2x-3y-5=0 , 4x-ky-9=0 find the value of k for which the
Answers
Answer:
The value of k for which the system of equations
2x + 3y = 5
4x + ky = 10
has infinite number of solutions, is
A. 1
B. 3
C. 6
D. 0
FORMULA USED
• 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 and
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.
Both the straight lines will be parallel if ;
a1/a2 = b1/b2 ≠ c1/c2
In this case , the system will have no solution.
Both the straight lines will intersect if ;
a1/a2 ≠ b1/b2
In this case , the system will have an unique solution.
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|| ✰✰ ANSWER ✰✰ ||
→ 2x + 3y - 5 = 0 = a1x + b1y + c1 = 0
→ 4x + ky - 10 = 0 = a2x + b2y + c2 = 0
we get,
→ a1 = 2 , b1 = 3 , c1 = (-5)
→ a2 = 4 , b2 = k , c2 = (-10)
As, we have to Find value of k , so that, the system will have infinitely many solutions.
So,
Both the straight lines will coincide and,
a1/a2 = b1/b2 = c1/c2
Putting values we get,
→ 2 / 4 = 3/k = (-5) /(-10)
→ 1/2 = 3/k = 1/2
Comparing first two ,
→ 1/2 = 3/k
Cross - Multiply ,
→ k = 2 * 3.
→ k = 6 .
Hence, value of k will be 6 Option (C), So that, The Equations has infinite number of solutions ..
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Answer:
2/4=3/k
1/2 = 3/k
k=6
I think it is help you