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The pair of linear equations 3x + 5y = 3, 6x + ky = 8 do not have any solution if –
  • a)
    k = 5
  • b)
    k = 10
  • c)
    k ≌ 10
  • d)
    k ≌ 5
Correct answer is option 'B'. Can you explain this answer?
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The pair of linear equations 3x + 5y = 3, 6x + ky = 8 do not have any ...
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Most Upvoted Answer
The pair of linear equations 3x + 5y = 3, 6x + ky = 8 do not have any ...
For having no solution -->
a1/a2=b1/b2 is not equal to c1/c2,
or,3/6=5/k is not equal to3/8,
3/6=5/k,
k=10,
that's why option 'B' is correct
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Community Answer
The pair of linear equations 3x + 5y = 3, 6x + ky = 8 do not have any ...
The coefficients of x and y in the two equations are proportional to each other.

In other words, if we can multiply one equation by a constant factor to obtain the other equation, then the system has no solution.

To check if this is the case for the given equations, we can compare their coefficients:

3x + 5y = 3

6x + ky = 8

If we multiply the first equation by 2, we get:

6x + 10y = 6

Comparing this to the second equation, we see that if k is equal to 10, then the two equations become equivalent.

Therefore, if k = 10, the pair of linear equations 3x - 5y = 3, 6x - 10y = 8 has no solution.
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The pair of linear equations 3x + 5y = 3, 6x + ky = 8 do not have any solution if –a)k = 5b)k = 10c)k 10d)k 5Correct answer is option 'B'. Can you explain this answer?
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