Ken has $60, and he wants to buy at least 8 plants for his garden. Plants A is $6, and Plants B is $10.

Question

A: Set up the inequality system.

B: Graph it.

Answer 1

Let A be the number of Plant A plants and B the number of Plant B plants.

First inequality: A+B≥8. The total cost of the plants = 6A+10B and cannot exceed $60, so the second inequality is 6A+10B≤60, which simplifies to 3A+5B≤30.

In the above graph Plant B is the vertical axis and Plant A the horizontal axis.

The pink line and shading represents the quantity of the plants while the blue line and shading represents the cost of the plants. The dark shading is the region where both inequalities are true. The intersection of the lines is where equality is satisfied, that is, A+B=8 and 6A+10B=60. This happens when A=5 and B=3. The rest of the dark shaded area above the horizontal axis contains all the values of A and B where both inequalities are satisfied.