Question

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

2x + (k – 2)y = k

6x + (2k – 1)y = 2k + 5

Answer 1

Given: 2x + (k – 2)y = k, 6x + (2k – 1)y = 2k + 5

Step-wise calculation:

1. Rewrite in standard form ax + by + c = 0:

2x + (k – 2)y – k = 0   ...(a₁ = 2, b₁ = k – 2, c₁ = –k) 

6x + (2k – 1)y – (2k + 5) = 0   ...(a₂ = 6, b₂ = 2k – 1, c₂ = –(2k + 5))

2. For infinitely many solutions the two equations must be proportional:

`a_1/a_2 = b_1/b_2 = c_1/c_2`

3. Compute `a_1/a_2`:

`a_1/a_2 = 2/6 = 1/3`

4. Set `b_1/b_2 = 1/3` and solve for k: 

`(k - 2)/(2k - 1) = 1/3`

⇒ 3(k – 2) = 2k – 1

⇒ 3k – 6 = 2k – 1

⇒ k = 5

5. Check `c_1/c_2` equals `1/3` at k = 5: 

`c_1/c_2 = k/(2k + 5)`

= `5/(2 xx 5 + 5)` 

= `5/15`

= `1/3`   ...(Consistent)

The system has infinitely many solutions when k = 5.