (iv) does (4,10) lie on it? (v) find its intersection with x=3.

School Level Solution

(i) Slope m=(y₂−y₁)/(x₂−x₁)=(6−0)/(2−0)=3.(ii) At x=0, y=0, so the y-intercept is 0. In the tank context, the water level starts at 0 cm.(iii)(a) At x=5, y=3(5)=15 cm. (b) For y=21, 21=3x, so x=7 minutes.(iv) At x=4 the line predicts y=3(4)=12, not 10. Therefore (4,10) does not lie on the line.(v) If x=3, then y=3×3=9. The intersection point is (3,9).

Use F=(9/5)C+32 to: (i) draw the graph using Celsius on x-axis and Fahrenheit on y-axis; (ii) convert 30°C; (iii) convert 95°F; (iv) find F at 0°C and C at 0°F; (v) find a temperature numerically equal in both scales.

School Level Solution

(i) The graph is a straight line. Useful points are: C=0 gives F=32; C=100 gives F=(9/5)(100)+32=212; C=−40 gives F=−40. Plot points (0,32), (100,212), and join them with a straight line.(ii) F=(9/5)(30)+32=54+32=86°F.(iii) 95=(9/5)C+32 ⇒ 63=(9/5)C ⇒ C=63×5/9=35°C.(iv) At C=0, F=32°F. At F=0, 0=(9/5)C+32 ⇒ C=−160/9≈−17.78°C.(v) Set F=C: C=(9/5)C+32 ⇒ −(4/5)C=32 ⇒ C=−40. Thus −40°C=−40°F.

Solve graphically the system 2x+y=6 and 2x−y−2=0.

School Level Solution

Rewrite the equations as y=6−2x and y=2x−2. For the first line, plot (0,6) and (3,0). For the second, plot (0,−2) and (1,0). Draw both straight lines; their intersection satisfies both equations. Algebraically checking the intersection: add 2x+y=6 and 2x−y=2 to get 4x=8, so x=2; then y=2. The graphical solution is (2,2).

Find the point of intersection of the two lines shown on the cover page.

School Level Solution

The cover gives the two lines 15x − 10y − 115 = 0 and 15x + 100y − 335 = 0.Rewrite and divide by 5:3x − 2y = 23 … (1)3x + 20y = 67 … (2)Subtract (1) from (2): 22y = 44, so y = 2.Substitute y = 2 into (1): 3x − 4 = 23 ⇒ 3x = 27 ⇒ x = 9.Check: 15(9) − 10(2) − 115 = 0 and 15(9) + 100(2) − 335 = 0.Therefore, the point of intersection is (9, 2).

Give a formula for the x-intercept of y=mx+c.

School Level Solution

At the x-intercept, y=0. Substitute into y=mx+c: 0=mx+c ⇒ mx=−c. If m≠0, x=−c/m. If m=0 and c≠0, the horizontal line never meets the x-axis; if m=c=0, the line is the x-axis itself and every point on it is an intercept.

Plan A costs ₹50 monthly plus ₹0.20 per call minute; Plan B costs ₹30 monthly plus ₹0.30 per minute. When is each cheaper and when are they equal?

School Level Solution

For m call minutes, A(m)=50+0.20m and B(m)=30+0.30m. Set equal: 50+0.20m=30+0.30m ⇒ 20=0.10m ⇒ m=200. At 200 min both cost ₹90. If m<200, Plan B is cheaper because it has the lower fixed fee; if m>200, Plan A is cheaper because its per-minute charge is lower.

How many lines (i) have a given slope? (ii) have a given slope and pass through a given point?

School Level Solution

(i) A fixed slope fixes only the direction. The line y=mx+c has that slope for every choice of intercept c, so there are infinitely many parallel lines.(ii) A fixed slope m and point (x₀,y₀) determine c uniquely: c=y₀−mx₀. Therefore exactly one line has that slope and passes through that point.

Find the values of p for which 4x+py+8=0 and 2x+2y+2=0 have a unique solution.

School Level Solution

Two lines have a unique intersection when their x and y coefficients are not proportional, i.e. a₁b₂−a₂b₁≠0. Here 4(2)−2p=8−2p≠0 ⇒ p≠4. When p=4 the coefficient ratios agree and the lines are parallel but not coincident (constants do not match), so there is no unique solution.

Find a and b so that (a+b)x−2by=5a+2b+1 and 3x−y=14 have infinitely many solutions.

School Level Solution

For coincident lines, all three ratios must agree. Write the first as (a+b)x−2by−(5a+2b+1)=0 and the second as 3x−y−14=0. The coefficient ratio requires (a+b)/3=(−2b)/(−1)=2b. Hence a+b=6b, so a=5b. The constant ratio gives (5a+2b+1)/14=2b. Substitute a=5b: (25b+2b+1)=28b ⇒ b=1; thus a=5. These values make the first equation 6x−2y=28, twice the second equation.

Find k so that x+2y=3 and (k−1)x+(k+1)y=k+3 represent coincident lines.

