find the value of k for which the pair of linear equation 3x+(k+1)y = 5k +1,(2k+1)x +15y = 63....
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Answers
Question :
Find the value of k for which the pair of Linear equation 3x + (k + 1)y = 5k + 1 , (2k + 1)x + 15y = 63 have infinitely many solutions .
Answer :
k = 4
Note:
★ A linear equation is two variables represent a straight line .
★ The word consistent is used for the system of equations which consists any solution .
★ The word inconsistent is used for the system of equations which doesn't consists any solution .
★ Solution of a system of equations : It refers to the possibile values of the variable which satisfy all the equations in the given system .
★ A pair of linear equations are said to be consistent if their graph ( Straight line ) either intersect or coincide each other .
★ A pair of linear equations are said to be inconsistent if their graph ( Straight line ) are parallel .
★ If we consider equations of two straight line
ax + by + c = 0 and a'x + b'y + c' = 0 , then ;
• The lines are intersecting if a/a' ≠ b/b' .
→ In this case , unique solution is found .
• The lines are coincident if a/a' = b/b' = c/c' .
→ In this case , infinitely many solutions are found .
• The lines are parallel if a/a' = b/b' ≠ c/c' .
→ In this case , no solution is found .
Solution :
Here ,
The given linear equations are ;
3x + (k + 1)y = 5k + 1
(2k + 1)x + 15y = 63
These equations can be rewritten as ;
3x + (k + 1)y - (5k + 1) = 0
(2k + 1)x + 15y - 63 = 0
Clearly ,
a = 3
a' = 2k + 1
b = k + 1
b' = 15
c = -(5k + 1)
c' = -63
For infinitely many solutions ,
a/a' = b/b' = c/c'
Thus ,
=> 3/(2k + 1) = (k + 1)/15 = -(5k + 1)/-63
=> 3/(2k + 1) = (k + 1)/15 = (5k + 1)/63
• Case1 :
=> 3/(2k + 1) = (k + 1)/15
=> 3×15 = (k + 1)(2k + 1)
=> 45 = 2k² + k + 2k + 1
=> 2k² + k + 2k + 1 - 45 = 0
=> 2k² + 5k - 44 = 0
=> 2k² + 11k - 8k - 44 = 0
=> k(2k + 11) - 4(2k + 11) = 0
=> (2k + 11)(k - 4) = 0
=> k = -11/2 , 4
• Case2 :
=> (k + 1)/15 = (5k + 1)/63
=> 63(k + 1) = 15(5k + 1)
=> 63k + 63 = 75k + 15
=> 75k - 63k = 63 - 15
=> 12k = 48
=> k = 48/12
=> k = 4
• Case3 :
=> 3/(2k + 1) = (5k + 1)/63
=> 3×63 = (2k + 1)(5k + 1)
=> 189 = 10k² + 2k + 5k + 1
=> 10k² + 2k + 5k + 1 - 189 = 0
=> 10k² + 7k - 188 = 0
=> 10k² + 47k - 40k - 188 = 0
=> k(10k + 47) - 4(10k + 47) = 0
=> (10k + 47)(k - 4) = 0
=> k = -47/10 , 4
Clearly ,
The common value of k in each case is 4 .