RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.4

RD Sharma Solutions Class 10 Chapter 3 Exercise 3.4

RD Sharma Class 10 Solutions Chapter 3 Ex 3.4 PDF Free Download

Exercise 3.4

 

Q.1: x + 2y + 1 = 0 and 2x – 3y – 12 = 0

 

Soln:

x+2y+1 =0 …………………………………….(i)

2x-3y-12=0………………………………….. (ii)

Here a1= 1 , b1= 2 , c1= 1

a2= 2 , b2= -3 , c2= -12

By cross multiplication method,

\(\frac{x}{-24+3}=\frac{-y}{-12-2}=\frac{1}{-3-4}\)

\(\frac{x}{-21}=\frac{-y}{-14}=\frac{1}{-7}\)

Now,

\(\frac{x}{-21}=\frac{1}{-7}\)

=x= 3

And,

\(\frac{-y}{-14}=\frac{1}{-7}\)

=y=-2

The solution of the given system of equation is 3 and -2 respectively.

 

Q.2: 3x + 2y + 25 = 0, 2x + y + 10 = 0

 

Soln:

3x+2y+25 =0 …………………………………….(i)

2x+y+10=0………………………………….. (ii)

Here a1= 3 , b1= 2 , c1= 25

a2= 2 , b2= 1 , c2= 10

By cross multiplication method,

\(\frac{x}{20-25}=\frac{-y}{30-50}=\frac{1}{3-4}\)

\(\frac{x}{-5}=\frac{-y}{-20}=\frac{1}{-1}\)

Now,

\(\frac{x}{-5}=\frac{1}{-1}\)

=x= 5

And,

\(\frac{-y}{-20}=\frac{1}{-1}\)

=y=-20

The solution of the given system of equation is 5 and -20 respectively.

 

Q.3: 2x + y = 35, 3x + 4y = 65

 

Soln:

2x+y= 35 ……………………………….(i)

3x+4y=65…………………………….. (ii)

Here a1= 2 , b1= 1 , c1= 35

a2= 3 , b2= 4 , c2= 65

By cross multiplication method,

\(\frac{x}{-65+140}=\frac{-y}{-130+105}=\frac{1}{8-3}\)

\(\frac{x}{75}=\frac{-y}{-25}=\frac{1}{5}\)

Now,

\(\frac{x}{75}=\frac{1}{5}\)

=x= 15

And,

\(\frac{-y}{-25}=\frac{1}{5}\)

=y=5

The solution of the given system of equation is 15 and 5 respectively.

 

Q.4: 2x – y – 6 = 0, x – y – 2 = 0

 

Soln:

2x-y= 6 ……………………………….(i)

x-y=2…………………………….. (ii)

Here a1= 2 , b1= -1 , c1= 6

a2= 1 , b2= -1 , c2= 2

By cross multiplication method,

\(\frac{x}{2-6}=\frac{-y}{-4+6}=\frac{1}{-2+1}\)

\(\frac{x}{-4}=\frac{-y}{2}=\frac{1}{-1}\)

Now,

\(\frac{x}{-4}=\frac{1}{-1}\)

=x= 4

And,

\(\frac{-y}{2}=\frac{1}{-1}\)

=y=2

The solution of the given system of equation is 4 and 2 respectively.

 

Q5: \(\frac{x+y}{xy}=2\), \(\frac{x-y}{xy}=6\)

 

Soln:

\(\frac{x+y}{xy}=2\)

= \(\frac{1}{x}+\frac{1}{y}=2\) ……………………………….. (i)

\(\frac{x-y}{xy}=6\)

= \(\frac{1}{x}-\frac{1}{y}=6\) ………………………………… (ii)

Taking \(\frac{1}{x}\) =u

Taking \(\frac{1}{y}\) =v

= u+v=2 …………………………… (iii)

= u-v= 6 ……………………………. (iv)

