# 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)

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)

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

=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.

#### Practise This Question

Let a,b and c be three non -zero vectors such that no two of these are collinear. If the vector a+2b is collinear with c and b+3c is collinear with a (λ being some non -zero scalar) then a+2b+6c equals