Story problem in algebra 2

Question

The negative of a certain number divided by three more than that number is less than four more than the number. Find the solutions that satisfy this inequality.

Answer 1

-x / (x+3) < x+4

-x/(x+3) - (x+4) < 0 <--- subtracts (x+4) from both sides

[-x - (x+4)(x+3)]/(x+3) < 0 <---- common denominator of (x+3)

[ -x - (x^2 + 7x + 12]/(x+3) < 0 <--- FOIL

[ -x -x^2 - 7x - 12 ] / (x+3) < 0 <--- distributive of -1 over the quadratic

[-x^2 - 8x - 12]/(x+3) < 0 <--- combines like terms

(x^2 + 8x + 12)/(x+3) > 0 <--- multiplies both sides by -1 switch the inequality

(x + 6)(x + 2 )/(x+3) > 0 <--- factors

Not one single step involves multiplying by
a potential negative number. It was avoided
like the plague!

Let f(x) = (x+5)(x+2)/(x+3)

We are interested in intervals of x, where
f(x) >0.

There is a vertical asymptope at x=-3.
The zeros are x=-2, and x=-5


The intervals are x<-5, -5 < x < -3, -3 < x < -2, x>-2

Now it's time for the sign table. Just to make double sure,
The last row is the ORIGINAL problem -x/(x+3) < x+4

Interval (x < -5) (-5 < x < -3) (-3 < x < -2) (x>-2)
------------------------------------------------------------------------
Test X -10 -4 -2.5 0
---------------------------------------------------------------------------
sign of f(x) - - - + - - + - + + + +
---------------------------------------------------------------------------
result - + - +
------------------------------------------------------------------------
Original problem -10/7<-6 -4 < 0 5 <1.5 0 < 4
FALSE TRUE FALSE



Notice that whenever f(x) is positive, the
original problem results in a true statement.

So (-5 < x < -3) or (x>-2)