How do you graph the inequality y<2-4x/3 on a graph? (not a number line)?

Question

i think your suppose to put random numbers in for x? but i still don't understand

Answer 1

Start with putting 1 in for x and then solve for y and those would be the first two coordinates (1,y). and then put in 2 for x and solve for y and keep going like this until you get to putting 6 in for x and that should be enough to graph the coordinates you get.

This is generally what you do for any of these kinds of problems. It&#039;s easier if you start with 1 when plugging numbers in for x.

Answer 2

No. You put x in for the random numbers.

Answer 3

firstly, change the &lt; to = and then draw the line.

i.e. draw the line y = 2 - 4x/3  (gradient of -4/3 with y-intercept of 2).

Because u drew the line y = ... then u know that ALL points on one side of the line DO satisfy the inequality, and ALL points on the other side of the line DONT satisfy the inequality. This is true for every inequality.

Then the trick is to pick any random point thats not on the line. So lets, for argument&#039;s sake, pick the point (10, 9) which just happens to be above the line rather than below it. Put x=10 and y=9 into the inequality y &lt; 2 - 4x/3. So:

9 &lt; 2 - 4*10/3

9 &lt; -11.33

This is false, so u shade the region on the OTHER side of the line (i.e. below the line). I hope this makes sense.

Technically since it is &#039;less than&#039; and not &#039;less than or equal to&#039; the line u drew should be a dashed line rather than a broken line.

In general:

-change the &lt; or &gt; to =

-draw the line y = ...

-pick a point above or below the line, but not on the line

-put it into the inequality

-if the result is true, u shade the region on the side of the line that the point IS in

-if the result is false, u shade the region on the side of the line that the point ISNT in

good luck