How do you graph 3x+y<5
To start, change the inequality sign to equal sign.
`3x + y lt 5`
`3x + y = 5`
Then, plot the equation above.
It's graph is:
The equation 3x+ y = 5 serves as a boundary between regions. That is why we have a broken line, not a solid one.
As shown above, there are two regions. The region above the line and below the line.
Then, let's take a point in each region and substitute the values to the inequality equation.
For region above the line:
Point (3,3) ===> 3x + y < 5
3(3) + 3 < 5
9 + 3 < 5
12 < 5
The resulting condition is False. Hence, the upper region of the line is not part of the graph of the inequality equation.
For region below the line:
Point (0,0) ===> 3x + y < 5
3(0) + 0 < 5
0 < 5
Since the resulting condition is TRUE, shade the region below the line to indicate that it is the graph of the inequality equation.
Answer: The shaded portion is the region below the line of `3x+y=5` . This shaded portion is the graph of the inequality equation `3x+ y lt 5` .
Now let f(x) = 5-3x
we can graph 5-3x
when x=0 then f(x) =5
when f(x) = 0 then x=5/3
So 5-3x graph goes through (0,5) and (5/3,0) cordinates.
If you draw a line across these points you will get the 5-3x graph.
What we want is y<5-3x
So the area below 5-3x line gives the required range for the answer.