RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.3

RD Sharma Solutions Class 10 Chapter 3 Exercise 3.3

RD Sharma Class 10 Solutions Chapter 3 Ex 3.3 PDF Free Download

Exercise 3.3

 

Solve the following system of equations:

 

Question 1: 11x + 15y + 23 = 0 and 7x – 2y – 20 = 0

 

Soln:

The given system of equation is

11x+15y+23=0 …………………………. (i)

7x-2y-20=0 ………………………………..(ii)

From (ii)

2y=7x-20

y= \(\frac{7x-20}{2}\) ……………………………… (iii)

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

= \(11x+15(\frac{7x-20}{2})+23=0\)

= \(11x+(\frac{105x-300}{2})+23=0\)

= \((\frac{22+105x-300+46}{2})=0\)

= 127x = 254 = x=2

Putting the value of x in the equation (iii)

= y= \(\frac{7(2)-20}{2}\)

y= -3

The value of x and y are 2 and -3 respectively.

 

Question 2: 3x – 7y + 10 = 0, y – 2x – 3 = 0

 

Soln:

The given system of equation is

3x-7y+10=0 …………………………. (i)

y-2x-3=0 ………………………………..(ii)

From (ii)

y-2x-3=0

y= 2x+3 ……………………………… (iii)

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

= 3x-7(2x+3)+10 =0

= 3x+14x-21+10=0

= -11x=11

= x=-1

Putting the value of x in the equation (iii)

= y= 2(-1)+3

y= 1

The value of x and y are -1 and 1 respectively.

 

Question 3: 0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8

 

Soln:

The given system of equation is

0.4x+0.3y=1.7

0.7x-0.2y=0.8

Multiplying both sides by 10

4x+3y=17 ……………………….. (i)

7x-2y=8 …………………………… (ii)

From (ii)

7x-2y=8

x=\(\frac{8+2y}{7}=0\) ……………………………… (iii)

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

= 4(\(\frac{8+2y}{7}=0\))+3y=17

= 32+29y=119

= 29y=87

= y=3

Putting the value of y in the equation (iii)

= x= \(\frac{8+2(3)}{7}=0\)

=x=\(\frac{14}{7}=0\)

= x= 2

The value of x and y are 2 and 3 respectively.

 

Question 4

\(\frac{x}{2}+y=0.8\)

\(\frac{7}{x+\frac{y}{2}}=10\)

 

Soln:

The given system of equation is

\(\frac{x}{2}+y=0.8\)

\(\frac{7}{x+\frac{y}{2}}=10\)

Therefore x+2y=1.6

\(\frac{14}{2x+y}=10\)

x+2y=1.6

7=10x+5y

Multiplying both sides by 10

10x+20y=16 ……………………….. (i)

10x+5y=7 …………………………… (ii)

Subtracting two equations we get,

15y=9

y=\(\frac{3}{5}\)

x=\(1.6-2(\frac{3}{5})\)

= \(1.6-(\frac{6}{5})\)

= \(\frac{2}{5}\)

The value of x and y are \(\frac{2}{5}\)  and\(\frac{3}{5}\)  respectively.

 

Question 5

7(y+3)-2(x+3) = 14

4(y-2)+3(x-3) = 2

 

Soln:

The given system of equation is

7(y+3)-2(x+3) = 14…………………………. (i)

4(y-2)+3(x-3) = 2………………………………..(ii)

From (i)

7y+21-2x-4=14

7y=14+4-21+2x

=y=\(\frac{2x-3}{5}\)

From (ii)

= 4y-8+3x-9=2

= 4y+3x-17-2=0

= 4y+3x-19=0 ……………..(iii)

Substituting the value of y in equation (iii)

=4(\(\frac{2x-3}{5}\))+3x-19=0

= 8x-12+21x-133=0

= 29x=145

= x=5

Putting the value of x in the above equation

= y = 1

The value of x and y are 5 and 1 respectively.

