10 Math LinearEquations

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Finish Line & Beyond

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

1. Two linear equations in the same two variables are called a pair of linear equations

in two variables. The most general form of a pair of linear equations is

a1 x + b1y + c 1 = 0

a2 x + b2y + c 2 = 0

where, a1, a2, b1, b2, c 1, c 2 are real numbers, such that 0,02

2

2

2

2

1

2

1 ≠+≠+ baba

2. A pair of linear equations in two variables can be represented, and solved, by the:

(i) graphical method

(ii) algebraic method

3. Graphical Method :

The graph of a pair of linear equations in two variables is represented by two lines.

(i) If the lines intersect at a point, then that point gives the unique solution of thetwo equations. In this case, the pair of equations is consistent.

(ii) If the lines coincide, then there are infinitely many solutions — each point on theline being a solution. In this case, the pair of equations is dependent (consistent).

(iii) If the lines are parallel, then the pair of equations has no solution. In this case,

the pair of equations is inconsistent.

4. Algebraic Methods : We have discussed the following methods for finding the solu-

tion(s) of a pair of linear equations :

(i) Substitution Method(ii) Elimination Method

(iii) Cross-multiplication Method

5. If a pair of linear equations is given by a1 x + b1y + c 1 = 0 and a2 x + b2y + c 2 = 0,then the following situations can arise :

(i)2

1

2

1

b

b

a

a≠

In this case, the pair of linear equations is consistent.

(ii)2

1

2

1

2

1

c

c

b

b

a

a≠=

In this case, the pair of linear equations is inconsistent.

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(iii)2

1

2

1

2

1

c

c

b

b

a

a==

In this case, the pair of linear equation is dependent and consistent.

6. There are several situations which can be mathematically represented by twoequations that are not linear to start with. But we alter them so that they are re-

duced to a pair of linear equations.

EXERCISE 1

1. Aftab tells his daughter, “Seven years ago, I was seven times as old asyou were then. Also, three years from now, I shall be three times as old as

you will be.” Represent this situation algebraically and graphically.

Answer: Let us assume that seven years ago Aftab’s age was x years and hisdaughter’s age was y.

So, as per the question y x 7=07 =−⇒ y x --------------------------------- (i)

The equation gives us the following table:

X Y

7 1

14 2

21 3

28 4

35 5

42 6

49 7

Now, three years from now means 10 years from 7 years back

Aftab age will be 10+ x

Daughter’s age will be 10+ y

As per question )10(310 +=+ y x

30310 +=+⇒ y x

203 +=⇒ y x0203 =−−⇒ y x ---------------------------- (ii)







The equation gives us following table:

X Y

23 126 2

29 3

32 4

35 5

38 6

41 7

0

10

20

30

40

50

60

1 2 3 4 5 6 7

Daughter's Age

F a t h e r ' s A g e

07 =−⇒ y x

0203 =−−⇒ y x

2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later,

she buys another bat and 2 more balls of the same kind for Rs 1300. Rep-resent this situation algebraically and geometrically.

Answer: Let us assume number of bats = x and that of balls = y. As per the ques-

tion we get following equations:

390063 =+ y x

y x 639003 −=⇒ y x 21300 −=⇒ ---------------------- (i)

13002 =+ y x

y x 21300 −=⇒ ---------------------- (ii)

As both equations are similar so there coincident lines in the graph, meaning there

can be infinite number of solutions to this problem.









130

135

140

145

150

155

160

1 2 3 4 5

Apples

G r a p e s

y=160-2x

y=150-2x

Since we get parallel lines in this graph so there will be no solution as equations are

inconsistent.

EXERCISE 2

1. Form the pair of linear equations in the following problems, and find their

solutions graphically.(i) 10 students of Class X took part in a Mathematics quiz. If the number of

girls is 4 more than the number of boys, find the number of boys and girlswho took part in the quiz.

