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Bunuel
For what values of k will the pair of equations \(3x + 4y = 12\) and \(kx + 12y = 30\) does not have a unique solution?

A. 12
B. 9
C. 7.5
D. 3
E. 2.5


uniique solution when two lines are parallel.

3/k = 4/12
k =9
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Bunuel
For what values of k will the pair of equations \(3x + 4y = 12\) and \(kx + 12y = 30\) does not have a unique solution?

There are some issues with the wording that might make the question more confusing than it needs to be -- the word "does" should not be there, the word "values" should say "value", and when we say an equation or system of equations "does not have a unique solution", that invariably means "has more than one solution". But that's not what happens here -- here, the two equations either have one solution or no solution, depending on the value of k. The equations could only have infinitely many solutions if one equation was an exact multiple of the other, but we'd need to multiply the first equation by 3 to go from '4y' to '12y', and then the constant term 12 from the first equation becomes 36, which doesn't match the 30 in the second equation. That proves that k = 9 anyway, since when k = 9, multiplying the first equation by 3, we learn that 9x + 12y = 36 which contradicts the second equation, which says 9x + 12y = 30, so no solutions can exist. But the word "unique" shouldn't appear in the question.
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There will not be a unique solution if lines are parallel. Non-parallel lines will always intersect and have a solution.

Eq1: a1x + b1y = c1
Eq2: a2x + b2y = c1

Two lines are parallel if a1/a2 = b1/b2

3/k = 4/12; k = 9
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Linear equations:

1. ax + by = c
2. ux + vy = w

If a/u = b/v = c/w
-> Infinite solutions
If a/u <>(not equal to) b/v
-> Unique Solution
If a/u = b/v <> c/w
-> No Solution

So, the above Questions asks for a value that isn't unique. So, we can plug in the values in the options and validate the equation.
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