Ncert Solutions For Class 10 Maths Ex 3.6

Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.6

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

Question 1:

 12a+13b=2

 13a+12b=136

 Solution:

 Let us assume 1a=mand1b=n, then the equation will change as follows.

m2+n3=23m+2n12=0   …….(1)

m3+n2=1362m+3n13=0   …….(2)

Now, Using cross-multiplication method, we get

m26(36)=n24(39)=194 m10=q15=15 m10=15andq15=15

m = 2 and n = 3

1a=2and1b=3 a=12andb=13

 

Question 2:

4a+3b=14

4a4b=23

 

Solution:

 Putting 1a=m in the given equation we get,

So, 4m + 3b = 14     => 4m + 3b – 14 = 0  ………..(1)

3m – 4b = 23     => 3m – 4b – 23 = 0  ………..(2)

By cross-multiplication, we get,

m6956=b42(92)=1169 m125=b50=125 m125=125andb50=125

m = 5 and b = -2

m = 1a = 5

a = 15

b = -2

 

Question 3:

7a2bab=5

7b2a=5 …..(i)

8a+7bab=15

8b+7a=15 …..(ii)

Solution:

 Substituting 1a=m in the given equation we get,

– 2m + 7n = 5     => -2 + 7n – 5 = 0  ……..(iii)

7m + 8n = 15     => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m105(40)=n3530=11649 m65=n65=165 m65=165andb65=165

m = 1 and n = 1

m = 1a = 1        n = 1a = 1

a = 1    b = 1

 

Question 4:

10a+b+2ab=4

15a+b5ab=2

Solution:

Substituting 1a+b=m and 1ab=n in the given equations, we get,

10m + 2n = 4      =>  10m + 2n – 4 = 0      …..(i)

15m – 5n = -2     =>   15m – 5n + 2 = 0    …..(ii)

Using cross-multiplication method, we get,

m420=n60(20)=15030 m16=n80=180 m16=180andb80=180

m = 15 and n = 1

m = 1a+b=15 and n = 1ab=1

a + b = 5  …..(iii)

and a – b = 1 …..(iv)

Adding equation (iii) and (iv), we get

2a = 6   => a = 3 …….(v)

Putting the value of a = 3 in equation (3), we get

b = 2

Hence, a = 3 and b = 2

 

Question 5:

2a+3b=2

4a9b=1

 

Solution:

 Substituting 1a=m and 1b=n in the given equations, we get

2m + 3n = 2 …….(i)

4m – 9n = -1 ……(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………..(iii)

Adding equation (ii) and (iii), we get

10m = 5

m = 12 ……(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×12+2n=2

3n = 1

n = 13

m = 1a=12

a=2

a = 4

n = 1b=13

b=3

b = 9

Hence, a = 4 and b = 9

 

Question 6:

5a1+1b2=2

     6a13b2=1

 

Solution:

 Substituting 1a1=m and 1b2=n in the given equations, we get,

5m + n = 2 ……(i)

6m – 3n = 1 …..(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 ……..(iii)

Adding (ii) and (iii) we get

21m = 7

m = 13

Putting this value in equation (i), we get

5×13+n=2 n=253=13 m=1a1=13a1=3a=4 n=1b2=13b2=3b=5

Hence, a = 4 and b = 5

 

Question 7:

6a + 3b = 6ab

=> 6b+3a=6 …..(i)

    2a + 4b = 5ab

=> 2b+4a=5 …….(ii)

Substituting 1a=m and 1b=n
3m + 6n – 6 = 0

4m + 2n – 5 = 0

 

By cross-multiplication method, we get

m30(12)=n24(15)=1624 m18=n9=118 m18=118andn9=118

m = 1 and n = 1/2

m = 1/a = 1 and n = 1/b = 1/2

a = 1                              b = 2

Hence, a = 1 and b = 2

 

Question 8:

13a+b+13ab=34

     12(3a+b)12(3ab)=18

 

Solution:

 substituting 13a+b=mand13ab=n in the given equations, we get

m + n = 3/4  ……(1)

m/2 – n/2 = -1/8

m – n = -1/4  ……(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

2m = 1/2

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

 

m = 13a+b=14

3a + b = 4  …….(3)

n = 13ab=12

3a – b = 2 ………(4)

Adding equations (3) and (4), we get

6a = 6

a = 1 ……..(5)

Putting in (3), we get

3(1) + b = 4

b = 1

Hence, a = 1 and b = 1

 

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