How do you solve #10x-6<=4x+42# and graph the solution on a number line?

First, subtract #4x# from both sides.

#10x - 6 - color(blue)(4x) le 4x + 42 - color(blue)(4x)#

#6x - 6 le cancel(4x) + 42 - cancel(color(blue)(4x)#

#6x - 6 le 42#

Next, add 6 to both sides.

#6x - 6 + color(red)6 le 42 + color(red)6#

#6x - cancel6 + cancelcolor(red)6 le 48#

#6x le 48#

Finally, divide both sides by 6.

#(6x)/color(limegreen)6 le 48/color(limegreen)6#

#(cancel6x)/cancelcolor(limegreen)6 le 8#

#x le 8#

This is the solution! To graph it on a number line, first consider what the solution is telling you: "x is less than or equal to 8"

We need to include 8 in our solution, and everything less than 8. Therefore, we will draw a solid dot at #x=8# on our number line, and then shade in everything to the left of it. The number line will look like this:

Final Answer

First, add #color(red)(6)# and subtract #color(blue)(4x)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#-color(blue)(4x) + 10x - 6 + color(red)(6) <= -color(blue)(4x) + 4x + 42 + color(red)(6)#

#(-color(blue)(4) + 10)x - 0 <= 0 + 48#

#6x <= 48#

Now, divide each side of the inequality by #color(red)(6)# to solve for #x# while keeping the inequality balanced:

#(6x)/color(red)(6) <= 48/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) <= 8#

#x <= 8#

To graph this we draw a solid circle at #8# on the number line. The circle is solid because the inequality operator contains an "or equal to" clause.

We will draw an arrow to the left of #8# because the inequality operator also contains a "less than" clause: