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**Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6**

Question 1.

Solve the following pairs of equations by reducing them to a pair of linear equations :

(i) 12x+13y=2

13x+12y=136

(ii) 2x√+3y√=2

4x√+9y√=−1

(iii) 4x+3y=14

3x−4y=23

(iv) 5x−1+1y−2=2

6x−1−3y−2=1

(v) 7x−2yxy=5

8x+7yxy=15

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

(vii) 10x+y+2x−y=4

15x+y+5x−y=−2

(viii) 13x+y+13x−y=34

12(3x+y)−12(3x−y)=−18

Solution:

(i) 12x+13y=2 ……………. (i)

13x+12y=136 ……………… (ii)

From eqn. (i),

12x+13y=2

3y+2x6xy=2

3y + 2x = 12xy

2x + 3y – 12xy = 0

2x + 3y = 12xy …………….. (i)

From eqn. (ii),

13x+12y=136

2y+3x6xy=136

6(2y + 3x) = 13 × 6xy

12y + 18x = 78xy

18x + 12y – 78xy = 0

18x + 12y = 78xy ……………… (ii)

Multiplying eqn. (i) by 4,

4(2x + 3y) = 4 × 12xy

8x + 12y = 78xy …………….. (iii)

From eqn. (ii) – eqn. (iii),

x=30xy10=3xy

xx=3y

1 = 3y

y=13

2x + 3y = 12xy

2x+3(13)=12×x×13

2x + 1 = 4x

2x – 4x = 1

-2x = -1

2x = 1

x=12

x=12,y=13

(ii) 2x√+3y√=2 ………….. (i)

4x√+9y√=−1 ………….. (ii)

From eqn (i)

2x√+3y√=2

2y√+3x√xy√=2

3x−−√+2xy−−√=2xy−−√ …………… (i)

From eqn (ii),

4x√+9y√=−1

4y√−9x√xy√=−1

−9x−−√+4y√=−xy−−√

9x−−√−3y√=xy−−√ ……………….. (ii)

Multiplying eqn (i) with 2

2(3x−−√+2y√)=2×2xy−−√

6x−−√+4y√=4xy−−√ ……………….. (iii)

By adding eqn (ii) + eqn (iii)

x−−√=5xy√15

x√x√=5y√15

1=y√3

3=y√y√=3

∴ y = 9

Substituting the value of ‘y’ in eqn (i)

∴ x = 4, y = 9

(iii) 4x+3y=14

3x−4y=23

by putting the value of By multiplying eqn. (i) by 4, multiplying eqn. (ii) by 3, and then adding them,

∴ p = 5

1x=p=5x=15

Substituting the value of ‘p’ in eqn. (i)

4p + 3y = 14

4 × 5 + 3y = 14

20 + 3y = 14

3y = 14 – 20

3y = -6

∴ y = -2

∴ x⋅=15,y=−2

(iv) 5x−1+1y−2=2

6x−1−3y−2=1

Let 1x−1=p and 1y−2=q

5(1x−1)+1y−2=2 ……………. (i)

6(1x−1)−3(1y−2)=1 …………… (ii)

5p + 1 = 2 …………… (iii)

6p – 3q = 1 …………… (iv)

By solving equation p=13,q=13

Substituting 1x−1 for p,

p=1x−1=13

x – 1 = 3

∴ x = 4

Substituting 1y−2 for q,

q=1y−2=13

y – 2 = 3

∴ y = 5

∴ x = 4, y = 5

(v) 7x−2yxy=5 ……………. (i)

8x+7yxy=15 ………………… (ii)

From eqn. (i)

7x−2yxy=5

7x – 2y = 5xy ………….. (i)

From eqn. (ii)

8x+7yxy=15

8x + 7y = 15xy ……………… (ii)

Eqm (i) × 7 and Eqn. (ii) × 2, we have

49x – 14y = 35xy ………….. (iii)

16x + 14y = 30xy ……………. (iv)

From eqn. (iii) + eqn. (iv)

∴ 65x65x=y

1 = y

∴ y = 1

Substituting the value of ‘y’ in eqn. (i)

7x – 2y = 5xy

7x – 2 (1) = 5 × x × 1

7x – 2 = 5x

7x – 5x = 2

2x = 2

∴ x = 1

∴ x = 1, y = 1

(vi) 6x + 3y = 6xy ………………. (i)

2x + 4y = 5xy ………………… (ii)

Multiplying eqn. (ii) by 3

3(2x + 4y = 5xy)

6x + 12y = 18xy ………………. (iii)

