When two resistors are connected in series across a 12-V battery, 1.1A flows. When they are connected in parallel across the same battery, 6.3A flows. What is the smaller resistor's value?

Answer 1

V/rparallel=6.35

V/rsum=1.2

rsum=12/1.2=you do it.
R1+R2=rsum=above

then r2=abovesum-r1

rparallel=12/6.35
or
r1r2/(r1+r2)=12/6.35

r1(abovesum-r1)=12/6.35 (r1+12-r1)
and that quickly solves as a quadratic.

check my math, I did it in my head.