School Level Solution

Write the first as x+2y−3=0 and the second as (k−1)x+(k+1)y−(k+3)=0. For coincident lines, the second coefficient triple must be a scalar multiple λ of (1,2,−3). Thus k−1=λ and k+1=2λ. Subtracting gives 2=k−1, so k=3 and λ=2. Check the constant: −(k+3)=−6=2(−3), so all three coefficients are proportional. Therefore k=3.

(i) Robot 1 starts from (0,0) and repeatedly moves 3 units right, then 4 units up. Robot 2 starts from (10,0) and repeatedly moves 1 unit right, then 2 units up. Do their traced paths intersect, and where? (ii) Robot 1 starts from (3,0) and repeatedly moves 5 units right, then 5 units up. Robot 2 starts from (7,0) and repeatedly moves 5 units right, then 3 units down. Do their traced paths intersect, and where?

School Level Solution

Trace the actual horizontal and vertical segments, including their endpoints; comparing only the robots’ overall directions would miss overlaps.(i) Robot 1 has horizontal segments at heights 0, 4, 8, …, with each segment 3 units long, and vertical segments 4 units long. Robot 2 has horizontal segments at heights 0, 2, 4, …, with each segment 1 unit long, and vertical segments 2 units long. Comparing segments at the same height or on the same vertical line gives these common parts:• Isolated intersection points: (24, 28) and (33, 44). • Shared horizontal segment at height 32: from (26, 32) to (27, 32). • Shared vertical segment at x = 27: from (27, 32) to (27, 34). • Shared horizontal segment at height 36: from (28, 36) to (29, 36). • Shared vertical segment at x = 30: from (30, 38) to (30, 40). • Shared horizontal segment at height 40: from (30, 40) to (31, 40).Thus, the paths share portions of five segments, besides the two isolated points. The endpoints listed for adjacent shared portions are intentionally repeated because the overlap changes from horizontal to vertical there.(ii) Robot 1’s first horizontal segment runs from (3, 0) to (8, 0). Robot 2’s first horizontal segment runs from (7, 0) to (12, 0). Hence they share the segment from (7, 0) to (8, 0). After leaving the x-axis, Robot 1 moves upward while Robot 2 moves downward, so their later vertical/horizontal steps do not meet.

At first Jacob’s watch reads a minutes after 2 o’clock. Fifteen minutes later it reads b minutes after 3 o’clock. If a=6b, what is the second time?

School Level Solution

From 2:a to 3:b, the elapsed time is (60−a)+b minutes. It is 15, so 60−a+b=15 ⇒ a−b=45. Given a=6b, 6b−b=45 ⇒ 5b=45 ⇒ b=9. The second reading is 3:09.

The digits of a two-digit number sum to 15. Reversing the digits gives a number 9 greater than the original. Find the number.

School Level Solution

Let the tens digit be x and units digit be y. x+y=15. The number is 10x+y; the reversed number is 10y+x. The difference condition is (10y+x)−(10x+y)=9 ⇒ 9(y−x)=9 ⇒ y−x=1. Add the two equations: x+y=15 and y−x=1 give 2y=16 ⇒ y=8, x=7. The number is 78; reversed 87 is 9 greater.

In cyclic quadrilateral ABCD, ∠A=(x+7)°, ∠B=(y+8)°, ∠C=(3y+23)° and ∠D=(4x+12)°. Find all four angles.

School Level Solution

Opposite angles of a cyclic quadrilateral are supplementary.(x+7)+(3y+23)=180 ⇒ x+3y=150 ...(1).(y+8)+(4x+12)=180 ⇒ 4x+y=160 ...(2).From (2), y=160−4x. Substitute in (1): x+3(160−4x)=150 ⇒ −11x=−330 ⇒ x=30, y=40. Therefore A=37°, B=48°, C=143°, D=132°. The opposite sums are 37+143=180 and 48+132=180.

A train takes 6 hours less for a journey when its speed is increased by 6 km/h, and 6 hours more if its speed is decreased by 4 km/h. Find its original speed and the distance.

School Level Solution

Let original speed be v km/h and distance be d km. Original time is d/v.With increased speed: d/v−d/(v+6)=6 ⇒ 6d/[v(v+6)]=6 ⇒ d=v(v+6).With decreased speed: d/(v−4)−d/v=6 ⇒ 4d/[v(v−4)]=6 ⇒ d=3v(v−4)/2.Equate the expressions for d: v(v+6)=3v(v−4)/2. Since v>4, divide by v: v+6=1.5v−6 ⇒ 0.5v=12 ⇒ v=24. Then d=24×30=720 km.

A father’s age equals the sum of the ages of his four children. After 20 years, the sum of the children’s ages will be twice the father’s age. Find the father’s present age.

School Level Solution

Let the father’s present age be F. The sum of the four children’s present ages is also F. After 20 years, each child ages by 20 years, so their combined age becomes F+4(20)=F+80. The father’s age becomes F+20. Given F+80=2(F+20), so F+80=2F+40 ⇒ F=40 years.

