Question

Find the value of k for which the following system of equations have infinitely many solution:

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k – 1)

Answer 1

Given: 2x + 3y = 2 and (k + 2)x + (2k + 1)y = 2(k – 1).

Step-wise calculation:

1. Write in standard form:

2x + 3y – 2 = 0 and (k + 2)x + (2k + 1)y – 2(k – 1) = 0

2. For infinitely many solutions the lines must be coincident.

So `2/(k + 2) = 3/(2k + 1)`

= `(-2)/(-2(k - 1))`

= `1/(k - 1)`

3. Solve `2/(k + 2) = 3/(2k + 1)`:

2(2k + 1) = 3(k + 2)

⇒ 4k + 2 = 3k + 6

⇒ k = 4

4. Check with the third ratio:

`1/(k - 1) = 1/(4 - 1) = 1/3` and `2/(4 + 2) = 2/6 = 1/3`

`3/(2 xx 4 + 1) = 3/9 = 1/3`, all equal

5. Note excluded values from denominators:

`k ≠ -2, -1/2, 1`   ...(None conflict with k = 4)

The system has infinitely many solutions when k = 4.