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, a1 = 2, b1 = 3, c1 = – 7;
And a2 = 2p, b2 = [p + q], c2 = – 28;
a1/a2 = 2/2p
b1/b2 = 3/ [p+q]
c1/c2 = ¼
Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.
a1/a2 = b1/b2 = c1/c2
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.
The given system of equations can be written as
2x + 3y - 7 = 0 ….[i]
2ax + [a + b]y – 28 = 0 ….[ii]
This system is of the form:
`a_1x+b_1y+c_1 = 0`
`a_2x+b_2y+c_2 = 0`
where, `a_1 = 2, b_1= 3, c_1= -7 and a_2 = 2a, b_2 = a + b, c_2= – 28`
For the given system of linear equations to have an infinite
number of solutions, we must have:
`[a_1]/[a_2] = [b_1]/[b_2] = [c_1]/[c_2]`
`⇒2/[2a] = 3/[a+b] = [−7]/[−28]`
`⇒ 2/[2a] =[ −7]/[−28 ]= 1/4 and 3/[a+b] = [−7]/[−28] = 1/4`
⇒ a = 4 and a + b = 12
Substituting a = 4 in a + b = 12, we get
4 + b = 12 ⇒ b = 12 – 4 = 8
Hence, a = 4 and b = 8.
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