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

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