 

Question 6

\(\frac{x}{7}+\frac{y}{3}=5\)

\(\frac{x}{2}-\frac{y}{9}=6\)

 

Soln:

The given system of equation is

\(\frac{x}{7}+\frac{y}{3}=5\)…………………………. (i)

\(\frac{x}{2}-\frac{y}{9}=6\)………………………………..(ii)

From (i)

\(\frac{x}{7}+\frac{y}{3}=5\)

=x=\(\frac{105-7y}{3}\)

From (ii)

\(\frac{x}{2}-\frac{y}{9}=6\)

=9x-2y=108 ………………………(iii)

Substituting the value of x in equation (iii) we get,

= 9(\(\frac{105-7y}{3}\))-2y=108

= 945-63y-6y=324

=945-324=69y

=69y=621

=y=9

Putting the value of y in the above equation

= x=\(\frac{105-7(9)}{3}\)

y= 14

The value of x and y are 5 and 14 respectively.

 

Question 7

\(\frac{x}{3}+\frac{y}{4}=11\)

\(\frac{5x}{6}-\frac{y}{3}=-7\)

 

Soln:

The given system of equation is

\(\frac{x}{3}+\frac{y}{4}=11\)…………………………. (i)

\(\frac{5x}{6}-\frac{y}{3}=-7\)………………………………..(ii)

From (i)

\(\frac{4x+3y}{12}=11\)

=4x+3y=132……………………(iii)

From (ii)

\(\frac{5x+2y}{6}=-7\)

=5x-2y=-42 ………………………(iv)

Let us eliminate y from the given equations. The co efficient of y in the equation (iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.

Multiplying equation (iii)*2 and (iv)*3 we get

= 8x+6y=264 ……………………..(v)

= 15x-6y=-126 ………………………(vi)

Adding equation (v) and (vi)

8x+15x=264-126

=23x=138

x=6

Putting the value of x in the equation (iii)

= 24+3y=132

=3y=108

y= 36

The value of x and y are 36 and 6 respectively.

 

Question 8

\(\frac{4}{x}+3y=8\)

\(\frac{6}{x}-4y=-5\)

 

Soln:

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

The new equation becomes

4u+3y=8……………………(i)

6u-4y=-5…………………….(ii)

From (i)

4u=8-3y

=u= \(\frac{8-3y}{4}=u\)

From (ii)

=6(\(\frac{8-3y}{4}=u\))-4y=-5

= \(\frac{3(8-3y)}{2}-4y=-5\)

= \(\frac{24-9y-8y}{2}=-5\)

= 24-17y=-10

= -17y=-34

= y=2

Putting y=2 in u = \(\frac{8-3y}{4}=u\) we get ,

=u=\(\frac{8-3(2)}{4}\)

=u=\(\frac{8-6}{4}\)

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

=x=u=2

So the Solution of the given system of equation is x=2 and y =2

 

Question 9

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

\(2y+\frac{x}{3}=5\)

 

Soln:

The given system of equation is:

\(x+\frac{y}{2}=4\) …………………….(i)

\(2y+\frac{x}{3}=5\)…………………….(ii)

From (i) we get,

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

= 2x+y=8

=y=8-2x

From (ii) we get,

x+6y=15 ………………(iii)

Substituting y=8-2x in (iii) , we get

= x+6(8-2x)=15

= x+48-12x=15

= -11x=15-48

= -11x=-33

=x=3

Putting x=3 in y 8-2x, we get

y=8-(2*3)

=y=8-6=2

The Solution of the given system of equation are x=3 and y=2 respectively.

 

Question 10

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

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

Soln:

The given system of equation is

x+2y= \(\frac{3}{2}\) ………………….(i)

2x+y= \(\frac{3}{2}\)……………………(ii)

Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.

