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

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

Exercise 3.4

Solve each of the following systems of equation by the method of cross-multiplication:

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

Solution:

Given equations are:

x+2y+1 =0 …………………………………….(i)

2x-3y-12=0………………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=3 and y=-2 respectively.

 

 

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

Solution:

Given equations are:

3x+2y+25 =0 …………………………………….(i)

2x+y+10=0………………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=5 and y=-20 respectively.

 

 

3. 2x + y = 35, 3x + 4y = 65

Solution:

Given equations are

2x+y= 35 ……………………………….(i)

3x+4y=65…………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=15 and y=5 respectively.

 

 

4. 2x – y – 6 = 0, x – y – 2 = 0

Solution:

Given equations are:

2x-y= 6 ……………………………….(i)

x-y=2…………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=4 and y=2 respectively.

 

 

5. \(\frac{x+y}{xy}=2\), \(\frac{x-y}{xy}=6\)

Solution:

Given equations are:

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

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

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

a2= 1 , b2= -1 , c2= -6

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

 

 

6. ax+by=a-b, bx-ay=a+b

Solution:

Given equations are:

ax+by=a-b……………………………….(i)

bx-ay=a+b…………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=1 and y=-1 respectively.

 

 

7. x+ay-b=0, ax-by-c=0

Solution:

Given equations are:

x+ay-b=0 ……………………………….. (i)

ax-by-c=0………………………………. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

 

 

8.ax+by=a2 and bx+ay=b2

Solution:

Given equations are:

ax+by=a2……………………………….(i)

bx+ay=b2…………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

 

 

9.

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

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

Solution:

Given equations are:

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

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

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

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

The given system of equations are :

5u-2v=-1

15u+7v =10

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

Substituting 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.

 

 

10.

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

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

Solution:

Given equations are:

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

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

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

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

The given system of equations becomes:

2u+3v=13 ……………………………… (i)

5u-4v=-2…………………………………. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

a1= 2, b1= 3 , c1= -13

a2= 5 , b2= -4 , c2= 2

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 x=\(\frac{1}{2}\) and y=\(\frac{1}{3}\) respectively.

 

 

11.

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

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

Solution:

Given equations are:

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

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

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

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

The given system of equations are :

57u+6v=5

38u+21v =9

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

Substituting the value of x in equation (i)

11+y=19

=y=8

∴, the solution of the given system of equation is x=11 and y=8 respectively.

 

 

12.

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

ax-by=a2-b2

Solution:

Given equations are:

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

ax-by=a2-b2

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

∴, the solution of the given system of equation are x=a and y=b respectively.

 

 

13.

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

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

Solution:

Given equations are:

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

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

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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  x=aand y=b2 respectively.

 

 

14.

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

ax+by=a2+b2

Solution:

Given equations are:

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

ax+by=a2+b2

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

Hence, 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  x=a and y=b respectively.

 

 

15.

2ax+3by=a+2b

3ax+2by=2a+b

Solution:

Given equations are:

2ax+3by=a+2b …………………………… (i)

3ax+2by=2a+b…………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

 

 

16. 

5ax+6by=28

3ax+4by=18

Solution:

Given equations are:

5ax+6by=28 …………………………. (i)

3ax+4by=18……………………………. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=\(\frac{2}{a}\) and y=\(\frac{3}{b}\).

 

 

17.

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

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

Solution:

Given equations are :

(a+2b)x+(2a-b)y=2 ………………………. (i)

(a-2b)x+(2a+b)y=3………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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.

 

 

18.

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

x+y=2a2

Solution:

Given 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

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x= \(\frac{a^{3}-b^{3}}{a}\) and y=\(\frac{a^{3}+b^{3}}{a}\) respectively.

 

 

19.

bx+cy=a+b

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

Solution:

Given equations are:

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)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

a1= b, b1= c , c1=-(a+b)

a2=2ab, b2=-2ac , c2=-2a(a-b)

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

 

 

20.

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

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

Soln.

Given 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

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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

 

 

21.

a2x+b2y=c2

b2x+a2y=d2

Solution:

The given system of equations are :

a2x+b2y=c2 ………………………………….. (i)

b2x+a2y=d2……………………………………… (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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

 

 

22. 

\(ax+by=\frac{a+b}{2}\)

3x+5y=4

Solution:

Given equations are:

\(ax+by=\frac{a+b}{2}\)

3x+5y=4

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

a2= 3 , b2 = 5 , c2= -4

By cross multiplication method

RD Sharma class 10 Maths chapter 3 exercise 3.4-1

 

 

23.

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

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

Solution:

Given equations are:

2(ax-by)+a+4b=0 …………………….. (i)

2(bx+ay)+b-4a=0……………………. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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 x=\(\frac{-1}{2}\) and y=2 respectively.

 

 

24.

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

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

Solution:

Given 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)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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 x=\(\frac{1}{2}\) and y=\(\frac{1}{3}\) respectively.

 

 

25.

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

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

Solution:

Given 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

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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 x=\(\frac{1}{a^{2}}\) and y=\(\frac{1}{b^{2}}\) respectively.

 

 

26.

mx-my=m2+n2

x+y=2m

Solution:

Given equations are:

mx-my=m2+n2………………………………… (i)

x+y=2m…………………………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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

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 x=m+n and y=m-n respectively.

 

 

27.

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

ax-by=2ab

Solution:

Given equations are:

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

ax-by=2ab …………………………….. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 equations are x=b and y=–a respectively.

 

 

28.

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

x+y-2ab=0

Soln:

Given equations are:

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

x+y-2ab=0…………………………………………. (ii)

The equations are of the form ax+by+c=0

We know that,

a1x+b1y+c1=0

a2x+b2y+c2=0

⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)

Here, According to the question,

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 x=ab and y=ab respectively.

Leave a Comment

Your email address will not be published. Required fields are marked *