Answer:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Here a1, b1, c1, a2, b2, c2 are all real numbers.
Note that, a12 + b12 ≠ 0, a22 + b22 ≠ 0
Explanation:
How many solutions does the following system have?
y = -2x – 4
y = 3x + 3
Solution:
Given y = -2x – 4
y = 3x + 3
Rewriting to the general form
-2x – y – 4 = 0
3x – y + 3 = 0
Comparing the coefficients,
(a1/a2) = -⅔
(b1/b2) = -1/-1 = 1
(a1/a2) ≠ (b1/b2)
Hence, this system of equations will have only one