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Parameter values for no, unique or infinitely many solutions (coefficient ratios)

This type of question, asked 9 times: 10 Sep 2024 S3, 12 Sep 2024 S2, 12 Sep 2024 S3, 13 Sep 2024 S2, 17 Sep 2024 S1, 17 Sep 2024 S2, 24 Sep 2024 S2, 25 Sep 2024 S1, 25 Sep 2024 S2. All 10 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    10 Sep 2024 S3

    70
    For what value of k will the lines 2x+7ky−8=02x + 7ky - 8 = 0 and x+y−9=0x + y - 9 = 0 have no solution?

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  2. Question 2

    12 Sep 2024 S2

    75
    For the following equations, what are the values of a and b to have infinitely many solutions? ax+by=2ax + by = 2 3x−(5−2a)y=63x - (5 - 2a)y = 6

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  3. Question 3

    12 Sep 2024 S3

    54
    Which of the following is the sum of the values of a and b if the equations 2x+y=a2x + y = a, 8x+by=128x + by = 12 have infinite solutions?

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  4. Question 4

    13 Sep 2024 S2

    69
    For which of the following values of a and b do the given equations, have NO solution? x - ay = 2 - a (1 - a)x + 6y = a + b

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  5. Question 5

    17 Sep 2024 S1

    65
    For what value of k, the system of equations kx - 15y + 7 = 0 and 7x - 21y - 19 = 0 has NO solution?

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  6. Question 6

    17 Sep 2024 S2

    53
    For what value of k, the system of equations 4x + 12y + 36 = 0 and 5x + ky + 45 = 0 has an infinite number of solutions?

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  7. Question 7

    24 Sep 2024 S2

    61
    For what value of K does the following system have an infinite number of solutions? x + 2Ky - 8 = 0 2x - y - 16 = 0

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  8. Question 8

    25 Sep 2024 S1

    71
    Find the values of 'a' and 'b' for which the system of equations 3x+y=3 and (a-b)x+ (a+b)y=3a+b-3 has infinite solutions.

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  9. Question 9

    25 Sep 2024 S2

    52
    The relation between K1K_1 and K2K_2 for which the system of linear equations K1x+3y=8K_1x + 3y = 8 and 4x+K2y=164x + K_2y = 16 represents coincident lines, is:

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  10. Question 10

    25 Sep 2024 S2

    74
    Find the value of 'k' for which the system of equations 4x+6y=74x+6y=7 and 6x+(k+4)y−21=06x+(k+4)y-21=0 has a unique solution.

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