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.
Find the value of k for which each of the following systems of linear equations has an infinite number of solutions: 2x+3y=7, k 1x+k+2y=3k.
Solution
The given system may be written as
2x+3y-7=0
(k−1)x+(k+2)y-3k=0
The given system of equation is of the form
a1x+b1y+c1 = 0
a2x+b2y+c2 = 0
Where, a1=2,b1=3,c1=−7
a2=k,b2=k+2,c2=3k
For unique solution,we have
a1a2=b1b2=c1c2
2k−1=3k+2=−7−3k
2k−1=3k+2 and 3k+2=−7−3k
⇒2k+4=3k−3 and 9k=7k+14
⇒k=7and k=7
Therefore, the given system of equations will have infinitely many solutions, if k=7.