By cross multiplication method,

\(\frac{u}{6-2}=\frac{-v}{6+2}=\frac{1}{-1-1}\)

\(\frac{u}{4}=\frac{-v}{8}=\frac{1}{-2}\)

Now,

\(\frac{u}{4}=\frac{1}{-2}\)

=u= -2

And,

\(\frac{-v}{-8}=\frac{1}{-2}\)

=v=4

\(\frac{1}{u}\) =x=\(\frac{-1}{2}\)

\(\frac{1}{v}\) =y=\(\frac{1}{4}\)

 

The solution of the given system of equation is \(\frac{-1}{2}\)  and \(\frac{1}{4}\) respectively.

 

Q.6: ax+by=a-b, bx-ay=a+b

 

Soln:

ax+by=a-b……………………………….(i)

bx-ay=a+b…………………………….. (ii)

Here a1= a , b1= b , c1= a-b

a2= b , b2= -a , c2= a+b

By cross multiplication method,

\(\frac{x}{-ab-b^{2}+ab-a^{2}}=\frac{-y}{-a^{2}-ab-b^{2}+ab}=\frac{1}{-a^{2}-b^{2}}\)

\(\frac{x}{-b^{2}-a^{2}}=\frac{-y}{-a^{2}-b^{2}}=\frac{1}{-a^{2}-b^{2}}\)

Now,

\(\frac{x}{-ab-b^{2}+ab-a^{2}}=\frac{1}{-a^{2}-b^{2}}\)

=x= 1

And,

\(\frac{-y}{-a^{2}-ab-b^{2}+ab}=\frac{1}{-a^{2}-b^{2}}\)

=y=-1

The solution of the given system of equation is 1 and -1 respectively.

 

Q.7: x+ay-b=0, ax-by-c=0

 

Soln:

x+ay-b=0 ……………………………….. (i)

ax-by-c=0………………………………. (ii)

Here a1= 1 , b1= a , c1= -b

a2= a , b2= -b , c2= -c

By cross multiplication method,

\(\frac{x}{-ac-b^{2}}=\frac{-y}{-c+ab}=\frac{1}{-a^{2}-b}\)

Now,

\(\frac{x}{-ac-b^{2}}=\frac{1}{-a^{2}-b}\)

=x= \(\frac{b^{2}+ac}{a^{2}+b}\)

And,

\(\frac{-y}{-c+ab}=\frac{1}{-a^{2}-b}\)

=y=\(\frac{-c+ab}{a^{2}+b}\)

The solution of the given system of equation is  \(\frac{b^{2}+ac}{a^{2}+b}\) and \(\frac{-c+ab}{a^{2}+b}\)  respectively.

 

Q8

ax+by=a2

bx+ay=b2

Soln:

ax+by=a2……………………………….(i)

bx+ay=b2…………………………….. (ii)

Here a1= a , b1= b , c1= a2

a2= b , b2= a , c2= b2

By cross multiplication method,

\(\frac{x}{-b^{2}+a^{2}}=\frac{-y}{-ab^{2}-a^{2}b}=\frac{1}{a^{2}-b^{2}}\)

Now,

\(\frac{x}{-b^{2}+a^{2}}=\frac{1}{a^{2}-b^{2}}\)

=x= \(\frac{a^{2}+ab+b^{2}}{a+b}\)

And,

\(\frac{-y}{-ab^{2}-a^{2}b}=\frac{1}{a^{2}-b^{2}}\)

=y=-\(\frac{-ab(a-b)}{(a-b)(a+b)}\)

The solution of the given system of equation is  \(\frac{a^{2}+ab+b^{2}}{a+b}\) and \(\frac{-ab(a-b)}{(a-b)(a+b)}\) respectively.

