Question 7751
Ok, here we go!

{{{(x+4)/(6x^2+5x-6) + x/(2x+3) = x/(3x-2)}}}  Factor the denominator of the 1st fraction.

{{{(x+4)/(2x+3)(3x-2) + x/(2x+3) = x/(3x-2)}}} Add the two fractions on the left side.  Common denominator is (2x+3)(3x-2), so multiply the 2nd fraction by (3x-2)/(3x-2)

{{{(x+4)/(2x+3)(3x-2) + (3x-2)(x)/(2x+3)(3x-2) = x/(3x-2)}}} Add 'em up.

{{{(x+4+(3x-2)x)/(2x+3)(3x-2) = x/(3x-2)}}} Simplify the left side.

{{{(3x^2 - x + 4)/(2x+3)(3x-2) = x/(3x-2)}}} Multiply both sides by (3x-2)

{{{(3x^2 - x + 4)/(2x+3) = x}}} Multiply both sides by (2x+3)

{{{3x^2 - x + 4 = 2x^2 + 3x}}} Simplify and set equal to zero.

{{{x^2 -4x + 4 = 0}}} Solve the quadratic equation by factoring.

{{{(x-2)(x-2) = 0}}} Apply the zero product principle.

x-2 = 0, then x = 2 

The solution is: a double root; x = 2

Check:

{{{(2+4)/(6(2^2)+5(2)-6)+ 2/(2(2)+3) = 2/(3(2)-2)}}}

{{{6/28 + 2/7 = 2/4}}}

{{{6/28 + 8/28 = 14/28}}}

{{{(6+8)/28 = 14/28}}}

{{{14/28 = 14/28}}}