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2*x*+ 3*y*= 7 and 2*p* + py = 28 – qy, if the pair of equations have infinitely many solutions. Find p and q.

Answer:

Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

Here, a_{1} = 2, b_{1} = 3, c_{1} = – 7;

And a_{2} = 2p, b_{2} = (p + q), c_{2} = – 28;

a_{1}/a_{2} = 2/2p

b_{1}/b_{2} = 3/ (p+q)

c_{1}/c_{2} = ¼

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

1/p = 3/(p+q) = ¼

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

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