How do you solve #2x + 7 \leq 19#?

3 Answers
Oct 8, 2017

#x <= 6#

Explanation:

First, minus the 7 from both sides of the equation and get this:
#2x <= 12#

Then divide both sides by 2 to get x on itself and get the answer:
#x <= 6#

Oct 8, 2017

#x<=6#

Explanation:

#2x+7≤19#
Subtract 7 from both sides.

#2x+7-7≤19-7#

#2x<=12#

#(2x)/(2x)<=12/(2x)#

#x<=6#

Oct 8, 2017

#x leq 6#

Explanation:

To start, pretend the#leq#symbol is really and#=#symbol, and then solve it like a normal algebra equation:

Start with #2x+7leq19#.

Replace#leq#with#=#:

#2x+7=19#

Subtract 7 from both sides to isolate #x#:

#2x=12#

Divide both sides by #2# to get #x# to be just #x#, with no coefficients or other terms:

#x=6#

Replace#=#with#leq#:

#xleq6#

(If you had multiplied or divided by a negative number, you would have had to flip the inequality sign to#geq#. But you didn't have to, so you're good.)