solve systems of equations substitution 2

​Let's say you have $1,000 to spend. (not much really). You have 2500 square feet to work with.

You think you would like to purchase some modules, each module has a specific size and cost.

Module A costs $200 per module and takes up 25 square feet

Module B costs $150 per module and takes up 50 square feet

​Let's say you have $1,000 to spend. (not much really). You have 2500 square feet to work with.

You think you would like to purchase some modules, each module has a specific size and cost.

Module A costs $200 per module and takes up 25 square feet

Module B costs $150 per module and takes up 50 square feet

​Let's look at the cost of each first and assume we will spend exactly $1000 (because we don't feel like dealing with inequalities yet)

​We now have a system of equations

​200A +150B = 1000

25A + 50B = 2500

​200A + 150B = 1000

25A +50B = 2500

​you wanted to . . .

​200A + 150B = 1000

25A +50B = 2500

​Substitute into the other equation

​200A + 150B = 1000

25A +50B = 2500

​Solve for the chosen variable

​200A + 150B = 1000

25A +50B = 2500

​Solve for the other variable

​Let's say you have $1,000 to spend. (not much really). You have 2500 square feet to work with.

You think you would like to purchase some modules, each module has a specific size and cost.

Module A costs $200 per module and takes up 25 square feet

Module B costs $150 per module and takes up 50 square feet

​Answer the question in the context of the problem. We found that to spend exactly $1000 and use exactly 2500 square space you will need 88 A modules and 6 B modules.

​Real problems get more complicated with multiple variables and constraints. Practice with the smaller ones to become comfortable with the bigger puzzles.

one last Rube before you go to practice