 

Q9

\(\frac{5}{x+y}-\frac{2}{x-y}=-1\)

\(\frac{15}{x+y}+\frac{7}{x-y}=-10\)

Soln:

Let \(\frac{1}{x+y}=\) =u

Let \(\frac{1}{x-y}=\) =v

The given system of equations are :

5u-2v=-1

15u+7v =10

Here a1= 5, b1= -2 , c1= 1

a2= 15 , b2= 7 , c2= -10

By cross multiplication method,

\(\frac{u}{20-7}=\frac{-v}{-50-15}=\frac{1}{35+30}\)

\(\frac{u}{13}=\frac{-v}{-65}=\frac{1}{65}\)

Now,

\(\frac{u}{13}=\frac{1}{-65}\)

=u= \(\frac{1}{5}\)

\(\frac{1}{u}=\) =x+y

=x+y=5 ………………………..(i)

And,

\(\frac{-v}{-65}=\frac{1}{-65}\)

=v=1

\(\frac{1}{v}=\) =x-y

=x-y =1 ………………………… (ii)

Adding equation (i) and (ii)

2x= 6

=x=3

Putting the value of x in equation (i)

3+y=5

=y=2

The solution of the given system of equation is 3 and 2 respectively.

 

Q10

\(\frac{2}{x}+\frac{3}{y}=13\)

\(\frac{5}{x}-\frac{4}{y}=-2\)

Soln:

Let \(\frac{1}{x}\) =u

Let \(\frac{1}{y}\) =v

The given system of equations becomes:

2u+3v=13 ……………………………… (i)

5u-4v=-2…………………………………. (ii)

By cross multiplication method,

\(\frac{u}{6-52}=\frac{-v}{4+65}=\frac{1}{-8-15}\)

\(\frac{u}{-46}=\frac{-v}{69}=\frac{1}{-23}\)

Now,

\(\frac{u}{-46}=\frac{1}{-23}\)

=u= 2

\(\frac{1}{u}=\) =\(\frac{1}{x}\)

=x=\(\frac{1}{2}\)

And,

\(\frac{-v}{69}=\frac{1}{-23}\)

=v=3

\(\frac{1}{v}\)= \(\frac{1}{y}\)

=y =\(\frac{1}{3}\)

The solutions of the given system of equations are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively.

 

Q11

\(\frac{57}{x+y}+\frac{6}{x-y}=5\)

\(\frac{38}{x+y}+\frac{21}{x-y}=9\)

Soln:

Let \(\frac{1}{x+y}=\) =u

Let \(\frac{1}{x-y}=\) =v

The given system of equations are :

57u+6v=5

38u+21v =9

Here a1= 57, b1= 6 , c1= -5

a2= 38 , b2=21 , c2= -9

By cross multiplication method,

\(\frac{u}{-54+105}=\frac{-v}{-513+190}=\frac{1}{1193-228}\)

\(\frac{u}{51}=\frac{-v}{-323}=\frac{1}{969}\)

Now,

\(\frac{u}{51}=\frac{1}{969}\)

=u= \(\frac{1}{19}\)

\(\frac{1}{u}\) =x+y

=x+y=19 ………………………..(i)

And,

\(\frac{-v}{-323}=\frac{1}{969}\)

=v=\(\frac{1}{3}\)

\(\frac{1}{v}=\) =x-y

=x-y =3 ………………………… (ii)

Adding equation (i) and (ii)

2x= 22

=x=11

Putting the value of x in equation (i)

11+y=19

=y=8

The solution of the given system of equation is 11 and 8 respectively.

 

Q12

\(\frac{x}{a}-\frac{y}{b}=2\)

ax-by=a2-b2

Soln:

a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\), Let c1=-2

a2=a ,  b2=-b,  c2=b2-a2

By cross multiplication method

= \(\frac{x}{\frac{b^{2}-a^{2}}{b}-2b}=\frac{-y}{\frac{b^{2}-a^{2}}{b}+2b}=\frac{1}{\frac{-b}{a}-\frac{a}{b}}\)

= \(\frac{x}{\frac{b^{2}-a^{2}-2b^{2}}{b}}=\frac{-y}{\frac{b^{2}-a^{2}+2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)

Now, \(\frac{x}{\frac{b^{2}-a^{2}-2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)

x= a

and, \(\frac{-y}{\frac{b^{2}-a^{2}+2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)

=y=b

Hence the solution of the given system of equation are a and b respectively.

