How do you solve the inequality #-5q+9>24#?

First, subtract #color(red)(9)# from each side of the inequality to isolate the #q# term while keeping the inequality balanced:

#-5q + 9 - color(red)(9) > 24 - color(red)(9)#

#-5q + 0 > 15#

#-5q > 15#

Next, divide each side of the inequality by #color(blue)(-5)# to solve for #q# while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator:

#(-5q)/color(blue)(-5) color(red)(<) 15/color(blue)(-5)#

#(color(blue)(cancel(color(black)(-5)))q)/cancel(color(blue)(-5)) color(red)(<) -3#

#q color(red)(<) -3#

Solving an inequality is almost exactly like solving an equality, and for the most part you can treat it as such while solving it, except for one additional rule: whenever you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. For example, #># would go to #<#, #<=# to #>=# and vice versa. If you wish to know why you must do this, read the next paragraph; otherwise, you may skip it.

The reason this rule arises is because of how the number line works. Observe that on the standard number line, numbers go smallest (#-oo#) to largest (#oo#) from left to right, with #0# in the exact center. If we write #a< b# we mean to say that #a# is farther to the right than #a#. But, if we consider #-a# and #-b#, we will notice that #-a < -b# is false because #-a# is farther to the right than #-b#.

Now we solve your inequality:

#-5q+9>24#.

First we subtract #9# from both sides to get,

#-5q+9-9>24-9 rArr -5q>15#.

Now we divide both sides by #-5#, flipping the inequality:

#(-5q)/-5>(15)/-5 rArr q<-3#.

#"isolate "-5q" by subtracting 9 from both sides"#

#rArr-5q> 24-9#

#rArr-5q> 15#

#"divide both sides by "-5#

#color(blue)"remember to reverse the sign as a consequence"#

#rArrq< -3#