Multiplying equation (i)*1 and (ii)*2

x+2y= \(\frac{3}{2}\) ……………………….(iii)

4x+2y=3 …………………………………………………….(iv)

Subtracting equation (iii) from (iv)

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

= 3x= x+2y= \(\frac{6-3}{2}\)

= 3x= \(\frac{3}{2}\)

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

Putting x=\(\frac{1}{2}\) in equation (iv)

4(\(\frac{1}{2}\))+2y=3

= 2+2y=3

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

The Solution of the system of equation is x=\(\frac{1}{2}\) and y=\(\frac{1}{2}\)

 

Question 11

\(\sqrt{2}x+\sqrt{3}y=0\)

\(\sqrt{3}x-\sqrt{8}y=0\)

 

Soln:

\(\sqrt{2}x+\sqrt{3}y=0\)………………………..(i)

\(\sqrt{3}x-\sqrt{8}y=0\)………………………..(ii)

From equation (i)

=x=\(\frac{-\sqrt{3}y}{\sqrt{2}}\) ……………..(iii)

Substituting this value in equation (ii) we obtain

\(\sqrt{3}(\frac{-\sqrt{3}y}{\sqrt{2}})-\sqrt{8}y=0\)

\(\frac{-3y}{\sqrt{2}}-2\sqrt{2}y=0\)

\(y(\frac{-3}{\sqrt{2}}-2\sqrt{2})=0\)

=y=0

Substituting the value of y in equation (iii) we obtain

=x=0

The value of x and y are 0 and 0 respectively.

 

Question 12

\(3x-\frac{y+7}{11}+2=10\)

\(2y-\frac{x+11}{7}=10\)

Soln:

The given system of equation is:

\(3x-\frac{y+7}{11}+2=10\) ………………..(i)

\(2y-\frac{x+11}{7}=10\)……………………..(ii)

From equation (i)

\(\frac{33x-y-7+22}{11}=10\)

=33x -y+15=110

=33x+15-110=y

= y= 33x-95

From equation (ii)

\(\frac{14+x+11}{7}=109\)

= 14y+x+11=70

= 14y+x=70-11

= 14y+x=59 ……………………..(iii)

Substituting y = 33x-95 in (iii) we get,

14(33x-95)+x=59

= 462x-1330+x=59

= 463x=1389

= x=3

Putting x=3 in y=33x-95 we get,

=y=33(3)-95

= 99-95 = 4

The Solution of the given system of equation is 3 and 4 respectively.

 

Question 13

\(2x-\frac{3}{y}=9\)

\(3x+\frac{7}{y}=2\)

Soln:

\(2x-\frac{3}{y}=9\)……………………………. (i)

\(3x+\frac{7}{y}=2\)…………………………… (ii)

Taking \(\frac{1}{y}=u\) the given equation becomes,

2x-3u=9 ………………………..(iii)

3x+7u=2………………………..(iv)

From (iii)

2x=9+3u

=x= \(\frac{9+3u}{2}\)

Substituting the value x= \(\frac{9+3u}{2}\) in equation (iv) we get,

3(\(\frac{9+3u}{2}\))+7u=2

= \(\frac{27+9u+14u}{2}=2\)

=27+23u=4

= u= -1

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

Putting u=-1 in =x= \(\frac{9+3u}{2}\) we get,

=x= \(\frac{9+3(-1)}{2}\)

= x=3

The Solution of the given system of equation is 3 and -1 respectively.

 

Question 14

0.5x+0.7y=0.74

0.3x+0.5y=0.5

Soln:

The given system of equation is

0.5x+0.7y=0.74………………………(i)

0.3x-0.5y=0.5 …………………………..(ii)

Multiplying both sides by 100

50x+70y=74 ……………………….. (iii)

30x+50y=50 …………………………… (iv)

From (iii)

50x=74-70y

x=\(\frac{74-70y}{50}=0\) ……………………………… (iii)

Substituting the value of y in equation (iv) we get,

= 30(\(\frac{74-70y}{50}=0\))+50y=50

= 222-210y+250y=250

= 40y=28

= y=0.7

Putting the value of y in the equation (iii)

= x= \(\frac{74-70(0.7)}{50}=0\)

=x=\(\frac{25}{50}=0\)

= x= 0.5

The value of x and y are 0.5 and 0.7 respectively.

