Given,
3/3x + 1/(x + 4) = 10/7x
1/x + 1/(x+4) = 10/7x
Because the first term on LHS has 'x' in the denominator and the second term in the LHS has '(x + 4)' in the denominator. So to get a common denominator, multiply and divide the first term with '(x + 4)' and the second term with 'x' as shown below
{(1/x)(x + 4)/(x + 4)} + {(1/(x + 4))(x/x)} = 10/7x
{(1(x + 4))/(x(x + 4))} + {(1x)/(x(x + 4))} = 10/7x
Now the common denominator for both terms is (x(x + 4)); so combining the numerators, we get,
{1(x + 4) + 1x} / {x(x + 4)} = 10/7x
(x + 4 + 1x) / (x(x + 4)) = 10/7x
(2x + 4) / (x(x + 4)) = 10/7x
In order to have the same denominator for both LHS and RHS, multiply and divide the LHS by '7' and the RHS by '(x + 4)'
{(2x+4) / (x(x + 4))} (7 / 7) = (10 / 7x) {(x + 4) / (x + 4)}
(14x + 28) / (7x(x + 4)) = (10x + 40) / (7x(x + 4))
Now both LHS and RHS have the same denominator. These can be cancelled.
∴14x + 28 = 10x + 40
14x - 10x = 40 - 28
4x = 12
x = 12/4
∴x = 3