How do you solve #6+ 8x < 2( - 3+ 4x )#?
1 Answer
Explanation:
Your goal here is to isolate
#6 + 8x < 2 * (-3 + 4x)#
Expand the parenthesis to get
#6 + 8x < 2 * (-3) + 2 * 4x#
#6 + 8x < - 6 + 8x#
Now, notice what happens when you subtract
#6 + color(red)(cancel(color(black)(8x))) - color(red)(cancel(color(black)(8x))) < -6 + color(red)(cancel(color(black)(8x))) - color(red)(cancel(color(black)(8x)))#
#6 < -6#
When is
Never, which implies that this inequality has no real solutions. In other words, regardless of the value of
#6 + 8x color(red)(cancel(color(black)(<))) 2 * (-3 + 4x), " "(AA) x in RR#
Therefore, you can say that