For what value of $k$ does the pair $2x + 3y = 7$ and $(k + 1)x + (2k - 1)y = 4k + 1$ have infinitely many solutions?

k = 2

Correct answerYour answerEvery value of

k

except 5Correct answerYour answer

k = 5

Correct answerYour answerNo value of

k

worksCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Infinitely many solutions needs $\frac{k + 1}{2} = \frac{2k - 1}{3} = \frac{4k + 1}{7}$ .
From the first two: $3(k + 1) = 2(2k - 1)$ , so $3k + 3 = 4k - 2$ and $k = 5$ .
Check the third ratio: $\frac{4(5) + 1}{7} = 3$ , and $\frac{6}{2} = \frac{9}{3} = 3$ . All three agree, so $k = 5$ .

Worth noting

“Every value except 5” is the condition for exactly one solution. $k = 2$ comes from writing $3(k + 1)$ as $3k + 1$ and $2(2k - 1)$ as $4k - 1$ . Always check the constant ratio too: if it disagreed, the answer would be no value.

Concept: Parameter for coincident lines, checking all three ratios
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsPairs of linear equations

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Ten questions from Linear and Quadratic Equations, at your own pace.

For what value of kkk does the pair 2x+3y=72x + 3y = 72x+3y=7 and (k+1)x+(2k−1)y=4k+1(k + 1)x + (2k - 1)y = 4k + 1(k+1)x+(2k−1)y=4k+1 have infinitely many solutions?

Explanation

For what value of $k$ does the pair $2x + 3y = 7$ and $(k + 1)x + (2k - 1)y = 4k + 1$ have infinitely many solutions?