How do you solve #(q - 12) 3

Question

How do you solve #(q - 12) 3 <5q + 2#?

Nathan Axel · Accepted Answer

See the entire solution process below:

First, expand the terms on the left side of the inequality: #(3 xx q) - (3 xx 12) < 5q + 2# #3q - 36 < 5q + 2# Next, add #color(red)(36)# and subtract #color(blue)(5q)# from each side of the inequality to isolate the #q# term while keeping the inequality balanced: #3q - 36 + color(red)(36) - color(blue)(5q) < 5q + 2 + color(red)(36) - color(blue)(5q)# #3q - color(blue)(5q) - 36 + color(red)(36) < 5q - color(blue)(5q) + 2 + color(red)(36)# #(3 - 5)q - 0 < 0 + 38# #-2q < 38# Now, divide each side of the inequality by #color(blue)(-2)# to solve for #q# while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative term we must also reverse the inequality term: #(-2q)/color(blue)(-2) color(red)(>) 38/color(blue)(-2)# #(color(blue)(cancel(color(black)(-2)))q)/cancel(color(blue)(-2)) color(red)(>) -19# #q > -19#

Julia Adams · Answer

undefined To solve the inequality $ (q - 12) \times 3 < 5q + 2 $, first distribute the 3 on the left side to get $ 3q - 36 < 5q + 2 $. Then, subtract $3q$ from both sides to isolate the variable $q$, which gives $-36 < 2q + 2$. Next, subtract 2 from both sides to get $-38 < 2q$. Finally, divide both sides by 2 to solve for $q$, resulting in $-19 < q$. So, the solution to the inequality is $q > -19$.

How do you solve #(q - 12) 3 &lt;5q + 2#?

How do you solve #(q - 12) 3 <5q + 2#?