How do you solve and graph #5x+7<=32#?

1 Answer
May 11, 2017

#x<=5#

Explanation:

First we have
#5x+7<=32#
So we subtract 7 from both sides,
#5x<=25#
And then we divide both sides by 5,
#x<=5#

To graph this on a TI calculator,

Press [y=]
Press [#x,T,theta,n#]
Now press [2nd] and then press [math]
Select the inequality you want, in this case the #<=#
Press enter
Then you type 5

Finally press graph!

So you will see a straight line at #y=1# from #-oo<##x<=5# and #y=0# from #5<##x< oo#.
The 1 indicates that the inequality is true and 0 indicates that the inequality is false. Just like how in computer language 1 is true and 0 is false.

Hope this helps :3