Question 1
Write a linear equation with a=3, b=0 and c=−1/5.
Steps only
Use ax+by+c=0.
3x+0y−1/5=0.
Answer
3x − 1/5 = 0.
CLASS 9 · GANITA MANJARI PART II · CHAPTER 13
Two Variables, One Line
Represent linear relations, interpret slope, and solve pairs of linear equations algebraically and graphically.
86 worked answersAll 6 exercise setsAll 16 end-of-chapter questions
What you will learn
- Write and graph linear equations in two variables.
- Calculate and interpret slope and intercepts.
- Solve and classify pairs of linear equations.
86 worked answers · short steps first, explanations when you need them
Exercise Set 13.1
Textbook pages 87
Question 2
Complete the standard-form coefficient table.
Steps only
y−15=2x → −2x+y−15=0: (a,b,c)=(−2,1,−15).
3y−2x=0 → −2x+3y+0=0: (−2,3,0).
5x=3y → 5x−3y+0=0: (5,−3,0).
x=8 → x+0y−8=0: (1,0,−8).
3y=1 → 0x+3y−1=0: (0,3,−1).
Answer
Rows: (−2,1,−15), (−2,3,0), (5,−3,0), (1,0,−8), (0,3,−1).
Question 3(i)
A notebook costs twice a pen. Is t=2p or p=2t correct?
Steps only
Notebook cost t is two times pen cost p.
Therefore t=2p.
Answer
Charlie’s representation t = 2p is correct.
Question 3(ii)
Two batsmen together scored 176. Is x+y=176 correct?
Steps only
x and y represent their separate scores.
Adding them gives the combined score.
Answer
Yes, x + y = 176 is correct.
Exercise Set 13.2
Textbook pages 93–94
Question 1
Is (4,3) a solution of 5x−6y=2?
Steps only
Substitute: 5(4)−6(3)=20−18=2.
LHS equals RHS.
Answer
Yes, (4,3) is a solution.
Question 2(i)
Give two solutions of 7x−3y=21.
Steps only
If y=0, x=3.
If x=0, y=−7.
Answer
Two solutions are (3,0) and (0,−7).
Question 2(ii)
Give two solutions of 2x+3y=5.
Steps only
For x=1, y=1.
For x=4, y=−1.
Answer
Two solutions are (1,1) and (4,−1).
Question 3
If (2,−1) solves 2mx+3y=7 and 4x+ny=−10, find m,n.
Steps only
4m−3=7 gives m=5/2.
8−n=−10 gives n=18.
Answer
m = 5/2 and n = 18.
Question 4(i)
Find two solutions of 5x+3y=7 in different quadrants.
Steps only
Choose and verify (2,−1), quadrant IV.
Choose and verify (−1,4), quadrant II.
Answer
(2,−1), quadrant IV; (−1,4), quadrant II.
Question 4(ii)
Find two solutions of 5x−3y=7 in different quadrants.
Steps only
Choose and verify (2,1), quadrant I.
Choose and verify (−1,−4), quadrant III.
Answer
(2,1), quadrant I; (−1,−4), quadrant III.
Question 4(iii)
Find two solutions of −5x+3y=7 in different quadrants.
Steps only
Choose and verify (1,4), quadrant I.
Choose and verify (−2,−1), quadrant III.
Answer
(1,4), quadrant I; (−2,−1), quadrant III.
Question 4(iv)
Find two solutions of −5x−3y=7 in different quadrants.
Steps only
Choose and verify (1,−4), quadrant IV.
Choose and verify (−2,1), quadrant II.
Answer
(1,−4), quadrant IV; (−2,1), quadrant II.
Question 5
Does C(2,3) lie on 3x−7y=21?
Steps only
3(2)−7(3)=6−21=−15, not 21.
So C is not on the line.
Answer
No. A point off the graph cannot satisfy the equation.
Question 6(i)
A linear equation in two variables has only one solution.
Steps only
False; it has infinitely many ordered-pair solutions.
