Question 95623
1. Solve the equation.
{{{t/(t-2)}}} = {{{7/(t-2)}}} + 2
:
WE can get rid of the denominators by multiplying equation by(t-2) like this:
(t-2){{{t/(t-2)}}} = (t-2){{{7/(t-2)}}} + 2(t-2)
:
Cancel out the denominators leaving you with:
 t = 7 + 2(t-2)
 t = 7 + 2t - 4: multiplied what's inside the brackets
 t = 2t + 3
 t - 2t = 3; subtracted 2t from both sides
 -t = 3
  t = -3; multiplied equation by -1, we always want the variable to be positive
:
Check solution in original equation:
{{{(-3)/(-3-2)}}} = {{{7/(-3-2)}}} + 2
:
{{{(-3)/(-5)}}} = {{{7/(-5)}}} + {{{10/5}}}
which is:
+{{{3/5}}} = {{{-7/5}}} + {{{10/5}}}
:
:
2. Solve for x.
x + 5 + {{{x/(x-5)}}} = {{{5/(x-5)}}}
Multiply equation by (x-5) to get rid of the denominators
x(x-5) + 5(x-5) + (x-5){{{x/(x-5)}}} = (x-5){{{5/(x-5)}}}
:
x^2 - 5x + 5x - 25 + x = 5
x^2 + x - 25 - 5 = 0
x^2 + x - 30 = 0; a quadratic equation:
Factors to:
(x+6)(x-5) = 0
x = -6
and
x = +5
:
Check solution using x = -6 in original equation:
-6 + 5 + {{{(-6)/(-6-5)}}} = {{{5/(-6-5)}}}
:
-1 + {{{(-6)/(-11)}}} = {{{5/(-11)}}}
:
{{{-11/11}}} + {{{(6)/(11)}}} = {{{-5/(11)}}}
:
I better point out to you that x=+5 is not a solution, it puts a 0 as the
denominator in the original equation, a big no-no in algebra