Answer: B) -48
Step-by-step explanation:
If the system of two linear equations [tex]a_1x+b_1y=c_1\ \&\ a_2+b_2y=c_2[/tex] has infinitely many solutions , then
[tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=k[/tex]
For the given system of linear equations : [tex]rx-sy=26\\4x+3y=13[/tex]
We , will have
[tex]\dfrac{r}{4}=\dfrac{-s}{3}=\dfrac{26}{13}=2\\\\\Rightarrow\ r= 4\times2=8\ \ \&\ \ s=2\times-3=-6[/tex]
The value of [tex]rs=(8)\times-6=-48[/tex]
Hence, the value of rs is -48.
Thus , the correct option is B) -48.
