Question 473699
<pre>
Is it supposed to be this

{{{x^(-1) + 3/(x+2) = 8&1/2}}} ?

or this

{{{(x^(-1) + 3)/(x+2) = 8&1/2}}} ?

When there aren't enough parentheses to show where numerators
or denominators begin and end, we can't tell what you mean, but
either go by standard rules or whatever rules we think you meant.

If it's this:

{{{x^(-1) + 3/(x+2) = 8&1/2}}}

{{{1/x + 3/(x+2) = 17/2}}}

Clear of fractions by multiplying through by LCD = 2x(x+2)

{{{2(x+2) + 6x = 17x(x+2)}}}

{{{2x+4 + 6x = 17x^2 + 34x}}}

{{{8x+4=17x^2+34x}}}

{{{0=17x^2+26x-4}}}

{{{17x^2+26x-4=0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

a=17, b=26, c=-4

{{{x = (-26 +- sqrt( 26^2-4*17*(-4) ))/(2*17) }}}

{{{x = (-26 +- sqrt(676+272 ))/34 }}}

{{{x = (-26 +- sqrt(948))/34 }}}

{{{x = (-26 +- 2sqrt(237))/34 }}}

{{{x = (2(-13 +- sqrt(237)))/34 }}}

{{{x = (-13 +- sqrt(237))/17 }}}
---------------------------------

If it's this:

{{{(x^(-1) + 3)/(x+2) = 8&1/2}}}

{{{(x^(-1) + 3)/(x+2) = 17/2}}}

Multiply through by LCD = 2(x+2)

{{{2(x^(-1) + 3) = 17(x+2)}}}

{{{2x^(-1) + 6 = 17x + 34}}}

{{{2/x + 6 = 17x + 34}}}

Multiply through by LCD = x

{{{2 + 6x = 17x^2 + 34x}}}

{{{0=17x^2 + 28x - 2}}}

{{{x = (-28 +- sqrt(28^2-4*17*(-2) ))/(2*17) }}}

{{{x = (-28 +- sqrt(784+136) )/34 }}}

{{{x = (-28 +- sqrt(920) )/34 }}}


{{{x = (-28 +- sqrt(4*230) )/34 }}}

{{{x = (-28 +- 2sqrt(230) )/34 }}}

{{{x = (2(-14 +- sqrt(230)) )/34 }}}

{{{x = (-14 +- sqrt(230)) /17 }}}

Edwin</pre>