 

Question 15

\(\frac{1}{7x}+\frac{1}{6y}=3\)

\(\frac{1}{2x}-\frac{1}{3y}=5\)

Soln:

\(\frac{1}{7x}+\frac{1}{6y}=3\) ………………………….. (i)

\(\frac{1}{2x}-\frac{1}{3y}=5\)……………………………. (ii)

Multiplying (ii) by \(\frac{1}{2}\) we get,

\(\frac{1}{4x}-\frac{1}{6y}=\frac{5}{2}\)……………………………. (iii)

Solving equation (i) and (iii)

\(\frac{1}{7x}+\frac{1}{6y}=3\) ………………………….. (i)

\(\frac{1}{4x}-\frac{1}{6y}=\frac{5}{2}\) ……………………………. (iii)

Adding we get,

\(\frac{1}{7x}+\frac{1}{6y}=3+ \frac{5}{2}\)

=x=\(\frac{1}{14}\)

When, x=\(\frac{1}{14}\) we get,

Using equation (i)

\(\frac{1}{7(\frac{1}{14})}+\frac{1}{6y}=3\)

= \(2+\frac{1}{6y}=3\)

= \(\frac{1}{6y}\)=1

= y=\(\frac{1}{6}\)

The Solution of the given system of equation is x=\(\frac{1}{14}\) and y=\(\frac{1}{6}\) respectively.

 

Question 16

\(\frac{1}{2x}+\frac{1}{3y}=2\)

 

\(\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6} \)

Soln:

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

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

\(\frac{u}{2}+\frac{v}{3}=2\)

\(\frac{3u+2v}{6}=2\)

3u+2v=12 ……………………..(i)

And, \(\frac{u}{3}+\frac{v}{2}=\ frac{13}{6} \)

=v=3

\(\frac{1}{u}\) =x =  \(\frac{1}{2}\)

\(\frac{1}{v}\) =y =  \(\frac{1}{3}\)

 

Question 17

\(\frac{15}{u}+\frac{2}{v}=17\)

\(\frac{1}{u}+\frac{1}{v}=\frac{36}{5} \)

Soln:

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

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

15x+2y=17 …………………………..(i)

x+y=\(\frac{36}{5} \)……………………….(ii)

From equation (i) we get ,

2y=17-15x

=y= \(\frac{17-15x}{2}\)

Substituting y=\(\frac{17-15x}{2}\) in equation (ii) we get,

= x+\(\frac{17-15x}{2}\)= \(\frac{36}{5}\)

= \(\frac{-13x+17x}{2}\)= \(\frac{36}{5}\)

= 5(-13x+17)=72

= -65x=-13

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

Putting x=\(\frac{1}{5}\)in equation (ii) , we get

\(\frac{1}{5}\) +y=\(\frac{36}{5}\)

= y=7

=v=\(\frac{1}{y}\)= \(\frac{1}{7}\)

The Solution of the given system of equation is 5 and \(\frac{1}{7}\) respectively.

 

Question 18

\(\frac{3}{x}-\frac{1}{y}=-9\)

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

Soln:

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

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

3u-v=-9…………………..(i)

2u+3v=5 ……………………….(ii)

Multiplying equation (i) *3 and (ii) *1 we get,

9u-3v=-27 ………………………….. (iii)

2u+3v=5 ……………………………… (iv)

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

9u+2u-3v+3v=-27+5

= u=-2

Putting u=-2 in equation (iv) we get,

2(-2)+3v=5

= 3v=9

= v=3

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

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

 

Question 19

\(\frac{2}{x}-\frac{3}{y}=\frac{9}{xy}\)

\(\frac{2}{x}+\frac{1}{y}=\frac{9}{xy} \)

Soln:

\(\frac{2}{x}-\frac{3}{y}=\frac{9}{xy}\) ……………………… (i)

\(\frac{2}{x}+\frac{1}{y}=\frac{9}{xy} \)…………………….. (ii)

Multiplying equation (i) adding equation (ii) we get,

2y+3x=9………………………..(iii)

4y+9x=21 ……………………….(iv)

From (iii) we get ,

3x= 9-2y

= x= \(\frac{9-2y}{3}\)

Substituting x=\(\frac{9-2y}{3}\) in equation (iv) we get

4x+9(\(\frac{9-2y}{3}\))=21

= 4y+3(9-2y) =21

= -2y=21-27

= y=3

Putting y=3 in x= \(\frac{9-2y}{3}\) we get,

=x= \(\frac{9-2(3)}{3}\)

=x=1

Hence the Solutions of the system of equation are 1 and 3 respectively.