From eqn. (i) – eqn. (ii)

9y = 9xy

9y9y=x

1 = x

∴ x = 1

Substituting the value of ‘x’ in eqn (i),

6x + 3y = 6xy

6 × 1 + 3y = 6 × 1 × y

6 + 3y = 6y

6y – 3y = 6

3y = 6

∴ y=63

∴ y = 2

∴ x = 1, y = 2

(viii) 10x+y+2x−y=4

15x+y+5x−y=−2

Let u=1x+y,v=1x−y then,

10u + 2v = 4 …………. (i)

15u – 5v = -2 …………. (ii)

Multiplying eqn. (i) by 5 and multiplying eqn. (ii) by 2 and then by adding

∴ u=15

Substituting the value of ‘u’ in eqn. (i)

10×15+2v=4

2 + 2v = 4

2v = 4 – 2

2v = 2

∴ v = 1

∴ u = 1/5 , v =1

1x+y=151x−y=1

∴ x + y = 5 x – y = 1

x + y = 5 …………….. (iii)

x – y = 1 …………….. (iv)

From eqn. (iii) + eqn. (iv)

x = 3

∴ Substituting the value of ‘x’ in (iii)

x + y = 5

3 + y = 5

y = 5 – 3

y = 2

∴ x = 3, y = 2

(viii) 13x+y+13x−y=34 ………….. (i)

12(3x+y)−12(3x−y)=−18 ……………. (ii)

Substituting the value of ‘p’ in eqn. (iii)

p+q=34

14+q=34

∴ q=34−14

∴ q=12

∴ p=13x+y=14

3x + y = 4 ……………. (v)

q=13x−y=12

3x – y = 2 ……………… (vi)

From eqn. (v) + eqn. (vi)

3x + y = 4

+3x – y = 2

6x = 6

∴ x=66=1

Substituting the value of ‘x’ in eqn. (v)

3x + y = 4

3 × 1 + y = 4

3 + y = 4

∴ y = 4 – 3

∴ y = 1

∴ x = 1, y = 1

Question 2.

Formulate the following problems as a pair of equations, and hence find their solutions :

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

Let speed of Ritu

in still water be x km/hr.

in current water be y km/hr

Speed of water while

upstream is (x + y) km/hr,

downstream is (x – y) km/hr,

2(x + y) = 20

x + y = 10 ………… (i)

2(x – y) = 4

x – y = 2 …………… (ii)

By Adding eqn. (i) and eqn. (ii),

∴ x=122=6

Substituting the value, x = 6 in eqn. (i),

x + y = 10

6 + y = 10

y = 10 – 6

∴ y = 4

∴ Speed of Ritu

in still water, x = 6 km/hr.

in current water, y = 4 km/hr.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days,, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. ‘

Solution:

Number of days required

for women be ’x’

for men be ‘y’

Work done by woman in 1 day =1x

Work done by man in 1 day =1y

4(2x+54)=1

2x+54=14 ………….. (i)

3(3x+6y)=1

3x+6y=13 ……………… (ii)

Let 1x=p and 1y=q

∴ 2p+5q=14

8p + 20q = 1 ……….. (iii)

3p+6q=13

9p + 18q = 1 ……………. (iv)

By Cross-multiplication

p−20−(−18)=q−9−(−8)=1144−180

p−2=q−1=1−36

If p−2=1−36 then p=−2−36=118

If q−1=1−36 then q=−1−36=136

p=1x=118∴x=18

q=1y=136∴y=36

∴ Number of days taken by one woman to complete work, x = 18

Number of days taken by one man to complete the work, y = 36.

(iii) Roohl travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

Let speed of the train be x’ km/hr. and Speed of the bus is ‘y’ km/hr.

60x+240y=4 …………… (i)

100x+200y=256 …………… (ii)

If 1x=p1y=q then

60p + 240q = 4 ……………… (iii)

100p+200q=256 ……………….. (iv)

Multiplying eqn. (iii) by 10, Mulitplying eqn. (iv) by 6 and then eqn. (iii) – eqn. (iv), we have

∴ q=151200

∴ q=180

Substituting the value of q=180 in eqn. (iii)

60p+240×180=4

60p + 3 = 4

60p =4 – 3

∴ p=160

∴ p=160=1x

∴ x = 60 km/hr.

∴ q=180=1y

∴ y = 80 km/hr.

∴ Speed of train, x = 60 km/hr.

Speed of bus, y = 80 km/hr.

**All Chapter KSEEB Solutions For Class 10 Maths**

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