How do you solve #\frac { 1} { x - 2} + 3= - \frac { 16} { x ^ { 2} + x - 6}#?

1 Answer
Jun 8, 2017

Answer:

#x=(-4+-sqrt388)/6#

Explanation:

Let's start off with the original equation:

#1/(x-2)+3=16/(x^2+x-6)#

We then factor #x^2+x-6# into a #(x+a)(x+b)# format:

#1/(x-2)+3=16/((x-2)(x+3)#

We see that in both fractions, there is an #(x-2)# term, so we multiply by #(x-2)# on both sides:

#(x-2)(1/(x-2)+3)=(x-2)(16/((x-2)(x+3)))#

Before we can cancel out the #x-2# terms however, we need to use the distributive property to expand out the left side:

#(x-2)(1/(x-2))+(x-2)(3)=(x-2)(16/((x-2)(x+3)))#

Now, we can cancel out the #(x-2)# terms:

#cancel(x-2)(1/cancel(x-2))+(x-2)(3)=cancel(x-2)(16/(cancel(x-2)(x+3)))#

That leaves:

#1+(x-2)(3)=16/(x+3)#

We use the distributive property again on the #(x-2)(3)# term:

#1+3x-6=16/(x+3)#

We go ahead and combine the #1# and the #-6# and end up with this:

#3x-5=16/(x+3)#

Next, we multiply #(x+3)# on both sides in order to have all the #x# terms on one side:

#(3x-5)(x+3)=16#

Using FOIL, we can expand out the left side and rewrite the equation as so:

#3x^2+4x-15=16#

We can then move all the terms to the left side and leave a #0# on the right:

#3x^2+4x-31=0#

Here, we can use the Quadratic Formula to obtain the value of #x#. Recalling that the Quadratic Formula is

#x=(-b+-sqrt(b^2-4ac))/(2a)#

and that the #a#, #b#, and #c# values in our equation are #3#, #4#, and #-31#, respectively, we can substitute those values into the formula to find #x#:

#x=-(4)+-sqrt((4)^2-4(3)(-31))/(2(3))#

#x=-4+-sqrt(16(-12)(-31))/6#

#x=-4+-sqrt(16+372)/6#

#x=-4+-sqrt388/6#
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Additional Calculations:

#x=-4+-sqrt388/6# results in two different equations for #x#.

#x=-4+sqrt388/6# and #x=-4-sqrt388/6#

This means that #x# has two possible values:

  • #x=-4+sqrt388/6#
    #x=(-4+19.6977156036)/6#
    #x=15.6977156036/6#
    #color(red)(x=2.61628593393)#

  • #x=-4-sqrt388/6#
    #x=(-4-19.6977156036)/6#
    #x=(-23.6977156036)/6#
    #color(blue)(x=-3.94961926727)#