Solve for x: 3/3x + 1/x+4 = 10/7x

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