For what value of k do the equations 3x + ky = 15 and 9x + 6y = 30 have no solution?
Step-by-step solution
Multiply the first equation by 3: 9x + 3ky = 45. For no solution with 9x + 6y = 30, the y-coefficients must match but the constants differ: 3k = 6, so k = 2, and 45 \neq 30.
Common mistake
Picking the value that makes the system identical (k = 2 with constants also matching) or confusing no-solution with infinitely-many.
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What is the answer to For what value of k do the equations 3x + ky = 15 and 9x + 6y = 30 have no solution?
Answer: B) 2. Multiply the first equation by 3: 9x + 3ky = 45. For no solution with 9x + 6y = 30, the y-coefficients must match but the constants differ: 3k = 6, so k = 2, and 45 \neq 30.
How do you solve it?
Multiply the first equation by 3: 9x + 3ky = 45. For no solution with 9x + 6y = 30, the y-coefficients must match but the constants differ: 3k = 6, so k = 2, and 45 \neq 30.
What mistake do most students make?
Picking the value that makes the system identical (k = 2 with constants also matching) or confusing no-solution with infinitely-many.
What SAT topic is this?
Algebra. Practice more free Algebra questions with instant feedback at Syntene Learn.