Question 105706
:
{{{((x-4))/((x-5))}}} - {{{3/((x+5))}}} = {{{10/((x-5)(x+5))}}}
:
We can get rid of the denominators if we multiply thru by (x-5)(x+5)
(x-5)(x+5)*{{{((x-4))/((x-5))}}} - (x-5)(x+5)*{{{3/((x+5))}}} = (x-5)(x+5)*{{{10/((x-5)(x+5))}}}
:
Cancel out the denominator and you have:
(x+5)(x-4) - 3(x-5) = 10
:
FOIL, then combine like terms on the left:
(x^2 + x - 20) - 3x + 15 = 10
:
x^2 + x - 3x - 20 + 15 - 10 = 0
:
x^2 - 2x - 15 = 0: a quadratic equation, we can factor
:
(x-5)(x+3) = 0
:
x = +5
and
x = -3
:
It's important to notice that x = +5 is NOT a solution, as it produces
division by 0 in the original equation
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The leaves us with a solution: x = -3
:
Substituting -3 for x in the original equation:
{{{((-3-4))/((-3-5))}}} - {{{3/((-3+5))}}} = {{{10/((-3-5)(3+5))}}}
:
{{{((-7))/((-8))}}} - {{{3/((2))}}} = {{{10/((-8)(2))}}}
:
{{{((7))/((8))}}} - {{{3/((2))}}} = {{{10/((-16))}}}
:
{{{((14))/((16))}}} - {{{24/((16))}}} = -{{{10/((16))}}}; confirms our solution
:
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