### Chapter 3: Pair of Linear Equations in Two Variables

Q
##### Pair of linear equations in two variables CBSE NCERT Solutions

Question:

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)

(ii)

(iii)

(iv)

(v)  7x - 2yxy = 57x - 2yxy = 5

8x + 7yxy = 158x + 7yxy  = 15

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

(vii)

(viii)

(i)                          … (1)

… (2)

Let  1/x = p and  1/y = q

Putting the values of  1/x  = p and  1/y = q in equation (1) and equation (2), we get

and

Multiplying both equations by 6, we get

3p + 2q = 12 and 2p + 3q = 13

3p + 2q – 12 = 0    ...........… (3)

and 2p + 3q – 13 = 0 ..........… (4) = 2 and q = 3

But 1/x = p and  1/y = q

Putting the value of p = 2 and q = 3 in 1/x = p and  1/y = q respectively, we get

=  1/2 and y =   1/3

(ii)             … (1)

… (2)

Let   and

Putting  and    in equation (1) and equation (2), we get

2p + 3q = 2   … (3)

4p − 9q = −1 … (4)

Multiplying equation (3) by 2 and subtracting it from equation (4), we get

4p − 9q  – 2 (2p + 3q) = -1 - 2(2)

4p − 9q  − 4p − 6q = -1 -4

− 15q = - 5

=

Putting the value of q in equation (3), we get

2p + 1 = 2

2p = 1

=

Putting the values of p =  and q =   in    = p and    = q respectively, we get

and

and

= 4 and y = 9

(iii)    + 3y = 14 … (1)

− 4y = 23 … (2)

Let   1/x = p

we get

4p + 3y = 14 … (3)

3p − 4y = 23 … (4)

Multiplying equation (3) by 3 and equation (4) by 4, we get

3 (4p + 3y - 14 =0 ) and 4(3p − 4y – 23 = 0)

12p + 9y – 42 = 0 … (6)      12p − 16y – 92 = 0 … (7)

Subtracting equation (7) from equation (6), we get

9y − (−16y) – 42 − (−92) = 0

25y + 50 = 0

= −5025 – 5025 = −2

Putting the value of y in equation (4), we get

4p + 3 (−2) = 14

4p – 6 = 14

4p = 20

= 5

Putting the value of p in equation (3), we get

1/x  = 5

x =  1/5

Therefore, x = 1/5  and y = −2

(iv)             … (1)

… (2)

Let   = p and   = q

Putting    = p and    = q in (1) and (2), we get

5p + q = 2

5p + q – 2 = 0 … (3)

And, 6p − 3q = 1

6p − 3q – 1 = 0 … (4)

Multiplying equation (3) by 3 and adding it to equation (4), we get

3 (5p + q − 2) + 6p − 3q – 1 = 0

15p + 3q – 6 + 6p − 3q – 1 = 0

21p – 7 = 0

= 1/3

Putting p = 1/3  in equation (3), we get

5 + 3q = 6

3q = 6 5 = 1

= 1/3

Putting the values of p = 1/3 and q = 1/3  in  = p and  = q respectively, we get

and

3 = − 1 and 3 = y – 2

= 4 and y = 5

(v) 7x − 2y = 5xy … (1)

8x + 7y = 15xy … (2)

Dividing both the equations by xy, we get

Let  1/x = p and  1/y = q

Putting    = p and   = q in equation (3) and equation (4), we get

7q − 2p = 5 … (5)

8q + 7p = 15 … (6)

From equation (5), we get

2p = 7q – 5

=

Putting value of p =    in (6), we get

16q + 49q – 35 = 30

65q = 30 + 35 = 65

= 1

Putting value of q in (5), we get

7 (1) − 2p = 5

2p = 2

= 1

Putting the value of p = 1 and q = 1 in p = 1/x  and q =  1/y  respectively, we get x = 1 and y = 1

(vi) 6x + 3y − 6xy = 0 … (1)

2x + 4y − 5xy = 0 … (2)

Dividing both the equations by xy, we get

....(3)

....(4)

Let   1/x = p and   1/y = q

Putting  1/x = p and  1/y = q  in equation (3) and equation (4), we get

6q + 3p – 6 = 0 … (5)

2q + 4p – 5 = 0 … (6)

From (5),

3p = 6 − 6q

= 2 − 2q

Putting this in equation (6), we get

2q + 4 (2 − 2q) – 5 = 0

2q + 8 − 8q – 5 = 0

− 6q = − 3  = 1/2

Putting the value of q in (p = 2 – 2q), we get

= 2 – 2() = 2 – 1 = 1

Putting the values of p = 1 and q = 1/2 in  1/x = p and   1/y = q respectively, we get x = 1 and y = 2.

(vii)                  … (1)

…(2)

Let    = p and    = q

Putting  = p and   = q  in equation (1) and equation (2), we get

10p + 2q = 4 … (3)

15p − 5q = −2 … (4)

From equation (3),

2q = 4 − 10p

= 2 − 5p … (5)

Putting q = 2 − 5p  in equation (4), we get

15p – 5 (2 − 5p) = −2

15p – 10 + 25p = −2

40p = 8  =   1/5

Putting the value of p = 1/5  in equation (5), we get

Putting values of p = 1/5 and q = 1 in  = p and   = q respectively, we get,       and

+ y = 5 … (6) and x – y = 1 … (7)

Adding equation (6) and equation (7), we get

2x = 6  = 3

Putting x = 3 in equation (7), we get

3 – y = 1

= 3 – 1 = 2

Therefore, x = 3 and y = 2

(viii)                                 … (1)

… (2)

Let   = p and    = q

Putting  = p and  = q in (1) and (2), we get

+ q =   and

4p + 4q = 3 … (3) and 4p − 4q = − 1 … (4)

Adding equation (3) and equation (4), we get

8p = 2  = 1/4

Putting value of p in equation (3), we get

1 + 4q = 3

4q = 3 – 1 = 2

= 1/2

Putting the value of p = 1/4  and q = 1/2 in    = p and  = q respectively, we get

and

3x + y = 4 … (5) and 3x – y = 2 … (6)

Adding equation (5) and equation (6), we get

6x = 6  = 1

Putting x = 1 in equation (5), we get

3 (1) + y = 4

= 4 – 3 = 1

Therefore, x = 1 and y = 1

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