Question 446103
{{{5/(2x+3) - 1/(x-6) = 0}}}
1) Factor the denominators. In this case, there is no way to factor further, but if one of them were (for example), {{{x^2 - 36}}}, you would factor it to (x+6)(x-6).
2) Find the least common denominator by multiplying the DIFFERENT factors of the denominators. In this case, it is (2x+3)(x-6).
3) Multiply the LCD found in Step 2 with the entire equation. Note that anything times 0 is 0 and that multiplying with the factor that is the denominator cancels out the denominator (for example, {{{(5/(2x+3))(2x+3)(x-6) = 5(x-6)}}}). => 5(x-6) - 1(2x+3) = 0
4) Distribute: 5x - 30 - 2x - 3 = 0
5) Add like terms: 3x - 33 = 0
6) Add 33 to both sides: 3x = 33
7) Divide both sides by 3: x = 11
8) Check for excluded values, which are values that make the denominator equal zero, by putting each denominator equal zero and solve for x to find them; the excluded values MUST NOT be included in your answer.
2x+3=0
2x=-3
{{{x=-3/2}}}
x-6=0
x=6
Neither of these are x=11, so x=11 is still a valid solution.
9) Therefore, x = {11}