How do you solve #(x+7)/(x-4)<0#?

4 Answers
Aug 8, 2017

Answer:

#-7ltxlt4#

Explanation:

For #(x+7)/(x-4)# to be less than #0#, the top part must be less than #0#, or negative.

An #x# value of #-7# will cancel the top part, but only give us #0#, anything less than #-7# will give a negative fraction, and therefore less than #0#.

Proof:
#(x+7)/(x-4)lt0#

#((x+7)cancel((x-4)))/cancel((x-4))lt0(x-4)#

#x+7lt0#

#x+7-7lt-7#

#xlt-7#

However, if top and bottom are negative, it will be positive, and any #x# value less than #-7# will give a positive value. However if #xlt4#, then the bottom value will be negative, giving a negative value.

#-7ltxlt4#

Aug 8, 2017

Answer:

The solution is #x in (-7,4)#

Explanation:

Let #f(x)=(x+7)/(x-4)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-7##color(white)(aaaaaa)##4##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x+7##color(white)(aaaa)##-##color(white)(aaa)##0##color(white)(aa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaa)##-##color(white)(aaaaaa)##-##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaa)##0##color(white)(aa)##-##color(white)(aa)##||##color(white)(aa)##+#

Therefore,

#f(x)<0# when #x in (-7,4)#

graph{(x+7)/(x-4) [-41.1, 41.14, -20.54, 20.55]}

Aug 8, 2017

Answer:

#x in (-7, 4)#

Explanation:

Given:

#(x+7)/(x-4) < 0#

Note that since the linear expressions #(x+7)# and #(x-4)# each occur once, the rational expression will change sign at the points #x=-7# and #x=4#. It has a vertical asymptote at #x=4# and intercepts the #x# axis at #x=-7#.

For large positive or negative values of #x#, the rational expression is positive, so the interval in which it is negative is precisely #(-7, 4)#

graph{(y-(x+7)/(x-4))(x-3.99+y*0.0001) = 0 [-19.55, 20.45, -10.12, 9.88]}

Aug 8, 2017

Answer:

The answer is #x in (-7;4)#. See explanation.

Explanation:

First we have to calculate the domain of the rational expression. As the denominator cannot be zero, the excluded values are:

#x-4 != 0 => x!=4#

Now we can solve the inequality.

#(x+7)/(x-4)<0#

We can change the rational inequality to quadratic inequality by multiplying it by the square of the denominator:

#(x+7)(x-4)<0#

If we graph the quadratic function:

graph{(x-4)*(x+7) [-36.52, 36.52, -18.22, 18.35]}

we see that it takes negative values for #x in (-7;4)#, so this interval is also the solution of the initial rational inequality