How would you graph y<1 and y>2x-3?

Answer 1

Draw the line y = 1; this is a horizontal line through y = 1.
Think about all the points below this line (since you have y < 1).

Draw the line y = 2x - 3; this has a y-intercept of -3 and a slope of 2, so it goes through (0, -3) and goes upward to (1, -1), (2, 1), etc.)
Think about all the points above this line (since you have y > 2x - 3).

Because the question says *and* you should only consider the points that are in both regions (below y = 1 and above y = 2x - 3). Shade in the intersection of these two regions.

I've attached a graph of what it should look like...

Answer 2

Well the y<1 is everything below the y=1.
y>2x-3 is everything above the y=2x-3

It is better to make a graph of this functions and then you will see the situation.
^
|
-----|--------------- <<< ooo|oooooooooooooo
-----------------------> <<< ooo|ooooooooooooo
|<<<<<<<
Above is the solution for y<1 (ooooo is that what I mean under everything). You make the same for y>2x-3.
Then the solution (y<1 and y>2x-3) would be everything between those 2 functions.

I don't know if this is helpful but it is the solution. It is easyer on the paper. I've seen on preview that my "picture" will be modified but I'll post it anyway.

Answer 3

for y < 1, you would graph the line y = 1 and then shade in everything where the y values are less than 1. (the line is straight across the graph where y =1, shade everything
underneath the line.)

y > 2x - 3
This is a diagonal line that goes upwards through (0,3) and (1.5, 0) you would shade in everything above and to the left of the line.

heres a good online graphing calculator:
http://www.eocc.edu/divisions/mat_div/gcalc.htm

Puzzling is right about the intersection and shading the area where they overlap, but you shade everything under the line of y = 1 since y < 1 (y is less than 1) not above.