Method 1
Comparing..
We have;
#x + 5y = 4#
#darr color(white)x darr color(white)(xx) darr#
#2x + by = c#
Simply without solving if we compare we should have;
#x + 5y = 4 rArr 2x + by = c#
Hence;
#x rArr 2x#
#+color(blue)5y rArr +color(blue)by#
Therefore, #b = 5#
#4 rArr c#
Therefore, #c = 4#
Method 2
Solving simultaneously..
Using Elimination Method!
#x + 5y = 4 - - - - - - eqn1#
#2x + by = c - - - - - - eqn2#
Multiplying #eqn1# by #2# and #eqn2# by #1#
#2 (x + 5y = 4)#
#1 (2x + by = c)#
#2x + 10y = 8 - - - - - - eqn3#
#2x + by = c - - - - - - eqn4#
Subtract #eqn4# from #eqn3#
#(2x - 2x) + (10y - by) = 8 - c#
#0 + 10y - by = 8 - c#
#10y - by = 8 - c#
But, #by = c - 2x#
Hence;
#10y - (c - 2x) = 8 - c#
#10y -c + 2x = 8 - c#
#10y + 2x = 8 -> "Equation"#
Same thing as #rArr 5y + x = 4#
Proof:
Substitute #eqn1# into the above equation..
#10y + 2[4 - 5y] = 8#
#10y + 8 - 10y = 8#
#0 = 0#
Hence;
#b = 5 and c = 4#