 

Q13

\(\frac{x}{a}+\frac{y}{b}=a+b\)

\(\frac{x}{a^{2}}+\frac{y}{b^{2}}=2\)

Soln:

Here, a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\), Let c1=-(a+b)

a2=\(\frac{1}{a^{2}}\),  b2=\(\frac{1}{b^{2}}\),  c2=-2

By cross multiplication method

=\(\frac{x}{\frac{-2}{b}+\frac{a}{b^{2}}+\frac{1}{b}}=\frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)

= \(\frac{x}{\frac{a-b}{b^{2}}}=\frac{-y}{\frac{-a-b}{a^{2}}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)

Now, \(\frac{x}{\frac{a-b}{b^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)

=x=a2

\(\frac{-y}{\frac{-a-b}{a^{2}}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)

=y=b2

The solution of the given system of equation are  aand b2 respectively.

 

Q14

\(\frac{x}{a}=\frac{y}{b}\)

ax+by=a2+b2

Soln:

Here, a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\),  c1=0

Here, a1=a ,  b2=b, Let c1=-(a2+b2)

By cross multiplication method

\(\frac{x}{\frac{a^{2}+b^{2}}{b}}= \frac{y}{\frac{a^{2}+b^{2}}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)

Now, \(\frac{x}{\frac{a^{2}+b^{2}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)

=x=a

And \( \frac{y}{\frac{a^{2}+b^{2}}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)

=y=b

The solution of the given system of equations are  a and b respectively.

 

Q15

2ax+3by=a+2b

3ax+2by=2a+b

Soln:

The given system of equation is

2ax+3by=a+2b …………………………… (i)

3ax+2by=2a+b…………………………….. (ii)

Here a1= 2a, b1= 3b , c1= -(a+2b)

a2= 3a , b2=2b , c2= -(2a+b)

By cross multiplication method

\(\frac{x}{-4ab+b^{2}}=\frac{-y}{-a^{2}+4ab}=\frac{1}{-5ab}\)

Now,

\(\frac{x}{-4ab+b^{2}}=\frac{1}{-5ab}\)

=x= \(\frac{4a-b}{5a}\)

And, \(\frac{-y}{-a^{2}+4ab}=\frac{1}{-5ab}\)

=y=\(\frac{4b-a}{5b}\)

The solutions of the system of equations are \(\frac{4a-b}{5a}\) and \(\frac{4b-a}{5b}\).

 

Q16

5ax+6by=28

3ax+4by=18

Soln:

The systems of equations are:

5ax+6by=28 …………………………. (i)

3ax+4by=18……………………………. (ii)

Here a1= 5a, b1= 6b , c1= -(28)

a2= 3a , b2=4b , c2= -(18)

By cross multiplication method

\(\frac{x}{4b}=\frac{-y}{-6a}=\frac{1}{2ab}\)

Now,

\(\frac{x}{4b}=\frac{1}{2ab}\)

=x= \(\frac{2}{a}\)

And, \(\frac{-y}{-6a}=\frac{1}{2ab}\)

=y=\(\frac{3}{b}\)

The solution of the given system of equation is \(\frac{2}{a}\) and \(\frac{3}{b}\).

 

Q17

(a+2b)x+(2a-b)y=2

(a-2b)x+(2a+b)y=3

Soln.

