well ,just check the equations using (1,3)
(A) A+3B = C and 2A-D + 6B+3E = F
since A+3B = C, that means 2C-D+3E = F
that's certainly possible.
What you have to look out for is a pair of equations that cannot both be true.
The system below has the solution of (1,3) where A, B, C, D, E, and F are all nonzero real numbers.
Ax+By=C, Dx+Ey=F
Which of the following systems would not have (1,3) as the solution?
A- Ax+By=C & (2A-D)x+(2B+E)y=C-2F
B- Ax+By=C & 7Dx+7Ey=7F
C- Ax+By=C & (A+D)x+(B+E)y=C+F
D=(A/2+D)x+(B/2+E)y=(C/2+F) & Dx+Ey=F
1 answer