Question 825484
{{{4/(2x^2 +3x) - (2x -5)/(6x +9)}}}
In order to add or subtract rational expressions (which are fractions with variables in the denominator)...<ol><li>We must, as always when adding/subtracting fractions, have the same denominator. So we start by finding the lowest common denominator (LCD):<ol><li>Factor each denominator. Notes:<ul><li>If a denominator does not factor any other way, factor out a 1 (or -1).</li><li>If a factor of one denominator is the exact opposite of one of another denominator's factors, then factor a -1 out of <u>one</u> of them. (This is make matching factors.)</li></ul></li><li>Pick one of the factored denominators.</li><li>Append any factors from the other factored denominator(s) that are not already present. (Note: If a factor appears multiple times in any denominator, then it should appear at least that many times in our LCD.)</li><li>When finished, the appended list of factors is the LCD (in factored form). If the LCD is correct then you should be able to see each denominator's factors somewhere in the LCD.</li></ol></li><li>For each fraction whose denominator is not already equal to the LCD, multiply its numerator and denominator by each factor of the LCD which is not already present. (Hint: Only multiply out the numerators. Leave the denominator in factored form.)</li><li>Now that the fractions all have a factored LCD as its denominator we can go ahead and add (or subtract) the fractions. This, of course, is done by adding (or subtracting) the numerators. The denominators remain the same.</li><li>Factor the numerator, if possible.</li><li>If there are any common factors between the numerator and denominator, cancel them. (This is why we have left the denominator in factored form until now. We were hoping the final fraction would reduce.)</li><li>If there are multiple factors in the numerator or denominator, multiply them out. (Note: Some teachers do not require this step.)</li></ol>Let's see this in action:
1. Find the LCD
1.1 Factor the denominators:
{{{2x^2 +3x = x(2x+3)}}}
{{{6x +9 = 3(2x+3)}}}
1.2 Pick a factored denominator.
It doesn't matter which one. I'll pick the first one:
x(2x+3)
1.3 Append missing factors.
The other denominator, 3(2x+3), has a factor of 3 that is not a factor of x(2x+3). So we will append it:
x(2x+3)*3
The other factor of 3(2x+3) is already present in x(2x+3). So we do not append another one. (If the other denominator had more than one factor of (2x+3), then we would append the additional, "missing" (2x+3)'s.)
1.4 The appended list is our LCD.
2. Make the denominators equal to the LCD.
The first denominator, x(2x+3), compared to the LCD is missing a factor of 3. So we multiply the numerator and denominator of that fraction by 3. The second denominator, 3(2x+3), is missing a factor of x. So we multiply that fraction's numerator and denominator by x:
{{{(4/(x(2x +3)))(3/3) - ((2x -5)/(3(2x +3)))(x/x)}}}
Multiplying out the numerators only:
{{{12/(x(2x +3)*3) - (2x^2 -5x)/(3(2x +3)*x)}}}
3. Add or subtract.
{{{(12-(2x^2 -5x))/(x(2x +3)*3)}}}
Note the parentheses! A common error is to forget them and then, in this case, only subtract the {{{2x^2}}} and not the -5x, too.
Simplifying and reordering the numerator into standard form:
{{{(-2x^2+5x+12)/(x(2x +3)*3)}}}
4. Factor the numerator.
Since factoring is easier with a positive leading coefficient, I'll start by factoring out -1:
{{{(-1)(2x^2-5x-12)/(x(2x +3)*3)}}}
And now the quadratic:
{{{(-1)(2x+3)(x-4))/(x(2x +3)*3)}}}
5. Cancel any common factors.
We can see that 2x+3 is a factor in both the numerator and denominator:
{{{(-1)cross((2x+3))(x-4))/(x*cross((2x +3))*3)}}}
leaving:
{{{(-1)(x-4))/(x*3)}}}
6. Multiply out remaining factors.
{{{(-x+4)/3x}}}