Question

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

Answer 1

`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)`