Question 5530
multiply the entire equation by (3x+5)(x)(2x+3) {{{(3x+5)(x)(2x+3)5/(3x+5) -1*(3x+5)(x)(2x+3)/(x) = 1*(3x+5)(x)(2x+3)/(2x+3)}}}
This will clear our denominators, because they now cancel:
{{{(x)(2x+3)5 -1*(3x+5)(2x+3) = 1*(3x+5)(x)}}}
clear the parentheses:
{{{5x*(2x+3)-1(6x^2+19x+15)=x(3x+5)}}}
{{{(10x^2+15x)-6x^2-19x-15=3x^2+5x}}}
combine like terms:
{{{4x^2-4x-15=3x^2+5x}}}
subtract 5x from both sides:
{{{4x^2+-9x-15=3x^2}}}
Subtract 3x^2 from both sides:
{{{x^2-9x-15=0}}}
Then use the quadratic equation:
*[invoke quadratic "x", 1, -9, -15]