Represent the following inequalities on a single graph

[ai93]

\(\displaystyle 3x+4y\le12\)
\(\displaystyle 3x+y\ge3\)
\(\displaystyle y\ge-1\)

I understand the how to plot these on a graph, just not sure how to solve these inequalities!

Do you have to solve for x or y?

 

[MarkFL]

When you plot the 3 lines corresponding to the 3 inequalities, you should find that the triangular region bounded by the lines (including the boundaries because the inequalities are weak) forms the solution set.

 

[ai93]

When you plot the 3 lines corresponding to the 3 inequalities, you should find that the triangular region bounded by the lines (including the boundaries because the inequalities are weak) forms the solution set.

I thought you would have to solve the inequalities first?

 

[MarkFL]

I thought you would have to solve the inequalities first?

Each inequality will have as its solution part of the $xy$-plane. That part of the plane where the three solutions all intersect will be the solution to the system of inequalities, and you should find this in the region bounded by the 3 lines corresponding to the 3 inequalities.

View attachment 3797

 

[ai93]

Each inequality will have as its solution part of the $xy$-plane. That part of the plane where the three solutions all intersect will be the solution to the system of inequalities, and you should find this in the region bounded by the 3 lines corresponding to the 3 inequalities.

View attachment 3797

We have to solve for \(\displaystyle y\) to determine the linear equations that serve as boundaries for the described region.

Would this be right?

3x+4y≤12
4y\(\displaystyle \le\)12-3(0)
4y\(\displaystyle \le\)12
y\(\displaystyle \ge\)3

3x+y\(\displaystyle \ge\)3
y\(\displaystyle \ge\)3-3(0)
y\(\displaystyle \ge\)3

y\(\displaystyle \ge\)-1
This can already be put on the graph?

 

[MarkFL]

First, let's look at the inequality:

\(\displaystyle 3x+4y\le12\)

The boundary will be the corresponding line:

\(\displaystyle 3x+4y=12\)

So, we need to graph this line, and an easy way is to fivide through by 12 so that it is in the two-intercept form:

\(\displaystyle \frac{x}{4}+\frac{y}{3}=1\)

Hence, we know the points $(4,0)$ and $(0,3)$ are on the line. Plot these points and then draw the line through them. Then, if we use the origin as a test point, we see that:

\(\displaystyle 3(0)+4(0)\le12\)

is true, so we know we want the points below this line (on the same side of the line as the origin). Can you use this same technique for the second inequality to determine its solution?