 

Question 20

\(\frac{1}{5x}+\frac{1}{6y}=12\)

\(\frac{1}{5x}+\frac{1}{6y}=8\)

Soln:

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

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

\(\frac{u}{5}+\frac{v}{6}=12\)

=\(\frac{6u+5v}{30}=12\)

= 6u+5v=360 …………….(i)

\(\frac{u}{3}+\frac{3v}{7}\)=8

=\(\frac{7u+9v}{21}\)=8

= 7u-9v=168 …………….(ii)

Let us eliminate v from the equation (i) and (ii) . multiplying equation (i) by 9 and (ii) by 5

54u+35u=3240+840

89u=4080

=u=\(\frac{4080}{89}\)

Putting u =\(\frac{4080}{89}\) in equation (i) we get,

6(\(\frac{4080}{89}\))+5v=360

= \(\frac{24480}{89}\)+5v=360

=5v=\(\frac{32040-24480}{89}\)

= v= \(\frac{7560}{89}\)

= v= \(\frac{7560}{5*89}\)

=v= \(\frac{1512}{89}\)

\(\frac{1}{u}\) =x=\(\frac{89}{4080}\)

\(\frac{1}{v}\) =y=\(\frac{89}{1512}\)

.

Question 27

\(\frac{6}{x+y}=\frac{7}{x-y}+3\)

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

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

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

Then, the given system of equation becomes,

6u=7v+3

6u-7v=3……………………….. (i)

And \(\frac{u}{2}=\frac{v}{3}\)

3u=2v

3u-2v=0 ……………………… (ii)

Multiplying equation (ii) by 2 and (i) 1

6u-7v=3

6u-4v=0

Subtracting v=-1 in equation (ii) ,we get

3u-2(-1)=0

3u+2=0

3u=-2

=u=  \(\frac{-2}{3}\)

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

x+y= \(\frac{-3}{2}\) …………………….(v)

and v=-1

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

x-y=-1……………………(vi)

Adding equation (v) and equation (vi) we get,

2x= \(\frac{-3}{2}\)-1

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

Putting x=\(\frac{-2}{3}\) in equation (vi)

=\(\frac{-5}{4}\)-y=-1

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

 

Question 28

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

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

Soln:

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

5xy= 6(x+y)

=5xy= 6x+6y ……………….(i)

And

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

xy=6(y-x)

=xy= 6y-6x ……………………(ii)

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

6xy= 6y+6y

6xy=12y

x= 2

Putting x=2 in equation (i) we get,

10y=12+6y

10-6y=12

4y=12

y=3

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

 

Question 29

\(\frac{22}{x+y}+\frac{15}{x-y}=5\)

\(\frac{55}{x+y}+\frac{45}{x-y}=14\)

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

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

Soln:

Then the given system of equation becomes:

22u+45v=5 ……………………….. (i)

55u+45v=14 ………………………(ii)

Multiplying equation (i)by 3 and (ii) by 1

66u+45v=15 ……………………… (iii)

55u+45v=14 ……………………… (iv)

Subtracting equation (iv) from equation (iii) , we get

66u-55u=15-14

= 11u=1

=u= \(\frac{1}{11}\)

Putting =u= \(\frac{1}{11}\) in equation (i) we get,

=2+15v=5

=15v=3

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

Now,

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

=x+y=11 ……………..(v)

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

= x-y=5 …………………..(vi)

Adding equation (v) and (vi) we get,

2x=16

=x=8

Putting the value of x in equation (v)

8+y=11

= y=3

The Solutions of the given system of equation are 8 and 3 respectively.

 

Question 30

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

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

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

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

Soln:

Then the given system of equation becomes:

5u-2v=-1 ……………………….. (i)

15u+7v=10 ………………………(ii)

Multiplying equation (i) by 7 and (ii) by 2

35u-14v=-7 ……………………… (iii)

30u+14v=20 ……………………… (iv)

Subtracting equation (iv) from equation (iii) , we get

-2v=-1-1

= -2v=-2

=v=1

Now,

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

=x+y=5 ……………..(v)

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

= x-y=1 …………………..(vi)

Adding equation (v) and (vi) we get,

2x=6

=x=3

Putting the value of x in equation (v)

3+y=5

= y=2

The Solutions of the given system of equation are 3 and 2 respectively.

