Question 162827: how would you graph
x+5=10
3x+4y<12
x+2y=4
y>-3, x>6
y=8/3+4
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! how would you graph
x+5=10
3x+4y<12
x+2y=4
y>-3, x>6
y=8/3+4
------------------
These are all 1st order equations (2 are inequalities), which means the graphs are linear, a straight line. All you need is 2 points, then draw a straight line thru them.
-----------
x+5 = 10
x = 5
No matter what the value of y, x is 5. Pick 2 points, say (5,0) and (5,5). Then draw a straight line thru them. It will be parallel to the y-axis.
-----------------
x+2y = 4
Find 2 points. The easiest way is to set x = 0 and solve for y, the set y = 0 and solve for x.
0 + 2y = 4
y = 2, so the point is (0,2)
x +2*0 = 4
x = 4, so the point is (4,0)
Connect them with a straight line.
--------------
y=8/3+4
y = 8/3 + 12/3
y = 20/3
Regardless of the value of x, y = 20/3
This is a line parallel to the x-axis 20/3 units above it.
---------------------
3x+4y<12
To graph an inequality, solve for the 2 points with an equal sign first.
Find 2 points. The easiest way is to set x = 0 and solve for y, the set y = 0 and solve for x.
3*0 + 4y = 12
y = 3, so the point is (0,3)
3x + 4*0 = 12
x = 3, so the point is (3,0)
Connect the 2 points with a straight line. This is a line with a slope of -1, that is, y is decreasing as x increases.
The area "below" the line is the area where 3x+4y is less than 12. The line is where it is equal, and above it it is greater than.
-------------------------
y>-3, x>6
Locate the point (6,-3). The area above and to the right of the point satisfies the 2 conditions.
|
|
|
