The pair of linear equations 13x+ky = k and 39x+6y = k+4 has infinitely many solutions if
Answers
Correct Question :- The pair of linear equations 13x+ky = k and 39x+6y = k+4 has infinitely many solutions if k is ?
ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-
• A linear equation in two variables represents a straight line in 2D Cartesian plane .
• If we consider two linear equations in two variables, say :-
➻ a1x + b1y + c1 = 0
➻ a2x + b2y + c2 = 0
Then :-
✪ Both the straight lines will coincide if :-
a1/a2 = b1/b2 = c1/c2
➻ In this case , the system will have infinitely many solutions.
➻ If a consistent system has an infinite number of solutions, it is dependent and consistent.
✪ Both the straight lines will be parallel if :-
a1/a2 = b1/b2 ≠ c1/c2.
➻ In this case , the system will have no solution.
➻ If a system has no solution, it is said to be inconsistent.
✪ Both the straight lines will intersect if :-
a1/a2 ≠ b1/b2.
➻ In this case , the system will have an unique solution.
➻ If a system has at least one solution, it is said to be consistent..
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Sᴏʟᴜᴛɪᴏɴ :-
→ 13x + ky - k = 0
→ 39x + 6y - (k+4) = 0
Comparing it with ,
→ a1x + b1y + c1 = 0
→ a2x + b2y + c2 = 0
we get,
→ a1 = 13
→ a2 = 39
→ b1 = k
→ b2 = 6
→ c1 = (-k)
→ c2 = -(k+4)
Since , The pair of linear equations has infinitely many solutions,
→ a1/a2 = b1/b2 = c1/c2
Putting Values we get,
→ 13/39 = k/6 = (-k)/-(k+4)
→ 1/3 = k/6 = k/(k+4)
Comparing First 2, Now, we get,
→ 1/3 = k/6
→ 3k = 6
→ k = 2 .
Putting This in Second , and Third Now, And, comparing them we get,
→ 2/6 = 2/(2+4)
→ 2/6 = 2/6 = Satisfy.
Hence, we can conclude That, for k = 2 , The pair of linear equations has infinitely many solutions..
Answer:
if A1/A2=b1/B2=c1/c2
Step-by-step explanation:
13/39=k/6
1/3=k/6
cross multiple
6/3=k
k=2