 

Question 31

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

\(\frac{9}{x+y}-\frac{4}{x-y}=1\)

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

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

Soln:

Then the given system of equation becomes:

3u+2v=2 ……………………….. (i)

9u+4v=1 ………………………(ii)

Multiplying equation (i) by 3 and (ii) by 1

6u+4v=4 ……………………… (iii)

9u-4v=1 ……………………… (iv)

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

45u=5

=u=3

Subtracting equation (iv) from equation (iii) , we get

2v=2-1

= 2v=1

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

Now,

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

=x+y=3 ……………..(v)

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

= x-y=2 …………………..(vi)

Adding equation (v) and (vi) we get,

2x=5

=x=\(\frac{5}{2}\)

Putting the value of x in equation (v)

\(\frac{5}{2}\)+y=11

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

The Solutions of the given system of equation are \(\frac{5}{2}\)  and \(\frac{1}{2}\)  respectively.

 

 

Question 32

\(\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=\frac{-3}{2}\)

\(\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{6}\)

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

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

Soln:

Then the given system of equation becomes:

\(\frac{u}{2}+\frac{5v}{3}=\frac{-3}{2}\)

\(\frac{3u+10v}{6}=\frac{-3}{2}\)

3u+10v=-9 ………………………..(i)

\(\frac{5u}{4}-\frac{3v}{5}=\frac{61}{60}\)

25u-12v=\(\frac{61}{3}\) ………………………(ii)

Multiplying equation (i) by 12 and (ii) by 10

36u+120v=-108 ……………………… (iii)

250u+120v=\(\frac{610}{3}\) ……………………… (iv)

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

36u+250u=\(\frac{610}{3}\)-108

=286u=\(\frac{286}{3}\)

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

Putting u=\(\frac{61}{3}\) in equation (i)

3(\(\frac{1}{3}\))+10v=-9

=v=-1

Now,

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

=x+2y=3 ……………..(v)

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

= 3x-2y=-1 …………………..(vi)

Putting x=\(\frac{1}{2}\) in equation (v) we get,

\(\frac{1}{2}\)+2y=3

=y=\(\frac{5}{4}\)

The Solutions of the given system of equation are \(\frac{1}{2}\)

And \(\frac{5}{4}\) respectively.

 

Question 34

x+y=5xy

3x+2y=13xy

Soln:

The given system of equations is:

x+y=5xy ………………….. (i)

3x+2y=13xy……………… (ii)

Multiplying equation (i) by 2 and equation (ii) 1 we get,

2x++2y=10xy ………………… (iii)

3x+2y= 13xy ……………………. (iv)

Subtracting equation (iii) from equation (iv) we get,

3x-2x=13xy-10xy

= x=3xy

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

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

Putting y = \(\frac{1}{3}=y\) in equation (i) we get,

=x+y=5(x)( \(\frac{1}{3}\))

= x+\(\frac{x}{3x}\)= \(\frac{5x}{3}\)

= 2x=1

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

Hence Solution of the given system of equation is \(\frac{1}{2}\) and \(\frac{1}{3}\)

 

Question 35

x+y=xy

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

Soln:

x+y=xy ………………………………….. (i)

\(\frac{x-y}{xy}=6\)………………….. (ii)

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

2x=2xy+6xy

=2x= 6xy

= y= x+y=xy

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

Putting =y=\(\frac{1}{4}\) in equation (i) , we get,

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

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

Hence the Solution of the given system of equation is = x=\(\frac{-1}{2}\)

And y=\(\frac{1}{4}\) respectively.

 

Question 36

2(3u-v)=5uv

2(u+3v)=5uv

Solution

2(3u-v)=5uv

= 6u-2v=5uv …………………. (i)

2(u+3v)=5uv

2u+6v=5uv …………………….. (ii)

Multiplying equation (i) by 3 and equation (ii) by 1 we get,

18u-6v=15uv …………………….. (iii)

2u+6v=5uv ………………………….. (iv)

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

18u+2u=15uv+5uv

= v=1

Putting v=1 in equation (i) we get,

6u-2=5u

=u=2

Hence the Solution of the given system of Solution of equation is 2 and 1 respectively.

