Question 992884
Note: the LCD is {{{x(x+3)}}}. In order to combine these fractions, we must get each denominator equal to the LCD.

<table border=1><tr><th>Step Number</th><th>Statement</th><th>Reason/explanation for given step</th></tr><tr><td>1</td><td>{{{(2x)/(x+3)-4/x}}}</td><td>Original expression (no reason needed)</td></tr><tr><td>2</td><td>{{{(x/x)*((2x)/(x+3))-4/x}}}</td><td>Multiply the first fraction by {{{x/x}}}</td></tr><tr><td>3</td><td>{{{(x*2x)/(x*(x+3))-4/x}}}</td><td>Use the idea that {{{(x/y)*(z/w) = (x*z)/(y*w)}}}</td></tr><tr><td>4</td><td>{{{(2x^2)/(x*(x+3))-4/x}}}</td><td>{{{x*2x}}} becomes {{{2x^2}}}</td></tr><tr><td>5</td><td>{{{(2x^2)/(x*(x+3))-(4/x)*((x+3)/(x+3))}}}</td><td>Multiply the second fraction by {{{(x+3)/(x+3)}}}</td></tr><tr><td>6</td><td>{{{(2x^2)/(x*(x+3))-(4(x+3))/(x(x+3))}}}</td><td>Use the idea that {{{(x/y)*(z/w) = (x*z)/(y*w)}}}</td></tr><tr><td>7</td><td>{{{(2x^2-4(x+3))/(x(x+3))}}}</td><td>Use the idea that {{{a/d-b/d=(a-b)/d}}} (notice the common denominator d)</td></tr><tr><td>8</td><td>{{{(2x^2-4x-12)/(x(x+3))}}}</td><td>Distribute</td></tr></table>

Final Answer: {{{(2x^2-4x-12)/(x(x+3))}}}

Note: an optional step is to distribute in the denominator turning {{{x(x+3)}}} into {{{x^2+3x}}}. So an alternative equivalent answer is {{{(2x^2-4x-12)/(x^2+3x)}}}