How do you solve #4(x – 3) + 4 <10 + 6x#?

2 Answers
Jun 20, 2017

Answer:

#x>9#

Explanation:

Expand the bracket on the LHS by multiplying #x# with #4# and #-3# with #4#.

#4x-12+4<10+6x#

#4x-8<10+6x#

#4x<18+6x#

#-2x<18#

Now, divide the entire equation by #-2#. But keep in mind the the equality sign FLIPS when you divide by a negative number.

#x>9#

Below is the graph of this inequalities.

graph{x>9 [-0.76, 72.3, 0.74, 37.27]}

Jun 20, 2017

Answer:

See a solution process below:

Explanation:

First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(4)(x - 3) + 4 < 10 + 6x#

#(color(red)(4) * x) - (color(red)(4) * 3) + 4 < 10 + 6x#

#4x - 12 + 4 < 10 + 6x#

#4x - 8 < 10 + 6x#

Next, subtract #color(red)(4x)# and #color(blue)(10)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#4x - 8 - color(red)(4x) - color(blue)(10) < 10 + 6x - color(red)(4x) - color(blue)(10)#

#4x - color(red)(4x) - 8 - color(blue)(10) < 10 - color(blue)(10) + 6x - color(red)(4x)#

#0 - 18 < 0 + (6 - color(red)(4))x#

#-18 < 2x#

Now, divide each side of the inequality by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#-18/color(red)(2) < (2x)/color(red)(2)#

#-9 < (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#

#-9 < x#

To state the solution in terms of #x# we can reverse or "flip" the entire inequality:

#x > -9#