Answer: Let us assume number of girls to be x and that of boys to be y. Then weget following equations:

4+= y x -------------------- (i)

The equation will give following table:

X Y

4 0

5 1

6 2

7 3

8 4

9 5

10=+ y x

y x −=⇒ 10 --------------- (ii)








X Y

10 09 1

8 2

7 3

6 4

5 5

0

2

4

6

8

10

12

1 2 3 4 5 6

Boys

G i r l s x=y+4

x=10-y

As both lines intersect at 7, so Number of Girls= 7 and that of Boys=3

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens

together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer: 5075 =+ y x

y x 7505 −=⇒ ------------------ (i)


y X

-15 31

-10 24

-5 17

0 10

5 3

10 -4

15 -11

8 -1.2

9 -2.6

10 -4







4657 =+ y x

y x 5467 −=⇒ ----------------- (ii)

The equation gives following table

y X

-9 13

-2 8

5 3

12 -2

19 -7

-15

-10

-5

0

5

10

15

20

25

30

35

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

Pencil

P e n 5x=50-7y

7x=46-5y

Both lines are intersecting at 3, so x= 3Putting the value of x in either of the equations we can get the value of y, which is

equal to 5. Also, 5 is the only value of y which gives similar values for x in both thetables.

So cost of one pencil = Rs. 3 and one pen = 5

Check: 5075 =+ y x

5035155735 =+=×+×⇒

2. On comparing the ratios2

1

2

1

2

1 ,c

cand

b

b

a

a, find out if the lines representing

the following pairs of linear equations intersect at a point, are parallel or co-

incident:

(i) 5 x – 4y + 8 = 0

7 x + 6y – 9 = 0







7

5

7

5

2

1 == x

x

a

a

64

64

2

1 −=−= y

ybb

9

8

9

8

2

1 −=−

=c

c

As it is clear that2

1

2

1

b

b

a

a≠ , so lines will be intersecting.

(ii) 9 x + 3y + 12 = 0

18 x + 6y + 24 = 0

2

1

18

9

2

1 ==a

a

2

1

6

3

2

1 ==b

b

2

1

24

12

2

1 ==c

c

It is clear that2

1

2

1

2

1

c

c

b

b

a

a== , so lines are coincidental.

(iii) 6 x – 3y + 10 = 02 x – y + 9 = 0

1

3

2

6

2

1 ==a

a

1

3

1

3

2

1 =−−

=b

b

9

10

2

1 =c

c

It is clear that 2

1

2

1

2

1

c

c

b

b

a

a

≠= , so lines will be parallel







3. On comparing the ratios2

1

2

1

2

1 ,c

cand

b

b

a

afind out whether the following pair

of linear equations are consistent, or inconsistent.

(i) 3 x + 2y = 5 ; 2 x – 3y = 7

2

3

2

1 =a

a

2

3

2

3

2

1 −=−

=b

b

Here,2

1

2

1

b

b

a

a≠ , so the pair of linear equations is consistent.

(ii) 2 x – 3y = 8 ; 4 x – 6y = 9

2

1

4

2

2

1 ==a

a

2

1

3

6

2

1 =−−

=b

b

9

8

9

8

2

1 =−−

=c

c

Here,2

1

2

1

2

1

c

c

b

b

a

a≠= , so the pair of linear equations is inconsistent.

(iii) 73

5

2

3=+ y x , 9 x – 10y = 14

6

1

92

3

2

1 =×

=a

a

6

1

103

5

2

1 −=−×

=b

b

Here,2

1

2

1

b

b

a

a≠ , so the pair of linear equations is consistent.







(iv) 5 x – 3y = 11 ; – 10 x + 6y = –22

2

1

10

5

2

1 −=−

=a

a

2

1

6

3

2

1 −=−

=b

b

2

1

22

11

2

1 −=−

=c

c

Here,2

1

2

1

2

1

c

c

b

b

a

a== , so the pair of linear equations is dependent and consistent.