The given system of equations are :

(a+2b)x+(2a-b)y=2 ………………………. (i)

(a-2b)x+(2a+b)y=3………………………….. (ii)

Here a1= a+2b, b1= 2a-b , c1= -(2)

a2= a-2b , b2=2a+b , c2= -(3)

By cross multiplication method:

\(\frac{x}{-2a+5b}=\frac{y}{a+10b}=\frac{1}{10ab}\)

Now, \(\frac{x}{-2a+5b}=\frac{1}{10ab}\)

=x=  \(\frac{5b-2a}{10ab}\)

And \(\frac{y}{a+10b}=\frac{1}{10ab}\)

=y=\(\frac{a+10b}{10ab}\)

The solution of the system of equations are =x=  \(\frac{5b-2a}{10ab}\)

And =y=\(\frac{a+10b}{10ab}\) respectively.

 

Q18

\(x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})\)

x+y=2a2

Soln:

The given systems of equations are:

\(x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})\)

x+y=2a2

From equation (i)

\(x(\frac{a^{2}+b^{2}-2ab+ab}{a-b})-y(\frac{a^{2}+b^{2}+2ab-ab}{a+b})\)

= \(x(\frac{a^{2}+b^{2}-ab}{a-b})-y(\frac{a^{2}+b^{2}+ab}{a+b})\) ……………….. (iii)

From equation (ii)

X+y-2a2=0

Here a1= \((\frac{a^{2}+b^{2}-ab}{a-b})\) , b1= –\((\frac{a^{2}+b^{2}+ab}{a-b})\) , c1=0

a2=1, b2=1 , c2=-2a2

By cross multiplication method:

\(\frac{x}{2a^{2}(\frac{a^{2}+b^{2}+ab}{a+b})}=\frac{-y}{(-2a^{2})(\frac{a^{2}+b^{2}-ab}{a-b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)

Now, \(\frac{x}{2a^{2}(\frac{a^{2}+b^{2}+ab}{a+b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)

=x=\(\frac{a^{3}-b^{3}}{a}\)

And  \(\frac{-y}{(-2a^{2})(\frac{a^{2}+b^{2}-ab}{a-b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)

=y=\(\frac{a^{3}+b^{3}}{a}\)

The solutions of the given system of equations are \(\frac{a^{3}-b^{3}}{a}\) and \(\frac{a^{3}+b^{3}}{a}\) respectively.

 

Q19

bx+cy=a+b

\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})=\frac{2a}{a+b}\)

Soln:

The system of equation is given by :

bx+cy=a+b ……………………………………. (i)

\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})=\frac{2a}{a+b}\)…..(ii)

From equation (i)

bx+cy-(a+b) =0

From equation (ii)

\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})-\frac{2a}{a+b}=0\)

= \(x(\frac{2ab}{(a-b)(a+b)})+y(\frac{2ac}{(b-a)(b+a)})-\frac{2a}{a+b}=0\)

= \(\frac{1}{a+b}(\frac{2abx}{a-b}-\frac{2acy}{a-b}-2a)=0\)

= \(\frac{2abx}{a-b}-\frac{2acy}{a-b}-2a=0\)

= 2abx-2acy-2a(a-b) =0 …………………………. (iv)

By cross multiplication

= \(\frac{x}{-4a^{2}c}=\frac{-y}{4ab^{2}}=\frac{-1}{4abc}\)

Now, \(\frac{x}{-4a^{2}c}=\frac{-1}{4abc}\)

=x=\(\frac{a}{b}\)

And,

= \(\frac{-y}{4ab^{2}}=\frac{-1}{4abc}\)

=y=\(\frac{b}{c}\)

The solution of the system of equations are \(\frac{a}{b}\) and \(\frac{b}{c}\)

 

Q20

(a-b)x+(a+b)y=2a2-2b2

(a+b)(x+y) =4ab

Soln.

