Question 191816
Solve for x???
{{{9*((x+2)/(x+3))^2-6*((x+2)/(x+3))+1 = 0}}}
Just to facilitate solving, let's temporarily substitute: {{{z = ((x+2)/(x+3))}}}, so now we have:
{{{9z^2-6z+1 = 0}}} Factor this quadratic equation.
{{{(3z-1)(3z-1) = 0}}} Apply the zero product rule:
{{{3z-1 = 0}}}, so...
{{{3z = 1}}} and...
{{{z = 1/3}}} Now re-substitute {{{z = ((x+2)/(x+3))}}} to get:
{{{((x+2)/(x+3)) = 1/3}}} Cross multiply.
{{{3(x+2) = (x+3)}}} Simplify.
{{{3x+6 = x+3}}} Subtract x from both sides.
{{{2x+6 = 3}}} Subtract 6 from both sides.
{{{2x = -3}}} Finally, divide both sides by 2.
{{{highlight(x = -3/2)}}}
Check:
{{{9*((x+2)/(x+3))^2-6*((x+3)/(x+2))+1 = 0}}} Substitute {{{x = -3/2}}}
{{{9*(((-3/2)+2)/((-3/2)+3))^2 - 6*(((-3/2)+2)/((-3/2)+3))+1 = 0}}}
{{{9*((1/2)/(3/2))^2-6*((1/2)/(3/2))+1 = 0}}} Simplify:
{{{9*(1/3)^2-6*(1/3)+1 = 0}}}
{{{9*(1/9)-6*(1/3)+1 = 0}}}
{{{1-2+1 = 0}}}
{{{0 = 0}}} OK!