# How do you graph 3x+y<5

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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` .**

3x+y<5

y<5-3x

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.**