# 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$\frac{1}{2a}+\frac{1}{3b}=2$

13a+12b=136$\frac{1}{3a}+\frac{1}{2b}=\frac{13}{6}$

Solution:

Let us assume 1a=mand1b=n$\frac{1}{a}=m\:and\:\frac{1}{b}=n$, then the equation will change as follows.

m2+n3=23m+2n12=0$\frac{m}{2}+\frac{n}{3}=2\:\Rightarrow 3m+2n-12=0$   …….(1)

m3+n2=1362m+3n13=0$\frac{m}{3}+\frac{n}{2}=\frac{13}{6}\:\Rightarrow 2m+3n-13=0$   …….(2)

Now, Using cross-multiplication method, we get

m26(36)=n24(39)=194$\frac{m}{-26-\left ( -36 \right )}=\frac{n}{-24-\left ( -39 \right )}=\frac{1}{9-4}$ m10=q15=15$\frac{m}{10}=\frac{q}{15}=\frac{1}{5}$ m10=15andq15=15$\frac{m}{10}=\frac{1}{5}\:and\:\frac{q}{15}=\frac{1}{5}$

m = 2 and n = 3

1a=2and1b=3$\frac{1}{a}=2\:and\:\frac{1}{b}=3$ a=12andb=13$a=\frac{1}{2}\:and\:b=\frac{1}{3}$

Question 2:

4a+3b=14$\frac{4}{a}+3b=14$

4a4b=23$\frac{4}{a}-4b=23$

Solution:

Putting 1a=m$\frac{1}{a}=m$ 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$\frac{m}{-69-56}=\frac{b}{-42-\left ( -92 \right )}=\frac{1}{-16-9}$ m125=b50=125$\frac{m}{-125}=\frac{b}{50}=\frac{-1}{25}$ m125=125andb50=125$\frac{m}{-125}=\frac{-1}{25}\:and\:\frac{b}{50}=\frac{-1}{25}$

m = 5 and b = -2

m = 1a$\frac{1}{a}$ = 5

a = 15$\frac{1}{5}$

b = -2

Question 3:

7a2bab=5$\frac{7a-2b}{ab}=5$

7b2a=5$\frac{7}{b}-\frac{2}{a}=5$ …..(i)

8a+7bab=15$\frac{8a+7b}{ab}=15$

8b+7a=15$\frac{8}{b}+\frac{7}{a}=15$ …..(ii)

Solution:

Substituting 1a=m$\frac{1}{a}=m$ 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$\frac{m}{-105-\left ( -40 \right )}=\frac{n}{-35-30}=\frac{1}{-16-49}$ m65=n65=165$\frac{m}{-65}=\frac{n}{-65}=\frac{1}{-65}$ m65=165andb65=165$\frac{m}{-65}=\frac{1}{-65}\:and\:\frac{b}{-65}=\frac{1}{-65}$

m = 1 and n = 1

m = 1a$\frac{1}{a}$ = 1        n = 1a$\frac{1}{a}$ = 1

a = 1    b = 1

Question 4:

10a+b+2ab=4$\frac{10}{a+b}+\frac{2}{a-b}=4$

15a+b5ab=2$\frac{15}{a+b}-\frac{5}{a-b}=-2$

Solution:

Substituting 1a+b=m$\frac{1}{a+b}=m$ and 1ab=n$\frac{1}{a-b}=n$ 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$\frac{m}{4-20 }=\frac{n}{-60-\left ( -20 \right )}=\frac{1}{-50-30}$ m16=n80=180$\frac{m}{-16}=\frac{n}{-80}=\frac{1}{-80}$ m16=180andb80=180$\frac{m}{-16}=\frac{1}{-80}\:and\:\frac{b}{-80}=\frac{1}{-80}$

m = 15$\frac{1}{5}$ and n = 1

m = 1a+b=15$\frac{1}{a+b}=\frac{1}{5}$ and n = 1ab=1$\frac{1}{a-b}=1$

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$\frac{2}{\sqrt{a}}+\frac{3}{\sqrt{b}}=2$

4a9b=1$\frac{4}{\sqrt{a}}-\frac{9}{\sqrt{b}}=-1$

Solution:

Substituting 1a=m$\frac{1}{\sqrt{a}}=m$ and 1b=n$\frac{1}{\sqrt{b}}=n$ 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$\frac{1}{2}$ ……(iv)

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

2×12+2n=2$2\times \frac{1}{2}+2n = 2$

3n = 1

n = 13$\frac{1}{3}$

m = 1a=12$\frac{1}{\sqrt{a}}=\frac{1}{2}$

a=2$\sqrt{a}=2$

a = 4

n = 1b=13$\frac{1}{\sqrt{b}}=\frac{1}{3}$

b=3$\sqrt{b}=3$

b = 9

Hence, a = 4 and b = 9

Question 6:

5a1+1b2=2$\frac{5}{a-1}+\frac{1}{b-2}=2$

6a13b2=1$\frac{6}{a-1}-\frac{3}{b-2}=1$

Solution:

Substituting 1a1=m$\frac{1}{a-1}=m$ and 1b2=n$\frac{1}{b-2}=n$ 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$\frac{1}{3}$

Putting this value in equation (i), we get

5×13+n=2$5\times \frac{1}{3}+n=2$ n=253=13$n=2-\frac{5}{3}=\frac{1}{3}$ m=1a1=13a1=3a=4$m=\frac{1}{a-1}=\frac{1}{3}\:\:\Rightarrow a-1=3\:\:\Rightarrow a=4$ n=1b2=13b2=3b=5$n=\frac{1}{b-2}=\frac{1}{3}\:\:\Rightarrow b-2=3\:\:\Rightarrow b=5$

Hence, a = 4 and b = 5

Question 7:

6a + 3b = 6ab

=> 6b+3a=6$\frac{6}{b}+\frac{3}{a}=6$ …..(i)

2a + 4b = 5ab

=> 2b+4a=5$\frac{2}{b}+\frac{4}{a}=5$ …….(ii)

Substituting 1a=m$\frac{1}{a}=m$ and 1b=n$\frac{1}{b}=n$
3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m30(12)=n24(15)=1624$\frac{m}{-30-\left ( -12 \right )}=\frac{n}{-24-\left ( -15 \right )}=\frac{1}{6-24}$ m18=n9=118$\frac{m}{-18}=\frac{n}{-9}=\frac{1}{-18}$ m18=118andn9=118$\frac{m}{-18}=\frac{1}{-18}\:and\:\frac{n}{-9}=\frac{1}{-18}$

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$\frac{1}{3a+b}+\frac{1}{3a-b}=\frac{3}{4}$

12(3a+b)12(3ab)=18$\frac{1}{2\left ( 3a+b \right )}-\frac{1}{2\left ( 3a-b \right )}=\frac{-1}{8}$

Solution:

substituting 13a+b=mand13ab=n$\frac{1}{3a+b}=m\:and\:\frac{1}{3a-b}=n$ 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$\frac{1}{3a+b}=\frac{1}{4}$

3a + b = 4  …….(3)

n = 13ab=12$\frac{1}{3a-b}=\frac{1}{2}$

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