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The value of k, for which the system of equations 3x – ky – 20 = 0 and 6x – 10y + 40 = 0 has no solution, is
- a)10
- b)6
- c)5
- d)3
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The value of k, for which the system of equations 3x – ky &ndash...
For no solution, a1/a2 = b1/b2 ≠ c1/c2
⇒ a1 = 3, b1 = -k and c1 = -20 for 3x – ky – 20 = 0
⇒ a2 = 6, b1 = -10 and c1 = 40 for 6x – 10y + 40 = 0
⇒ a1/a2 = b1/b2
⇒ 3/6 = -k/ (-10)
⇒ k = 5
Checking if b1/b2 ≠ c1/c2
⇒ 5/ (-10) (= 2) ≠ -20/40 (= -2)
∴ The equations have no solution for k = 5
⇒ a1 = 3, b1 = -k and c1 = -20 for 3x – ky – 20 = 0
⇒ a2 = 6, b1 = -10 and c1 = 40 for 6x – 10y + 40 = 0
⇒ a1/a2 = b1/b2
⇒ 3/6 = -k/ (-10)
⇒ k = 5
Checking if b1/b2 ≠ c1/c2
⇒ 5/ (-10) (= 2) ≠ -20/40 (= -2)
∴ The equations have no solution for k = 5
Most Upvoted Answer
The value of k, for which the system of equations 3x – ky &ndash...
Since the two equations are the same, we can write the system of equations as:
3x + 4y = 7
3x + 4y = k
To find the value of k that makes the system consistent, we need to find the value of k that makes the two equations intersect.
If the two equations are the same, then the lines represented by the equations are parallel and will never intersect. Therefore, there is no value of k that makes the system consistent.
In conclusion, there is no value of k that makes the system of equations consistent.
3x + 4y = 7
3x + 4y = k
To find the value of k that makes the system consistent, we need to find the value of k that makes the two equations intersect.
If the two equations are the same, then the lines represented by the equations are parallel and will never intersect. Therefore, there is no value of k that makes the system consistent.
In conclusion, there is no value of k that makes the system of equations consistent.
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The value of k, for which the system of equations 3x – ky – 20 = 0 and 6x – 10y + 40 = 0 has no solution, isa)10b)6c)5d)3Correct answer is option 'C'. Can you explain this answer?
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