Answer
False; it has infinitely many ordered-pair solutions.
Question 6(ii)
Its graph always passes through the origin.
Steps only
False; it passes through the origin only when the constant term is 0.
Answer
False; it passes through the origin only when the constant term is 0.
Question 6(iii)
It can never have rational solutions.
Steps only
False; for example (1,1) solves x+y=2.
Answer
False; for example (1,1) solves x+y=2.
Question 6(iv)
x=3 is a valid linear equation in two variables.
Steps only
True; write x+0y−3=0.
Answer
True; write x+0y−3=0.
Question 6(v)
2x+3y=7 has infinitely many solutions.
Steps only
True; each chosen x gives a corresponding y.
Answer
True; each chosen x gives a corresponding y.
Question 6(vi)
(1,2) solves 2x+3y=7.
Steps only
False; 2(1)+3(2)=8, not 7.
Answer
False; 2(1)+3(2)=8, not 7.
Question 7(i)
Compare 3x+4y=7 and 6x+8y=14.
Steps only
The second equation is exactly 2 times the first.
Dividing it by 2 returns the first equation.
Answer
They have the same infinitely many solutions and the same graph.
Question 7(ii)
Show ax+by=c and kax+kby=kc have the same solutions for k≠0.
Steps only
Multiplying a true first equation by k gives the second.
Dividing the second by non-zero k gives the first.
Answer
The two equations have exactly the same solution set.
Exercise Set 13.3
Textbook pages 102–103
Question 1
Rank ski hills A(60/70), B(60/110), C(80/100) from least to greatest steepness.
Steps only
A slope=60/70=6/7≈0.857.
B=60/110=6/11≈0.545; C=80/100=0.8.
Answer
B, C, A.
Question 2
A ramp rises 2.5 m over a run of 4 m.
Steps only
Slope=rise/run=2.5/4=0.625.
Answer
5/8, or 0.625.
Question 3(i)
Find the slope of the horizontal line.
Steps only
Using the marked grid points, rise/run = 0.
Answer
Slope = 0.
Question 3(ii)
Find the slope of the down one per right one line.
Steps only
Using the marked grid points, rise/run = -1.
Answer
Slope = -1.
Question 3(iii)
Find the slope of the up 4 while right 6 line.
Steps only
Using the marked grid points, rise/run = 2/3.
Answer
Slope = 2/3.
Question 3(iv)
Find the slope of the vertical line.
Steps only
Run is 0, so rise/run would divide by zero.
Answer
Slope = undefined.
Question 4
A ramp must rise 1 cm per 12 cm run. Find the run for an 18 cm rise.
Steps only
run/rise=12/1.
Run=18×12=216 cm.
Answer
216 cm, or 2.16 m.
Exercise Set 13.4
Textbook pages 105
Question 1(i)
Two movie tickets and three snack boxes cost ₹850.
Steps only
Ticket cost=x, snack cost=y.
2x+3y=850.
Answer
2x + 3y = 850.
Question 1(ii)
Four movie tickets and one snack box cost ₹1100.
Steps only
4x+y=1100.
Answer
4x + y = 1100.
Question 2(i)
A taxi has fixed charge x and per-km charge y; 6 km costs ₹122.
Steps only
Fare=fixed charge+distance×rate.
x+6y=122.
Answer
x + 6y = 122.
Question 2(ii)
The same taxi costs ₹160 for 8 km.
Steps only
x+8y=160.
Answer
x + 8y = 160.
Question 3
Adult tickets cost ₹150, child tickets ₹100; 200 attendees pay ₹25,000.
Steps only
Count equation: x+y=200.
Money equation: 150x+100y=25000.
Answer
x + y = 200 and 150x + 100y = 25,000.
Exercise Set 13.5
Textbook pages 116–118
Question 1(i)
Two integers have sum 5 and difference −21.
Steps only
x+y=5, x−y=−21.
Add: 2x=−16, so x=−8; y=13.
