Question 575593
ok im going to guess u meant:
{{{(x/2)-((2)/(x+1))=1}}}
instead of
x.2/(2.x+1)=1
.
{{{(x/2)-((2)/(x+1))=1}}}
multiply by a number that will eliminate the denominators accross the equation
in this case the number is {{{2(x+1)}}}
{{{(2(x+1)(x/2))-(2(x+1))((2)/(x+1))=1}}} use cross cancelation when distributing
{{{(x+1)(x)-4=1}}} distribute the x
{{{x^2+x-4=1}}} put in quadratic formula
{{{x^2+x-5=0}}}
*[invoke quadratic "x", 1, 1, -5]