(v) 823

4

=+y x ; 2 x + 3y = 12

3

2

23

4

2

1 =×

=a

a

3

2

2

1 =b

b

3

2

12

8

2

1 =−−

=c

c

In this case,2

1

2

1

2

1

cc

bb

aa == , so the pair of linear equation is dependent and consist-

ent.

4. Which of the following pairs of linear equations are consistent/inconsist-

ent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2 x + 2y = 10

2

1

2

1 =a

a

21

2

1 =bb

2

1

10

5

2

1 =−−

=c

c









(ii) x – y = 8, 3 x – 3y = 16

3

1

2

1 =a

a

3

1

2

1 =b

b

2

1

16

8

2

1 =−−

=c

c

Here,2

1

2

1

2

1

c

c

b

b

a


(iii) 2 x + y – 6 = 0, 4 x – 2y – 4 = 0

2

1

2

1

=a

a

2

1

2

1 =b

b

2

3

4

6

2

1 =−−

=c

c

Here,2

1

2

1

2

1

c

c

b

b

a


(iv) 2 x – 2y – 2 = 0, 4 x – 4y – 5 = 0

21

42

2

1 ==aa

2

1

2

1 =b

b

5

2

2

1 =c

c

Here,2

1

2

1

2

1

c

c

b

b

a








5. Half the perimeter of a rectangular garden, whose length is 4 m more

than its width, is 36 m. Find the dimensions of the garden.

Answer: Let us assume Length = x and breadth =y

Then x=y+4

Perimeter = 2(x+y)

3642

Pr =++=+=⇒ x x y ximeter

3642 =+⇒ x

324362 =−=⇒ x

16232 =÷=⇒ xHence, y=16-4=12So, Length = 16 m and Breadth = 12 m

6. Given the linear equation 2 x + 3y – 8 = 0, write another linear equation

in two variables such that the geometrical representation of the pair soformed is:

(i) intersecting lines (ii) parallel lines (iii) coincident lines

Answer: (i) As you know that if 2

1

2

1

b

b

a

a≠ , then the lines will be intersecting, so let

us put a number for a2 and b2 in such a way which satisfies this condition.

Next possible equation can be 0723 =−+ y x

(ii) As you know that if 2

1

2

1

2

1

c

c

b

b

a

a

≠= , then the lines will be parallel, so the possible

equation can be as follows:

0864 =−+ y x

(iii) As you know that if 2

1

2

1

2

1

c

c

b

b

a

a== , then the lines will be coincidental, so the pos-

sible equation can be as follows:

01664 =−+ y x







EXERCISE 3

1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14, x-y=4

Answer: 14=+ y x

y x −=⇒ 14

Substituting this value of x in the second equation we get

414 =−− y y

4214 =−⇒ y

122 =⇒ y

6=⇒ y

8=⇒ x

(ii) s – t = 3, 623

=+t s

Answer: s = 3+t

Substituting the value of s in the second equation we get

623

3=+

+ t t

66

326=

++⇒

t t

3656 =+⇒ t

305 =⇒ t

6=⇒ t Hence, s = 3+6=9

(iii) 3 x – y = 3, 9 x – 3y = 9

Answer: y=3x-3

Putting this value of y in the second equation we get9x-3(3x-3)=9

9999 =+−⇒ x xIt is clear that this equation will have infinitely many solutions. It is important to

note that here2

1

2

1

2

1

c

c

b

b

a

a==







(iv) 0.2 x + 0.3y = 1.3

0.4 x + 0.5y = 2.3

Answer: 2x+3y=13 (after multiplying the equation by 10)3y=13-2x

Or, y =3

213 x−

Putting this value of y in the second equation we get

233

21354 =

−

+x

x (equation 2 is also multiplied like equation 1)

