Reference no: EM131025191

1.)How are these examples similar? How are they different? Post a similar to example 2,3 and 4.

Post for classmate to solve.

2.) Translate. The first row of the table and the fourth sentence of the problem tell us that a total of 63 pupae was received. Thus we have one equation:

p+q=63

Since each pupa of morpho granadensis costs $4.15 and p pupae were received, 4.15p is the cost for the morpho granadensis species. Similarly, 1.50q is the cost of the battus polydamus species. From the third row of the table and the information in the statement of the problem, we get a second equation:

4.15p+1.50q=147.50

We can multiply by 100 on both sides of this equation in order to clear the decimals. This gives us the following system of equations as a translation:

p+q=63 (1)
415p+150q=14,750 (2)

3.) Solve. We decide to use the elimination method to solve the system. We eliminate q by multiplying equation (1) by -150 and adding it to equation (2).

-150p-150q = -9450 Multiplying equation(1) by -150
415p+150q = 14750
265p = 5300---------------adding
P = 20-------------------solving for

To find q, we substitute 20 for p in equation (1) and solve for q:

p+q=63
20+q=63
q=43

We obtain (20,43), or p=20,q=43.

4.) Check We check in the original problem. Remember that p is the number of pupae of morpho granadensis and q is the number of pupae of battus polydamus.

Number of pupae; p+q=20+43=63
Cost of morpho granadensis: $4.15p=4.15(20)= $83.00
Cost of battus polydamus: $1.50q=1.50(43)=64.50
Total:= $147.50