Question

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Chp-Variation
STD-VIII
Question
↓↓↓

If p=7, q=28, so, p=11, q=?
​

Answer 1

Answer:

2x+3y=9

(p+q)x+(2p−q)y=3(p+q+1)

Here,

a

2

a

1

=

p+q

2

,

b

2

b

1

=

2p−q

3

,

c

2

c

1

=

3(p+q+1)

9

For a pair of linear equations to have infinitely many solutions:

a

2

a

1

=

b

2

b

1

=

c

2

c

1

So, we need,

p+q

2

=

2p−q

3

=

3(p+q+1)

9

or,

p+q

2

=

2p−q

3

=>2(2p−q)=3(p+q)

=>4p−2q=3p+3q

=>p=5q....(i)

Also,

2p−q

3

=

3(p+q+1)

9

=>9(p+q+1)=9(2p−q)

=>p+q+1=2p−q

=>2p−p=q+q+1

=>p=2q+1

Substituting(i),wehave,

5q=2q+1

=>q=

3

1

Also,p=5q=5(

3

1

)=

3

5

∴p=

3

5

andq=

3

1

Answer 2

Answer:

44

Step-by-step explanation:

p=7, q=28, so, p=11, q=?