The given system of equations are :

(a-b)x+(a+b)y=2a2-2b2 ………………………….. (i)

(a+b)(x+y) =4ab …………………………. (ii)

From equation (i)

(a-b)x+(a+b)y-2a2-2b2 =0

= (a-b)x+(a+b)y-2(a2-b2) =0

From equation (ii)

(a-b)x+(a-b)y-4ab=0

Here, a1= a-b , b1 = a+b , c1=-2(a2+b2)

Here, a2= a+b , b2 = a+b , c2=-4ab

By cross multiplication method

\(\frac{x}{2(a+b)(a^{2}-b^{2}+2ab)}=\frac{-y}{2(a-b)(a^{2}+b^{2})}=\frac{1}{-2b(a+b)}\)

Now,

\(\frac{x}{2(a+b)(a^{2}-b^{2}+2ab)}=\frac{1}{-2b(a+b)}\)

=x= \(\frac{2ab-a^{2}+b^{2}}{b}\)

And, \(\frac{-y}{2(a-b)(a^{2}+b^{2})}=\frac{1}{-2b(a+b)}\)

=y=\(\frac{(a-b)(a^{2}+b^{2})}{b(a+b)}\)

The solution of the system of equations are \(\frac{2ab-a^{2}+b^{2}}{b}\) and \(\frac{(a-b)(a^{2}+b^{2})}{b(a+b)}\) respectively.

 

Q21

a2x+b2y=c2

b2x+a2y=d2

Soln:

The given system of equations are :

a2x+b2y=c2 ………………………………….. (i)

b2x+a2y=d2……………………………………… (ii)

 

Here, a1= a2 , b1 = b2 , c1=-c2

Here, a2= b2 , b2 = a2 , c2=-d2

By cross multiplication method

= \(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{-y}{-a^{2}d^{2}+b^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)

Now,

\(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)

=x= \(\frac{a^{2}c^{2}-b^{2}d^{2}}{a^{4}-b^{4}}\)

And, = \(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{-y}{-a^{2}d^{2}+b^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)

=y= \(\frac{a^{2}d^{2}-b^{2}c^{2}}{a^{4}-b^{4}}\)

The solution of the given system of equations are \(\frac{a^{2}c^{2}-b^{2}d^{2}}{a^{4}-b^{4}}\) and \(\frac{a^{2}d^{2}-b^{2}c^{2}}{a^{4}-b^{4}}\) respectively.

 

Q23

2(ax-by+a+4b=0

2(bx+ay)+b-4a=0

Soln:

The given system of equation may be written as :

2(ax-by+a+4b=0 …………………….. (i)

2(bx+ay)+b-4a=0……………………. (ii)

Here, a1= 2a , b1 = -2b , c1=a+4b

Here, a2= 2b , b2 = 2a , c2=b-4a

By cross multiplication method

=\(\frac{x}{-2b^{2}+8ab-2a^{2}-8ab}=\frac{-y}{2ab-8a^{2}-2ab-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)

= \(\frac{x}{-2b^{2}-2a^{2}}=\frac{-y}{-8a^{2}-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)

Now, \(\frac{x}{-2b^{2}-2a^{2}}=\frac{1}{4a^{2}+4b^{2}}\)

=x= \(\frac{-1}{2}\)

And, \(\frac{-y}{-8a^{2}-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)

=y=2

The solution of the given pair of equations are \(\frac{-1}{2}\) and 2 respectively.

 

Q24

6(ax+by)=3a+2b

6(bx-ay) = 3b-2a

Soln:

The systems of equations are

6(ax+by)=3a+2b ………………………… (i)

6(bx-ay) = 3b-2a …………………………. (ii)

From equation (i)

6ax+6by-(3a+2b)=0  ……………………… (iii)

From equation (ii)

6bx-6ay-(3b-2a) =0 …………………………… (iv)

Here, a1= 6a , b1 = 6b , c1=-(3a+2b)

Here, a2= 6b , b2 = -6a , c2=-(3b-2a)

By cross multiplication method

\(\frac{x}{-18(a^{2}+b^{2})}=\frac{-y}{12(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)

Now, \(\frac{x}{-18(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)

=x= \(\frac{1}{2}\)

And , \(\frac{-y}{12(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)

=y=\(\frac{1}{3}\)

The solution of the given pair of equations are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively.

