Question

For which values of p and q. will the following pair of linear equations have infinitely many solutions? 3x + 4y = 6
(3p + 2q) * x + (5p + 2q) * y = (4p + 6q - 2)​

Answer 1

Answer:

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

Was this

Answer 2

Solution :-

we know that, A linear equation in two variables represents a straight line in 2D Cartesian plane .

• If we consider two linear equations in two variables, say :-

➻ a1x + b1y + c1 = 0

➻ a2x + b2y + c2 = 0

Then , Both the straight lines will coincide if :-

• a1/a2 = b1/b2 = c1/c2
• In this case , the system will have infinitely many solutions.

So , comparing given pair of linear equations 3x + 4y = 6 and (3p + 2q) * x + (5p + 2q) * y = (4p + 6q - 2) with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 we get :-

• a1 = 3
• a2 = (3p + 2q)
• b1 = 4
• b2 = (5p + 2q)
• c1 = (-6)
• c2 = -(4p + 6q - 2)

then, for infinite solutions,

→ a1/a2 = b1/b2 = c1/c2

→ 3/(3p + 2q) = 4/(5p + 2q) = (-6)/(-1)(4p + 6q - 2)

comparing first two we get,

→ 3(5p + 2q) = 4(3p + 2q)

→ 15p + 6q = 12p + 8q

→ 15p - 12p = 8q - 6q

→ 3p = 2q ------- Eqn.(1)

comparing second and third,

→ 4(4p + 6q - 2) = 6(5p + 2q)

→ 16p + 24q - 8 = 30p + 12q

→ 16p - 30p + 24q - 12q = 8

→ 12q - 14p = 8

→ 6 * 2q - 14p = 8

putting value from Eqn.(1),

→ 6 * 3p - 14p = 8

→ 18p - 14p = 8

→ 4p = 8

→ p = 2

putting value of p in Eqn.(1),

→ 3 * 2 = 2q

→ q = 3 .

therefore, value of p and q is 2 and 3 respectively .

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