 

 

Question 37

\(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\)

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

Let \(\frac{1}{3x+2y}\)=u

Let \(\frac{1}{3x-2y}\)=v

Soln:

Then the given system of equation becomes:

2u+3v=\(\frac{-17}{5}\) ………………………..(i)

5u-v=2 ……………………… (ii)

Multiplying equation (ii) by 3

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

15u-2u=\(\frac{-17}{5}\)+5

=13u=\(\frac{13}{5}\)

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

Putting u=\(\frac{1}{5}\) in equation (i)

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

=v=1

Now,

\(\frac{1}{3x+2y}\)=u

=3x+2y=5 ……………..(iv)

\(\frac{1}{3x-2y}\)=v

= 3x-2y=1 …………………..(v)

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

6x=6

=x=1

Putting the value of x in equation (v) we get,

3+2y=5

= y=1

The Solutions of the given system of equation are 1 and 1 respectively.

 

Question 38

\(\frac{44}{x+y}+\frac{36}{x-y}=4\)

\(\frac{55}{x+y}-\frac{40}{x-y}=13\)

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

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

Soln:

Then the given system of equation becomes:

44u+30v=10 ……………………….. (i)

55u+40v=13 ………………………(ii)

Multiplying equation (i) by 4 and (ii) by 3

176u+120v=40 ……………………… (iii)

165u+120v=39 ……………………… (iv)

Subtracting equation (iv) from (iii) we get,

176-165u=40-39

=u=\(\frac{1}{11}\)

Putting the value of u in equation (i)

44(\(\frac{1}{11}\))+30v=10

= 4+30v=10

=30v=6

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

=x+y=11……………..(v)

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

= x-y=5 …………………..(vi)

Adding equation (v) and (vi) we get,

2x=16

=x=8

Putting the value of x in equation (v)

8+y=11

= y=3

The Solutions of the given system of equation are 8 and 3 respectively.

 

Question 40

\(\frac{10}{x+y}+\frac{2}{x-y}=4\)

\(\frac{15}{x+y}-\frac{5}{x-y}=-2\)

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

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

Soln:

Then the given system of equation becomes:

10p+2q=4 ……………………….. (i)

15p-5q=-2 ………………………(ii)

Multiplying equation (i) by 4 and (ii) by 3

176u+120v=40 ……………………… (iii)

165u+120v=39 ……………………… (iv)

Using cross multiplication method we get,

\(\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}\)

\(\frac{p}{-16}=\frac{1}{-80}\)

\(\frac{q}{-80}=\frac{1}{-80}\)

p= \(\frac{1}{5}\) and q=1

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

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

x+y=5…………………….. 3

x-y=1 ……………………..4

Adding equation 3 and 4 we get,

x=3

Substituting the value of x in equation 3 we get,

y=2

The Solution of the given system of Solution is 3 and 2 respectively.

 

 

Question 41

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

\(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}= \frac{-1}{8}\)

Let \(\frac{1}{x-1}\)=p

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

Soln:

Then the given system of equation becomes:

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

\(\frac{6}{x-1}-\frac{3}{y-2}=1\)……………………………………. 2

Can be written as 5p+q=2 ………………………………… 3

6p-3q=1 ………………………………………. 4

Equation 3 and 4 from a pair of linear equation in the general form. Now, we  can use any method to solve these equations.

We get p=\(\frac{1}{3}\)

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

Substituting the \(\frac{1}{x-1}\) for p , we have

\(\frac{1}{x-1}\) = \(\frac{1}{3}\)

x-1 = 3

x= 4

\(\frac{1}{y-2}\) = \(\frac{1}{3}\)

y-2 =3

y=5

The Solution of the required pair of equation is 4 and 5 respectively.

