(2/5) + 4/(10x + 5) = 7/(2x + 1)
Note that
4/(10x + 5) = 4/[5(2x + 1)] = (4/5)/(2x + 1)
So we have
(2/5) + (4/5)/(2x + 1) = 7(2x + 1) mutiply through by 5 to get rid of the fractions
2 + 4/(2x + 1) = 35/(2x + 1) subtract 4/(2x + 1) from both sides
2 = 35/(2x + 1) - 4/(2x + 1) simplify
2 = 31/(2x + 1) cross-multiply
2(2x + 1) = 31 simplify
4x + 2 = 31 subtract 2 from both sides
4x = 29 divide by 4 on both sides
x = 29/4
(2/5) + 4/(10x + 5) = 7/(2x + 1)
Note that
4/(10x + 5) = 4/[5(2x + 1)] = (4/5)/(2x + 1)
So we have
(2/5) + (4/5)/(2x + 1) = 7(2x + 1) mutiply through by 5 to get rid of the fractions
2 + 4/(2x + 1) = 35/(2x + 1) subtract 4/(2x + 1) from both sides
2 = 35/(2x + 1) - 4/(2x + 1) simplify
2 = 31/(2x + 1) cross-multiply
2(2x + 1) = 31 simplify
4x + 2 = 31 subtract 2 from both sides
4x = 29 divide by 4 on both sides
x = 29/4