Pa.Linear Eq Ex-3.1 |
Pa.Linear Eq Ex-3.2 |
Pa.Linear Eq Ex-3.3 |
Pa.Linear Eq Ex-3.5 |
Pa.Linear Eq Ex-3.6 |
Pa.Linear Eq Ex-3.7 |

Solve the following pair of linear equations by the elimination method and the substitution method:

**Answer
1** :

Solutions:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13

By the method of Substitution:

From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

-5y = -15

y = 3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

**Answer
2** :

Solution:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

Solution:

Let us assume, present age of Nuri is x

And present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

Age of Sonu is 20 years.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Manna has 10 notes of Rs.50 and 15 notes of Rs.100.

Solution:

Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

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