 

Question 42

\(\frac{7x-2y}{xy}=5\)

\(\frac{8x+7y}{xy}=15\)

Soln:

\(\frac{7}{y}-\frac{2}{x}=5\)…………………………..1

\(\frac{8}{y}+\frac{7}{x}=15\)……………………………2

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

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

The given equation s reduce to:

-2p+7q=5

= -2p+7q-5=0 ……………………………….. 3

7p+8q=15

= 7p+8q-15=0 …………………………… 4

Using cross multiplication method we get,

\(\frac{p}{-105-(-40)}=\frac{q}{-30-35}=\frac{1}{-16-49}\)

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

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

p=1 and q=1

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

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

x=1 and y=1

 

Question 43

152x-378y=-74

-378x+152y=-604

Soln:

152x-378y=-74 ……………………………. 1

-378x+152y=-604 …………………………. 2

Adding the equations 1 and 2 , we obtain

-226x-226y=-678

=x+y=3 ……………………….. 3

Subtracting the equation 2 from equation 1, we obtain

530x+530y=530

x-y=1 …………………………….4

Adding equations 3 and 4 we obtain,

2x=4

= x=2

Substituting the value of x in equation 3 we obtain y=1

 

Question 44

99x+101y=409

101x+99y=501

Soln:

The given system of equation are :

99x+101y=409 ………………………….1

101x+99y=501…………………………….. 2

Adding equation 1 and 2 we get ,

99x+101x+101y+99y= 49+501

= 200(x+y) = 1000

=x+y=5 ……………………….. 3

Subtracting equation 1 from 2

101x-99x+99y-101y = 501-499

= 2(x-y)=2

= x-y = 1 ………………………………. 4

Adding equation 3 and 4 we get,

2x=6

= x= 3

Putting x=3 in equation 3 we get,

3+y =5

=y=2

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

 

Question 45

23x-29y=98

29x-23y=110

Soln:

23x-29y=98 ………………………………1

29x-23y=110 ………………………………… 2

Adding equation 1 and 2 we get,

= 6(x+y)=12

= x+y = 2 ………………………3

Subtracting equation 1 from 2 we get,

52(x-y) = 208

=x-y = 4 ……………………………. 4

Adding equation 3 and 4 we get,

2x= 6

= x= 3

Putting the value of x in equation 4

3+y=2

=y= -1

The Solution of the given system of equation is 3 and -1 respectively.

 

Question 46

x-y+z=4

x-2y-2z=9

2x+y+3z=1

Soln:

x-y+z=4 …………………………..1

x-2y-2z=9…………………………..2

2x+y+3z=1……………………………3

From equation 1

z=4-x+y

z= -x+y+4

Subtracting the value of the z in equation 2 we get,

x-2y-2(-x+y+4) =9

= x-2y+2x-2y-8=8

= 3x-4y= 17 …………………………….. 4

Subtracting the value of z in equation 3, we get,

2x+y+3(-x+y+4) =1

= 2x+y +3x+3y+12 =1

= -x+4y=-11

Adding equation 4 and 5 we get,

3x-x-4y+4y=17-11

= 2x=6

= x= 3

Putting x=3 in equation 4, we get,

9-4y=17

= -4y= 17-9

= y = -2

Putting x= 3 and y=-2 in z= -x+y+4 , we get,

Z= -3-2+4

= -1

The Solution of the given system of equation are 3 , -2 and -1 respectively.

 

 

Question 47

x-y+z=4

x+y+z=2

2x+y-3z =0

Soln:

x-y+z=4 ………………………………….1

x+y+z=2……………………………………..2

2x+y-3z =0………………………………………3

From equation 1

=z= -x+y+4

Substituting z= -x+y+4 in equation 2 , we get ,

=x+y+(-x+y+4) = 2

= x+y-x+y+4 = 2

= 2y =2

= y= 1

Substituting the value of z in equation 3

2x+y-3(-x+y+4) =0

= 2x+y+3x-3y-12 =0

= 5x-2y = 12 ………………………………. 4

Putting the y= -1 in equation 4

5x-2(-1) = 12

5x =10

=x=2

Putting x=2 and y = -1 in z =-x+y+4

Z= -2-1+4

= 1

The Solution of the given system of equations are 2 , -1 and 1 respectively.