How do you solve #(x+68)/(x+8)>=5#?

2 Answers
May 7, 2017

Answer:

#x in (-8, 7]#

Explanation:

Given:

#(x+68)/(x+8) >= 5#

Subtract #(x+68)/(x+8)# from both sides to get:

#0 >= 5-(x+68)/(x+8) = (5(x+8)-(x+68))/(x+8) = (4x-28)/(x+8) = (4(x-7))/(x+8)#

Note that the right hand side is a function which is continuous and non-zero except at #x=-8# and #x=7#. The function changes sign at each of these two points.

When #x=-8# the denominator is #0# so the right hand side is undefined. So #-8# is not part of the solution set.

When #x=7# the numerator is #0# and the inequality is satisfied. So #7# is part of the solution set.

When #x < -8# or #x > 7# then the signs of the numerator and denominator are the same, so the quotient is positive and the inequality is not satisfied.

When #x in (-8, 7)# the numerator is negative, the denominator is positive and the quotient is negative. So the inequality is satisfied.

So the solution set is:

#x in (-8, 7]#

May 7, 2017

Answer:

The solution is #x in (-8,7]#

Explanation:

We cannot do crossing over.

So, we simplify the inequality

#(x+68)/(x+8)>=5#

#(x+68)/(x+8)-5>=0#

#((x+68)-5(x+8))/(x+8)>=0#

#(x+68-5x-40)/(x+8)>=0#

#(28-4x)/(x+8)>=0#

#(4(7-x))/(x+8)>=0#

Let #f(x)=(4(7-x))/(x+8)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-8##color(white)(aaaaaaaa)##7##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+8##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##7-x##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

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

Therefore,

#f(x)>=0# when #x in (-8,7]#