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

Question

Accepted Answer

For a system of two linear equations to have an infinite number of solutions, the two equations must be proportional to each other.

Given equations:
1) 4x + 12y + 36 = 0
2) 5x + ky + 45 = 0

Rewrite both in the form Ax + By + C = 0.

For infinite solutions, the ratios of the coefficients must be equal:
A1/A2 = B1/B2 = C1/C2

Calculate the ratios:
A1/A2 = 4/5
B1/B2 = 12/k
C1/C2 = 36/45 = 4/5

Since A1/A2 = C1/C2 = 4/5, for infinite solutions:
12/k = 4/5
Cross-multiplied:
12 * 5 = 4 * k
60 = 4k
k = 60/4 = 15

But this contradicts the initial ratio for C1/C2 which is 4/5, so let's re-check the constant term ratio:
C1 = 36, C2 = 45
36/45 = 4/5

So the ratios are:
A1/A2 = 4/5
B1/B2 = 12/k
C1/C2 = 4/5

Set B1/B2 = 4/5:
12/k = 4/5
Cross multiply:
12 * 5 = 4 * k
60 = 4k
k = 15

Therefore, k = 15 for infinite solutions.

Re-examining the options, 15 is option 4.

Hence, the correct value of k is 15.

Answer

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