Answer
The integers are −8 and 13.
Question 1(ii)
Two numbers differ by 26 and one is three times the other.
Steps only
Let x=3y and x−y=26.
2y=26, so y=13 and x=39.
Answer
39 and 13.
Question 1(iii)
Seven bats and six balls cost ₹8880; three bats and five balls cost ₹4000.
Steps only
7b+6c=8880; 3b+5c=4000.
Elimination gives b=1200 and c=80.
Answer
Bat ₹1200; ball ₹80.
Question 1(iv)
Trips of 10 km and 15 km cost ₹155 and ₹220. Find fixed charge, rate and 25 km fare.
Steps only
f+10r=155; f+15r=220.
Subtract:5r=65, so r=13 and f=25.
25 km fare=25+25×13=₹350.
Answer
Fixed charge ₹25; ₹13/km; 25 km costs ₹350.
Question 1(v)
Adding 2 to numerator and denominator gives 9/11; adding 3 gives 5/6.
Steps only
11(x+2)=9(y+2) → 11x−9y=−4.
6(x+3)=5(y+3) → 6x−5y=−3.
Solving gives x=7,y=9.
Answer
The fraction is 7/9.
Question 1(vi)
(x+1)/(y−1)=1 and x/(y+1)=1/2.
Steps only
x−y=−2 and 2x−y=1.
Subtract to get x=3, then y=5.
Answer
The fraction is 3/5.
Question 1(vii)
Five years ago Nuri was thrice Sonu; in ten years Nuri will be twice Sonu.
Steps only
N−5=3(S−5) → N−3S=−10.
N+10=2(S+10) → N−2S=10.
Solve: S=20,N=50.
Answer
Nuri is 50 years old and Sonu is 20.
Question 1(viii)
Digit sum is 9 and nine times the number is twice its reversal.
Steps only
x+y=9.
9(10x+y)=2(10y+x) → y=8x.
Then 9x=9, so x=1,y=8.
Answer
The number is 18.
Question 1(ix)
₹2000 is given in 25 notes of ₹50 and ₹100.
Steps only
x+y=25; 50x+100y=2000.
Divide money equation by 50: x+2y=40.
Subtract to get y=15,x=10.
Answer
10 notes of ₹50 and 15 notes of ₹100.
Question 1(x)
Seven days cost ₹27 and five days cost ₹21 with a fixed first-three-day fee.
Steps only
f+4r=27; f+2r=21.
Subtract:2r=6, so r=3; f=15.
Answer
Fixed charge ₹15 and extra-day charge ₹3.
Question 2
Ten quiz students include four more girls than boys.
Steps only
b+g=10; g−b=4.
The two lines intersect at (b,g)=(3,7).
Answer
3 boys and 7 girls.
Question 3(i)
Classify 5x−4y+8=0 and 7x+6y−9=0.
Steps only
Coefficient comparison: 5/7≠−4/6.
Answer
The lines intersect at one point.
Question 3(ii)
Classify 9x+3y+12=0 and 18x+6y+24=0.
Steps only
Coefficient comparison: 9/18=3/6=12/24.
Answer
The lines are coincident.
Question 3(iii)
Classify 6x−3y+10=0 and 2x−y+9=0.
Steps only
Coefficient comparison: 6/2=(−3)/(−1)≠10/9.
Answer
The lines are parallel.
Question 4(i)
Determine solutions for x+y=5 and 2x+2y=10.
Steps only
The graphs are coincident.
Infinitely many solutions: every point on x+y=5.
Answer
Infinitely many solutions: every point on x+y=5.
Question 4(ii)
Determine solutions for x−y=8 and 3x−3y=16.
Steps only
The graphs are parallel.
No solution.
Answer
No solution.
Question 4(iii)
Determine solutions for 2x+y−6=0 and 4x−2y−4=0.
Steps only
The graphs are intersecting.
Unique solution (2,2).
Answer
Unique solution (2,2).
Question 4(iv)
Determine solutions for 2x−2y−2=0 and 4x−4y−5=0.
Steps only
The graphs are parallel.
No solution.
Answer
No solution.
Question 5
Half-perimeter is 36 m and length is 4 m more than width.
Steps only
l+w=36; l=w+4.
2w+4=36, so w=16 and l=20.
Answer
Length 20 m and width 16 m.
Question 6(i)
Pair 2x+3y−8=0 with an equation giving intersecting lines.
Steps only
One valid choice is x+y=0.
Answer
x+y=0.
Question 6(ii)
Pair 2x+3y−8=0 with an equation giving parallel lines.
Steps only
One valid choice is 4x+6y−1=0.
Answer
4x+6y−1=0.
Question 6(iii)
Pair 2x+3y−8=0 with an equation giving coincident lines.
Steps only
One valid choice is 4x+6y−16=0.
Answer
4x+6y−16=0.
Question 7
Solve 9x+7y=107 and 7x+9y=101.
Steps only
Add:16x+16y=208 → x+y=13.
Subtract:2x−2y=6 → x−y=3.
Thus x=8,y=5.
Answer
Citron costs 8 units; wood-apple costs 5 units.
Question 8
Five pencils and seven pens cost ₹50; seven pencils and five pens cost ₹46.
Steps only
Add:12p+12q=96 → p+q=8.
Subtract gives q−p=2.
So pencil p=3 and pen q=5.
Answer
Pencil ₹3; pen ₹5.
Question 9
A cat on a stool reaches 85 cm; stool top is 25 cm above the cat when it stands beside it.
Steps only
Let stool height s and cat height c.
s+c=85 and s−c=25.
Add:2s=110, so s=55.
Answer
The stool is 55 cm high.
End-of-Chapter Exercises
Textbook pages 119–121
Question 1(i)
Find the slope of y=3x through (0,0),(2,6).
Steps only
m=(6−0)/(2−0)=6/2=3.
Answer
Slope = 3.
Question 1(ii)
Find and interpret the y-intercept of y=3x.
Steps only
At x=0, y=0.
Answer
The y-intercept is 0 at (0,0); the starting water level is 0 cm.
Question 1(iii)(a)
Find water height after 5 minutes for y=3x.
Steps only
y=3(5)=15.
Answer
15 cm.
Question 1(iii)(b)
When does y=3x reach 21 cm?
Steps only
21=3x, so x=7.
Answer
7 minutes.
Question 1(iv)
Does (4,10) lie on y=3x?
Steps only
For x=4 the line requires y=12, not 10.
Answer
No.
Question 1(v)
Where do y=3x and x=3 intersect?
Steps only
Set x=3 in y=3x.
y=9.
Answer
(3,9).
Question 2(i)
Graph Celsius against Fahrenheit.
Steps only
Use F=(9/5)C+32.
Plot convenient points such as (0,32), (10,50), (−40,−40) and join them.
Answer
A straight line through (0,32), (10,50) and (−40,−40).
Question 2(ii)
Convert 30°C.
Steps only
F=(9/5)(30)+32=54+32.
Answer
86°F.
Question 2(iii)
Convert 95°F.
Steps only
95=(9/5)C+32; 63=(9/5)C.
C=35.
Answer
35°C.
Question 2(iv)
Convert both zero readings.
Steps only
0°C gives F=32°F.
0°F gives C=(5/9)(−32)=−160/9≈−17.78°C.
Answer
0°C = 32°F; 0°F ≈ −17.78°C.
Question 2(v)
Find the same numerical temperature.
Steps only
Set F=C: C=(9/5)C+32.
−(4/5)C=32, so C=−40.
Answer
−40°C = −40°F.
Question 3
Solve 2x+y=6 and 2x−y−2=0.
Steps only
Add the equations:4x=8, so x=2.
Then y=2.
Answer
The graphs intersect at (2,2).
Question 4
Find the intersection of the lines on the cover page.
