How do I graphically solve an inequality?

Question

I'm on my last final of the semester and I need to know how I can graphically solve an inequality!!! e.g 3y - 4 > 2x
Any tips or examples would be great.
Thanks so much!!!!

Answer 1

If you are able to graph equations you should not have any problem graphically solving an inequality.

The example you gave is linear. You should be able to rearrange it to get it in y=ax+b form. (Except in this case it will be y>ax+b)

Graph the line represented by that y=ax+b.
If the inequality is y>ax+b then you know that everything ABOVE that line is a solution to your inequality.... shade it in.
If y
If y is "greater than or equal to" or "less than or equal to" your solution also INCLUDES all points on the line.

*If your inequality is not also "equal to" the line use a dotted line when graphing. If the line is included in the solution draw a solid line.

Answer 2

3y - 4 > 2x
First solve for y:
3y> 2x+4
y> (2/3)x +4/3
Now graph the equation y = (2/3)x +4/3
This will be a line that passes through (0,4/3) and (-2,0). Draw the line as a dashed line to indicate that points on thi line are not included. Now shade the area above the dashed line and you will have shown the area which includes all numbers satisfying the inequality.

If you are not sure if you should shade the area below the line or above the line , just pick a point above the line and test it to see if it satisfies the inequality. If it does, then shade above. If not shade below.

For example the point (0,3) is definitely above the line.So we would have 3(3)-4>2(0) or 9-4>0 or 5>0. This is obviously true so shade above the line..

Hope this helped.

Incidentally if the inequality had been 3y - 4 =or > 2x then you would draw the line solidly.

Answer 3

First, circulate the words to the denominator and alter the powers to useful powers: a million/ (6^x) > a million/ (8^x) next, multiply the two facets via (8^x): 8^x / 6^x > a million Now take the log of the two area: log (8^x / 6^x) > log a million log (8^x) - log (6^x) > 0 (log rule of branch and log a million = 0) Now carry the powers out in front of the logs (potential rule of logs): x log 8 - x log 6 > 0 ingredient out an x: x (log 8 - log 6) > 0 Divide the two facets via (log 8 - log 6) x > 0 So, this works for any effectual fee of x.

Answer 4

first graph 3y-4=2x
now
3y-4>2x
y>(2x+4)/3
this is satisfied by anything above the line
Y< values are below the line

Answer 5

3y-4> 2x => 3y > 2x+4 => y > (2x+4)/3

You Need To Draw The Chart Of y = (2x +4)/3
and then hatch the upper side of the chart .
for example for y> 2x we have:

|xxxxxxxxx/
|xxxxxxxx/
|xxxxxxx/
|xxxxxx/
|xxxxx/ (y=2x)
|xxxx/
|xxx/
|xx/
|x/
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Answer 6

i wonder if there is only 1 equation or 2.or try giving values to x and y and plot the graph.

Answer 7

That depends.