Answer: There are infinitely many solutions.
Step-by-step explanation:
Given, Two lines, A and B, are represented by the following equations:
Line A: 3x + 3y = 12
Line B: x + y = 4
By comparing to the equations [tex]a_1x+b_1x=c_1[/tex] and [tex]a_2x+b_2x=c_2[/tex] respectively , we have
[tex]a_1=3,\ b_1=3\ \ \&\ c_1=12\\\\a_2=1,\ b_2=1\ \ \&\ c_2=4[/tex]
Now , [tex]\dfrac{a_1}{a_2}=\dfrac{3}{1},\ \dfrac{b_1}{b_2}=\dfrac{3}{1},\ \&\ \dfrac{c_1}{c_2}=\dfrac{12}{4}=\dfrac{3}{1}[/tex]
i.e. [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}[/tex]
It implies the gives lines are co-incident (linearly dependent).
That means it has infinitely many solutions.
So, the correct answer is "There are infinitely many solutions.".