69106512 =−+⇒ x x

69652 =+⇒ x

42 =⇒ x

2=⇒ x

33

413 =−=⇒ y

(v) 032 =+ y x , 083 =− y x

Answer: y x 32 −= y x2

3−=⇒

Putting this value of x in equation 2 we get

082

33 =−−× y y 08

2

3=−−⇒ y y 0

2

163=

−− y y0

2

43=

−−⇒

y y

043 =−−⇒ y y y y 43 =−⇒Since coefficients of y are different on both LHS and RHS, so only possible value of

y=0Hence, x=0

2. Solve 2 x + 3y = 11 and 2 x – 4y = – 24 and hence find the value of ‘m’ for

which y = mx + 3.

Answer: y x 3112 −=Substituting in equation 2 we get

244311 −=−− y y

24711 −=−⇒ y

3524117 =+=⇒ y

5=⇒ y

453112 −=×−=⇒ x

2−=⇒ x







Now putting values of x and y in equation 3 we get

325 +−×= m

2352 =−=−⇒ m

1−=⇒ m

3. Form the pair of linear equations for the following problems and find theirsolution by substitution method.

(i) The difference between two numbers is 26 and one number is threetimes the other. Find them.

Answer: Let us assume one of the numbers = x and another number = y

Then as per question 26=− y x ---------------- (i)

y x 3= -------------------------------------------- (ii)

Substituting the value of x in equation (i)

263 =− y y

262 =⇒ y

13=⇒ y

39313 =×=⇒ x

(ii) The larger of two supplementary angles exceeds the smaller by 18 de-

grees. Find them.

°=+ 180 y x ------------- (i)

°=− 18 y x -------------- (ii)

°+=⇒ 18 y x

Substituting in equation (i) we get°=°++ 18018 y y

°=°−°=⇒ 162181802 y

°=⇒ 81 y

°=°+°=⇒ 991881 x

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later,

she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and eachball.

Answer: 380067 =+ y x ----------- (i)

175053 =+ y x --------------------- (ii) y x 517503 −=⇒

3

51750 y x

−=⇒

Substituting in equation (i) we get

380063

)51750(7=+

− y

y







11400183512250 =+−⇒ y y

114001712250 =−⇒ y

850114001225017

=−=⇒y

50=⇒ y

5003

2501750=

−=⇒ x

Hence, Cost of Bat = Rs. 500

Cost of Ball = Rs. 50

(iv) The taxi charges in a city consist of a fixed charge together with thecharge for the distance covered. For a distance of 10 km, the charge paid is

Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are thefixed charges and the charge per km? How much does a person have to pay

for travelling a distance of 25 km?

Answer: Let us assume the fixed charge = xAnd variable charge = y

As per question:

10510 =+ y x ------------------- (i)

15515 =+ y x ------------------- (ii)

y x 15155 −=⇒Substituting in equation (i) we get

1051015155 =+− y y 1055155 =−⇒ y 501051555 =−=⇒ y 10=⇒ y

51015155 =×−=⇒ x

So for 25 km the fair = 25510255 =×+

(v) A fraction becomes11

9, if 2 is added to both the numerator and the de-

nominator. If, 3 is added to both the numerator and the denominator it be-

comes6

5. Find the fraction.

Answer: Let us assume the numerator = x and denominator = y

As per question:

119

22 =++ y

x

1892211 +=+⇒ y x

4119 =−⇒ x y --------------- (i)

6

5

3

3=

++

y

x

155186 +=+⇒ y x







365 =−⇒ x y --------------- (ii)

365 +=⇒ x y

536 +=⇒ x y

Substituting in equation (i) we get

4115

2754=−

+ x

x

20552754 =−+⇒ x x

2027 =−⇒ x

7=⇒ x

95

376=

+×=⇒ y

Hence, required fraction = 9

7

(vi) Five years hence, the age of Jacob will be three times that of his son.

Five years ago, Jacob’s age was seven times that of his son. What are theirpresent ages?