In-text and Think and Reflect(12 questions)

Why does every point (x,y) satisfying 2x + 3y = 6 lie on the graph of the equation? Give three different solutions and explain why a non-vertical, non-horizontal linear equation has infinitely many solutions.

School Level Solution

Substitution verifies each point: 2(0)+3(2)=6; 2(3)+3(0)=6; 2(−3)+3(4)=6. More generally, for any chosen x, y=(6−2x)/3 gives a solution. Since x can take infinitely many real values, so can the ordered pairs. The graph is a straight line containing all these points.

Explain why a line in standard form ax + by + c = 0 passes through the origin when c = 0.

School Level Solution

Substitute x=0 and y=0 into ax+by+c=0. If c=0, the left side is a(0)+b(0)+0=0, so the origin is a solution and lies on the line (provided a and b are not both zero).

How many solutions can a pair of linear equations in two variables have? Explain the cases represented by intersecting, parallel and coincident lines.

School Level Solution

A solution is an ordered pair lying on both lines. Two distinct intersecting lines meet at exactly one point, giving a unique solution. Distinct parallel lines never meet, so there is no solution. Coincident lines have every point in common, so there are infinitely many solutions.

Solve the in-text paratha-and-dahi problem: one paratha and two bowls of dahi cost ₹90; two parathas and seven bowls of dahi cost ₹260. Find each price.

School Level Solution

Let p be the price of a paratha and d the price of a bowl of dahi. Then p+2d=90 and 2p+7d=260. Double the first equation: 2p+4d=180. Subtract it from the second equation: 3d=80, so d=80/3. Then p=90−160/3=110/3. Check: 110/3+160/3=90 and 220/3+560/3=260. The fractional-rupee values follow from the data as stated.

If two angles of a triangle are x° and y°, with y = 4x, and the third angle is 50°, find x and y.

School Level Solution

The angle sum gives x+y+50=180, so x+y=130. Since y=4x, 5x=130 and x=26°. Therefore y=4×26=104°. Check: 26°+104°+50°=180°.

How many solutions do 10x − 5y = 20 and 4x − 2y = 8 have? Give a rule-based explanation.

School Level Solution

Divide the first equation by 5 to get 2x−y=4. Divide the second by 2 to get 2x−y=4. They represent the same line, so every point on that line satisfies both equations; there are infinitely many solutions.

Give an example of a pair of linear equations with no solution and explain the reason.

School Level Solution

Multiplying x−y=10 by 10 gives 10x−10y=100, but the other equation requires 10x−10y=101. The same expression cannot equal 100 and 101 simultaneously. The lines have the same slope but different intercepts, so they are distinct parallel lines.

Prove that two distinct lines with equal slopes are parallel using y = mx + d₁ and y = mx + d₂.

School Level Solution

Suppose a point (x,y) lies on both lines. Then y=mx+d₁ and y=mx+d₂. Subtracting gives d₁=d₂. This contradicts d₁≠d₂. Therefore the lines have no point in common. Their equal slopes mean they have the same direction, so they are distinct parallel lines.

The line AB in the figure passes through A=(2,0) and B=(4,3). (i) Find the point on AB whose x-coordinate is 8. (ii) Find the point on AB whose y-coordinate is 6.

School Level Solution

The slope is (3−0)/(4−2)=3/2. Using point A=(2,0), the line is y=(3/2)(x−2). (i) For x=8, y=(3/2)(6)=9, so the point is (8,9). (ii) Set y=6: 6=(3/2)(x−2), so 4=x−2 and x=6. The point is (6,6).

For two linear equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, explain how to classify the pair as having a unique solution, no solution, or infinitely many solutions without dividing by a coefficient that might be zero.

School Level Solution

The determinant D=a₁b₂−a₂b₁ checks whether the x- and y-coefficient pairs are proportional without dividing by a possibly zero coefficient. If D≠0, the two lines have different slopes and intersect once. If D=0, compare the constants as well: if (a₁,b₁,c₁) and (a₂,b₂,c₂) are proportional, the equations describe the same line and have infinitely many solutions. If the x- and y-coefficients are proportional but the constants are not in the same ratio, the lines are distinct and parallel, so there is no solution. This formulation also handles zero coefficients.

Give three pairs of linear equations with no solution and state the common rule.

School Level Solution

(i) The left sides are equal but the constants differ. (ii) The second equation has twice the x- and y-coefficients of the first, but its constant 7 is not twice 1. (iii) The two distinct vertical lines x=2 and x=5 never meet. Rule: proportional x- and y-coefficients with non-proportional constants give distinct parallel lines and no solution.

Describe another way to solve a pair of linear equations besides elimination or substitution, and mention one limitation.

School Level Solution

Plot each linear equation on the same coordinate plane using two or more points. If the lines intersect, the intersection coordinates solve both equations. Parallel lines indicate no solution; coincident lines indicate infinitely many. Graphing is useful visually, but a hand-drawn graph may not locate an exact fractional intersection, so algebra is usually better for an exact answer.

Other Chapters

1 2 3 4 5 6 7 8Part II9 10 11 12 13 14