LINR 3 | Lesson 2 | Supplement 4

Beans in multiple representations

Let’s look at a table of values…

Let \(x=\) pounds of rice, \(y=\) pounds of beans.

Table

x(lbs rice) y(lbs beans) $ Servings
0 0 0 ok 0 no
0 10 15 ok 120 no
0 20 30 ok 240 no
0 40 60 ok 480 no
0 50 75 ok 600 ok
0 60 90 720 ok

Graph

This is getting tedious…so let’s move on to equations…,

Equations

Each lb of rice costs $2, so rice cost is \(2x\).

Each lb of beans, cost $1.5, so beans cost is \(1.5y\).

Total cost is \(2x+1.5y\) and we want that to be no more than $100.

\(2x+1.5\leq100\) dollars

Similarly, we can serve 8 people, for every lb of rice, 8x servings.

And we can serve 12 people for every lb of beans, \(12y\) servings.

We want to have at least 500 servings or

\(8x+12y\leq100\) servings

We also can’t have negative pounds or beans or rice so,

\(x\geq0, \, y\geq0\)

Let’s use desmos.com to graph these inequalities:

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