Answer: Let us assume five years ago son’s age = x and Jacob’s age = y

Then x y 7= ------------- (i)

Son’s age five years from now = x+10 and Jacob’s age will be = y+10

So, 303)10(310 +=+=+ x x y

203 +=⇒ x y -------------- (ii)

Substituting in equation (i) x x 7203 =+

204 =⇒ x

5=⇒ x

35=⇒ y

Present age of Jacob = 40 yearsPresent age of son = 10 years







EXERCISE 4

1. Solve the following pair of linear equations by the elimination method

and the substitution method:(i) x + y = 5 and 2 x – 3y = 4

Answer: Multiplying equation (i) by 2 and subtracting equation (ii) from this as fol-

lows:

650

432

1022

=+

=−=+

y

y x

y x

5

6=⇒ y , substituting in equation (i)

55

6

=+ x

5

19

5

625=

−=⇒ x

(ii) 3 x + 4y = 10 and 2 x – 2y = 2

Answer: Multiplying equation (i) by 2 and equation (ii) by 3 and subtracting latter

from former as follows:

14140

666

2086

=+

=−=+

y

y x

y x


1043 =+ x

63 =⇒ x

2=⇒ x

(iii) 3 x – 5y – 4 = 0 and 9 x = 2y + 7

Answer: Multiplying equation (i) by 3 and subtracting equation (ii) as follows:

03100

0729

04159

=+−

=−−=−−

y

y x

y x

310 =⇒ y10


042

33 =−− x

02

836=

−−⇒

x







116 =⇒ x

6

11=⇒ x

(iv) 13

2

2−=+

y xand 3

3=−

y x

Answer: Simplify equation (i)

13

2

2−=+

y x

16

43−=

+⇒

y x

643 −=+⇒ y x ----------------- (iii)Simplify equation (ii)

33

=−y

x

93 =−⇒ y x ------------------- (iv)

Subtracting equation (iv) from (iii) as follows:

1550

93

643

−=+

=−−=+

y

y x

y x

3−=⇒ y , substituting in equation (iii)

6123 −=− x

63 =⇒ x

2=⇒ x

2. Form the pair of linear equations in the following problems, and find their

solutions (if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a

fraction reduces to 1. It becomes2

1, if we only add 1 to the denominator.

What is the fraction?

Answer: 111 =

++

y x

11 +=+⇒ y x

y x =⇒ ------------ (i)

2

1

1=

+ y

x







⇒ 12 += y x ------------ (ii)

Substituting value of x from equation (i) in equation (ii)

12 += y y

1=⇒ y

1=⇒ x

(ii) The sum of the digits of a two-digit number is 9. Also, nine times this

number is twice the number obtained by reversing the order of the digits.Find the number.

Answer: Let two digits of the number are x and y.

9=+ y x

y x −=⇒ 9 ---------------- (i)

Then First number = y x +10 ( x is at 10s place and y is at unit’s place)

Reversing the number we get a number which can be written as x y +10

As per question:

)10(2)10(9 x y y x +=+ x y y x 220990 +=+⇒ )9(2209)9(90 y y y y −+=+−⇒ y y y y 21820990810 −+=+−⇒ 181881810 +=−⇒ y y 7921881099 =−=⇒ y

899792 =÷=⇒ y

189 =−=⇒ xSmaller number = 18 and larger number = 81

162281918 =×=×

(iii) Meena went to a bank to withdraw Rs 2000. She asked the cashier to

give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how

many notes of Rs 50 and Rs 100 she received.

Answer: 25=+ y x

y x −=⇒ 25 ------------ (i)

200010050 =+ y x

2000100)25(50 =+−⇒ y y

2000100501250 =+−⇒ y y

7501250200050 =−=⇒ y

15=⇒ y Number of hundred rupees notes

10=⇒ x Number of fifty rupees notes.


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10 Math LinearEquations