Steps only
The downloadable chapter/front-matter PDF supplied for this edition shows no coordinate lines on its title cover.
Read the crossing coordinate from the printed outer cover if your copy contains the graph.
Answer
The coordinate cannot be determined from the current official digital cover because the referenced lines are absent.
Question 5
Find the x-intercept of y=mx+c.
Steps only
At the x-axis, y=0.
0=mx+c gives x=−c/m.
Answer
The x-intercept is (−c/m,0), for m≠0.
Question 6(i)
When is Plan A cheaper?
Steps only
50+0.20m < 30+0.30m.
20 < 0.10m, so m>200.
Answer
For more than 200 minutes.
Question 6(ii)
When is Plan B cheaper and when are costs equal?
Steps only
Plan B is cheaper for m<200.
At m=200 both cost ₹90.
Answer
Plan B is cheaper below 200 minutes; they are equal at 200 minutes.
Question 7(i)
How many lines have a given slope?
Steps only
Changing the y-intercept shifts the line without changing its slope.
Answer
Infinitely many lines.
Question 7(ii)
How many lines have a given slope and pass through a given point?
Steps only
The point fixes the intercept once the slope is fixed.
Answer
Exactly one line.
Question 8
For which p do 4x+py+8=0 and 2x+2y+2=0 have a unique solution?
Steps only
Unique means 4/2 ≠ p/2.
2 ≠ p/2, so p≠4.
Answer
All real p except p = 4.
Question 9
Find a,b so (a+b)x−2by=5a+2b+1 and 3x−y=14 coincide.
Steps only
Let the first be k times the second: a+b=3k and 2b=k.
Also 5a+2b+1=14k.
Solving gives k=2,b=1,a=5.
Answer
a = 5 and b = 1.
Question 10
Find k so x+2y=3 and (k−1)x+(k+1)y=k+3 coincide.
Steps only
Require k+1=2(k−1), giving k=3.
Then the second equation is 2x+4y=6, twice the first.
Answer
k = 3.
Question 11(i)
Robot paths have slopes 4/3 from (0,0) and 2 from (10,0). Find intersection.
Steps only
Paths: y=(4/3)x and y=2(x−10).
Solve: (4/3)x=2x−20, so x=30,y=40.
Answer
Yes; they intersect at (30,40).
Question 11(ii)
Robot paths have slope 1 from (3,0) and −3/5 from (7,0).
Steps only
Paths: y=x−3 and y=−(3/5)(x−7).
Solve to get x=9/2 and y=3/2.
Answer
Yes; they intersect at (4.5,1.5).
Question 12
At 2 o’clock the display is a minutes past; 15 minutes later it is b minutes past 3, with a=6b.
Steps only
Crossing 3 o’clock gives a+15=60+b, so a−b=45.
With a=6b:5b=45, so b=9.
Answer
The second time was 3:09.
Question 13
Digit sum is 15; reversal exceeds the number by 9.
Steps only
x+y=15.
(10y+x)−(10x+y)=9 gives y−x=1.
Solve x=7,y=8.
Answer
The number is 78.
Question 14
A=x+7, B=y+8, C=3y+23, D=4x+12. Find all angles.
Steps only
Opposite angles sum 180: x+3y=150 and 4x+y=160.
Solve x=30,y=40.
A=37,B=48,C=143,D=132.
Answer
∠A=37°, ∠B=48°, ∠C=143°, ∠D=132°.
Question 15
A train saves 6 h at v+6 and loses 6 h at v−4. Find speed and distance.
Steps only
Let normal time t and distance d=vt. The first condition gives t=v+6.
The second gives 2t−3v+12=0.
Solve v=24 km/h, t=30 h, d=720 km.
Answer
Speed 24 km/h; distance 720 km.
Question 16
A father’s age equals the sum of four children’s ages. In 20 years their sum will be twice his age.
Steps only
Let both current quantities be F.
After 20 years children sum F+80; father is F+20.
F+80=2(F+20), so F=40.
Answer
The father is 40 years old.