Question -
Answer -
Solution:
Given pair of linearequations is
2x + 3y = 7
2px + py = 28 тАУ qy
or 2px + (p + q)y тАУ 28= 0
On comparing with ax +by + c = 0,
We get,
Here, a1┬а=2, b1┬а= 3, c1┬а= тАУ 7;
And a2┬а=2p, b2┬а= (p + q), c2┬а= тАУ 28;
a1/a2┬а=2/2p
b1/b2┬а=3/ (p+q)
c1/c2┬а=┬╝
Since, the pair ofequations has infinitely many solutions i.e., both lines are coincident.
a1/a2┬а=b1/b2┬а= c1/c2
1/p = 3/(p+q) = ┬╝
Taking first and thirdparts, we get
p = 4
Again, taking last twoparts, we get
3/(p+q)┬а= ┬╝
p + q = 12
Since p = 4
So, q = 8
Here, we see that thevalues of p = 4 and q = 8 satisfies all three parts.
Hence, the pair ofequations has infinitely many solutions for all values of p = 4 and q = 8.