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