Q. 14.0( 302 Votes )

# Solve the followi

(i)

Let and , then the equations becomes.

Using cross-multiplication method, we obtain,

p = 2 and q = 3

Note: These questions can also be solved by elimination method and the substitution method.
In the elimination method, the coefficient of one variable in both equations is made the same by multiplying the equation and the variable is eliminated.
In the substitution method, the value of one variable is calculated in terms of another variable by equation one and then that value is put into another equation.

(ii) Putting in the given equations, we obtain

2p +3q = 2........(i)

4p - 9q = -1 .......(ii)

Multiplying equation (1) by 3,

we obtain 6p + 9q = 6....... (iii)

Adding equation (ii) and (iii), we obtain

10p = 5

Putting in equation (i), we obtain

=

= 3q = 1

=

p =

=

q =

=

Hence, x = 4 and y = 9.

(iii) Putting

= 4p + 3y = 14

= 4p + 3y - 14 = 0.....................(i)

And, 3p - 4y = 23

= 3p - 4y -23 = 0.......................(ii)

By cross- multiplication , we get,

=

=

Now,

=

=

Also, p =

Hence, x =

(iv) Putting

= 5p + q = 2....................(i)

= 6p - 3q = 1 .....................(ii)

Now, multiplying equation (i) by 3 we get,

= 15p + 3q = 6..............(iii)

Adding equations (ii) and (iii)

21 p = 7

= p =

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

=

=

= q =

We know that,

p =

= 3 = x - 1

= x = 4

And, q =

= y - 2 = 3

= y = 5

Hence, x = 4 and y = 5

(v)

=

=

=

Putting  in (i) and (ii) to get,

7q - 2p = 5 ................(iii)

8q + 7p = 15 .................(iv)

multiplying equation (iii) by 7 and equation (iv) by 2 . we get,

49q - 14p = 35.................(v)

16q + 14p = 30..................(vi)

After adding equations (v) and (vi) . we get,

65q = 65

= q = 1

Putting value of q in equation (iv) , we get,

8 + 7p = 15

= 7p = 15 - 8 = 7

= p = 1

Now,

p =

q =

Hence , x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

Putting

6q + 3p - 6 = 0

2q + 4p - 5 = 0

By cross multiplication method , we get,

=

=

After comparing we get,

p = 1 and q =

Now ,

p =

Hence, x = 1 and y = 2

(vii) Putting

10p + 2q - 4 = 0....................(i)

15p - 5q +2 = 0 ..................,,(ii)

By applying cross multiplication method , we get,

=

=

After comparing we get,

p =

Now,

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

2x = 6

= x =

Putting value of equation (iii) we get,

y = 2

Hence, x = 3 and y = 2

(viii) Putting

p + q =

p - q =

Adding (i) and (ii) we get,

2p =

= p =

Putting value of p in (ii) we get,

=

= q =

Now,

p =

q =

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

6x = 6

= x = 1

Putting value of x in equation (iii) we get,

3(1) + y = 4

= y = 1

Hence, x = 1 and y = 1

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