Ex-3.3 Pair Of Linear Equations In Two Variables (Part - 1), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Solve the following system of equations:

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

Sol. The given system of equation is

11x + 15y + 23 = 0 …………………………. (i)

7x − 2y − 20 = 0 ………………………………..(ii)

From (ii)

2y = 7x − 20

 ……………………………… (iii)

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

=  11x +  + 23 = 0

=  11x +  + 23 = 0

= 127x = 254 = x = 2

Putting the value of x in the equation (iii)

y = − 3

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

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

Sol. 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

Sol. 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

 ……………………………… (iii)

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

= 32 + 29y = 119

= 29y = 87

= y = 3

Putting the value of y in the equation (iii)

= x = 14/7 = 0

= x = 2

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

Question 4. x/2 + y = 0.8

Sol. The given system of equation is

x/2 + y = 0.8

Therefore x + 2y = 1.6

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 = 35

x = 1.6 − 2(3/5)

=  1.6 − (6/5)

=  2/5

The value of x and y are 2/5  and3/5  respectively.

Question 5. 7(y + 3) − 2(x + 3) = 14

4(y − 2) + 3(x − 3) = 2

Sol. 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

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)

  + 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

x/7 + y/3 = 5

x/2 − y/9 = 6

Sol. The given system of equation is

x/7 + y/3 = 5…………………………. (i)

x/2 − y/9 = 6………………………………..(ii)

From (i)

From (ii)

x/2 − y/9 = 6

= 9x − 2y = 108 ………………………(iii)

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

= 945 − 63y − 6y = 324

= 945 − 324 = 69y

= 69y = 621

= y = 9

Putting the value of y in the above equation

y = 14

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

Question 7

x/3 + y/4 = 11

5x/6 − y/3 = − 7

Sol. The given system of equation is

x/3 + y/4 = 11…………………………. (i)

5x/6 − y/3 = − 7………………………………..(ii)

From (i)

= 4x + 3y = 132……………………(iii)

From (ii)

= 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. 4/x + 3y = 8

6/x − 4y = − 5

Sol. taking1/x = u

The new equation becomes

4u + 3y = 8……………………(i)

6u − 4y = − 5…………………….(ii)

From (i)

4u = 8 − 3y

From (ii)

= 24 − 17y = − 10

= − 17y = − 34

= y = 2

Putting y = 2 in    = u we get ,

= u = 2/4

= x = u = 2

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

Question 9. x + y/2 = 4

2/y + x/3 = 5

Sol. The given system of equation is:

x + y/2 = 4 …………………….(i)

2/y + x/3 = 5…………………….(ii)

From (i) we get,

= 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 =  3/2

2x + y =  3/2

Sol. The given system of equation is

x + 2y =  3/2 ………………….(i)

2x + y =  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 =  3/2 ……………………….(iii)

4x + 2y = 3 …………………………………………………….(iv)

Subtracting equation (iii) from (iv)

4x − x + 2y − 2y = 3 − x + 2y =  3/2

= 3x = x + 2y =  

= 3x =  3/2

= x = 1/2

Putting x = 1/2 in equation (iv)

4(1/2) + 2y = 3

= 2 + 2y = 3

= y =  1/2

The Solution of the system of equation is x = 1/2 and y = 1/2

Question 11. √2x + √3y = 0

√3x − 8–√y = 0

Sol. √2x + √3y = 0………………………..(i)

√3x − 8–√y = 0………………………..(ii)

From equation (i)

 ……………..(iii)

Substituting this value in equation (ii) we obtain

= 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 −  + 2 = 10

2y −  = 10

Sol. The given system of equation is:

3x −  + 2 = 10 ………………..(i)

2y −  = 10……………………..(ii)

From equation (i)

= 33x − y + 15 = 110

= 33x + 15 − 110 = y

= y = 33x − 95

From equation (ii)

= 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 − 3/y = 9

3x + 7/y = 2

Sol. 2x − 3/y = 9……………………………. (i)

3x + 7/y = 2…………………………… (ii)

Taking 1y = u the given equation becomes,

2x − 3u = 9 ………………………..(iii)

3x + 7u = 2………………………..(iv)

From (iii)

2x = 9 + 3u

Substituting the value x =   in equation (iv) we get,

= 27 + 23u = 4

= u = − 1

= y = 1/u  = − 1

Putting u = − 1 in = x =  we get,

= 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

Sol. 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

……………………………… (iii)

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

 0) + 50y = 50

= 222 − 210y + 250y = 250

= 40y = 28

= y = 0.7

Putting the value of y in the equation (iii)

= x = 25/50 = 0

= x = 0.5

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

Question 15. 1/7x + 1/6y = 3

1/2x − 1/3y = 5

Sol. 1/7x + 1/6y = 3 ………………………….. (i)

1/2x − 1/3y = 5……………………………. (ii)

Multiplying (ii) by 1/2 we get,

1/4x − 1/6y = 5/2……………………………. (iii)

Solving equation (i) and (iii)

1/7x + 1/6y = 3 ………………………….. (i)

1/4x − 1/6y = 5/2 ……………………………. (iii)

Adding we get,

1/7x + 1/6y = 3 + 5/2

= x = 1/14

When, x = 1/14 we get,

Using equation (i)

=  1/6y = 1

= y = 1/6

The Solution of the given system of equation is x = 1/14 and y = 1/6 respectively.

Question 16. 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Sol. Let 1/x  = u

Let 1/y  = v

3u + 2v = 12 ……………………..(i)

And, u/3 + v/2 =  frac136

= v = 3

1/u  = x =   1/21/v  = y =   1/3