Find the value of ‘p’ and ‘q’ for which the following pair of

1.	Find the value of ‘p’ and ‘q’ for which the following pair of linear equations has infinite number of solutions. 2x + 3y = 7 , 2px +(p + q)y = 28
Answer» $\huge\bf{Answer:-}$ Given:- The equations are - 2x + 3y = 7 2px + ( p + q )y = 28 To Find :- The value of p and q Solution:- 2x + 3y = 7 2px + ( p + q )y = 28 Now, Comparing both the equations to the standard form ax + by + c = 0 we get, 2x + 3y -7 = 0 ......(i) 2px + ( p + q )y - 28 = 0 .....(ii) Here , in eq (i) a1 = 2 , b 1= 3 , c1 = -7 In eq(ii) a 2= 2P , b 2= (p + q) , C2 = - 28 For infinitely many solution , ⇒ $\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$ ⇒ $\frac{2}{2p} = \frac{3}{p+q} =\frac{-7}{-28}$ so, ⇒ $\frac{2}{2p} = \frac{3}{p+q}$ Now , by cross multiplication , ⇒2p + 2q = 6p ⇒2p - 6p = -2q ⇒-4p = -2q ⇒2p = q ..... ( iii) Now, $\frac{3}{p+q} = \frac{-7}{-28}$ Again, By cross multiplication 7p + 7q = 84 p+ q = 84 /7 p + q = 12 .....( iv) Now putting the value of q that we FIND out in eq (iii) in eq( iv) , we get p+ q = 12 p + 2p = 12 3p = 12 p = 12 / 3 p = 4 Therefore, the value of p is 4 Now , we get 2p = q in eq (iii) so , putting the value of p in eq (iii) 2 × 4 = q q = 8 Hence the value of p is 4 and q is 8 .

Find the value of ‘p’ and ‘q’ for which the following pair of linear equations has infinite number of solutions. 2x + 3y = 7 , 2px +(p + q)y = 28

Answer»

$\huge\bf{Answer:-}$

Given:-

The equations are -

2x + 3y = 7
2px + ( p + q )y = 28

To Find :-

The value of p and q

Solution:-

2x + 3y = 7

2px + ( p + q )y = 28

Now, Comparing both the equations to the standard form ax + by + c = 0 we get,

2x + 3y -7 = 0 ......(i)

2px + ( p + q )y - 28 = 0 .....(ii)

Here , in eq (i)

a1 = 2 , b 1= 3 , c1 = -7

In eq(ii)

a 2= 2P , b 2= (p + q) , C2 = - 28

For infinitely many solution ,

⇒ $\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$

⇒ $\frac{2}{2p} = \frac{3}{p+q} =\frac{-7}{-28}$

so,

⇒ $\frac{2}{2p} = \frac{3}{p+q}$

Now , by cross multiplication ,

⇒2p + 2q = 6p

⇒2p - 6p = -2q

⇒-4p = -2q

⇒2p = q ..... ( iii)

Now,

$\frac{3}{p+q} = \frac{-7}{-28}$

Again, By cross multiplication

7p + 7q = 84

p+ q = 84 /7

p + q = 12 .....( iv)

Now putting the value of q that we FIND out in eq (iii) in eq( iv) , we get

p+ q = 12

p + 2p = 12

3p = 12

p = 12 / 3

p = 4

Therefore, the value of p is 4

Now , we get 2p = q in eq (iii)

so , putting the value of p in eq (iii)

2 × 4 = q

q = 8

Hence the value of p is 4 and q is 8 .