 

Q25

\(\frac{a^{2}}{x}-\frac{b^{2}}{y}=0\)

\(\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b\)

Soln:

The given systems of equations are

\(\frac{a^{2}}{x}-\frac{b^{2}}{y}=0\)

\(\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b\)

Taking \(\frac{1}{x}\) =u

Taking \(\frac{1}{y}\)=v

The pair of equations becomes:

a2u-b2v=0

a2bu+b2av-(a+b) =0

Here, a1= a2 , b1 = -b2 , c1=0

Here, a2= a2b , b2 = b2a , c2=-(a+b)

By cross multiplication method

= \(\frac{u}{b^{2}(a+b)}=\frac{v}{a^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)

Now, \(\frac{u}{b^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)

=x= \(\frac{1}{a^{2}}\)

And, \(\frac{v}{a^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)

=y= \(\frac{1}{b^{2}}\)

The solution of the given pair of equations are \(\frac{1}{a^{2}}\) and  \(\frac{1}{b^{2}}\) respectively.

 

Q26

mx-my=m2+n2

x+y=2m

Soln:

mx-my=m2+n2………………………………… (i)

x+y=2m…………………………………………….. (ii)

Here, a1= m , b1 = -n , c1=-(m2+n2)

Here, a2= 1 , b2 = 1 , c2=-(2m)

By cross multiplication method

\(\frac{x}{(m+n)^{2}}=\frac{-y}{-m^{2}+n^{2}}=\frac{1}{m+n}\)

Now, \(\frac{x}{(m+n)^{2}}=\frac{1}{m+n}\)

=x= m+n

And, \(\frac{-y}{-m^{2}+n^{2}}=\frac{1}{m+n}\)

=y=m-n

The solutions of the given pair of equations are m+n and m-n respectively.

 

Q27

\(\frac{ax}{b}-\frac{by}{a}=a+b\)

ax-by=2ab

Soln:

The given pair of equations are:

\(\frac{ax}{b}-\frac{by}{a}=a+b\) ……………………….. (i)

ax-by=2ab …………………………….. (ii)

Here, a1= \(\frac{a}{b}\) , b1 = – \(\frac{b}{a}\) , c1=-(a+b)

Here, a2= a , b2 = – b, c2=-(2ab)

By cross multiplication method

= \(\frac{x}{b(b-a)}=\frac{-y}{a(-a+b)}=\frac{1}{b-a}\)

Now, \(\frac{x}{b(b-a)}=\frac{1}{b-a}\)

=x=b

And , \(\frac{-y}{a(-a+b)}=\frac{1}{b-a}\)

=y=-a

The solution of the given pair of equations are b and –a respectively.

 

Q28

\(\frac{b}{a}x+\frac{a}{b}y-(a^{2}+b^{2})=0\)

X+y-2ab=0

Soln:

\(\frac{b}{a}x+\frac{a}{b}y-(a^{2}+b^{2})=0\)…………………….. (i)

X+y-2ab=0…………………………………………. (ii)

Here, a1= \(\frac{b}{a}\) , b1 =  \(\frac{a}{b}\) , c1=-(a2+b2)

Here, a2= 1 , b2 = -1, c2=-(2ab)

By cross multiplication method

 

= \(\frac{x}{b^{2}-a^{2}}=\frac{-y}{-b^{2}+a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)

Now, \(\frac{x}{b^{2}-a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)

=x=ab

And, \(\frac{-y}{-b^{2}+a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)

=y=ab

The solutions of the given pair of equations are ab and ab respectively.


Practise This Question

Sanky got 59.25 marks in the final exams. Passing marks for the exam is 65 marks. How much more marks should Sanky have scored to pass the exam?