# Systems of Equations

Systems of equations can be solved by graphing or by using substitution or elimination. What are the pros and cons of each method? Which method do you like best? Why? What circumstances would cause you to use a different method?

Review examples 2, 3, and 4 in section 8.4 of the text. How does the author determine what the first equation should be? What about the second equation? How are these examples similar? How are they different?

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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 multiply equation (1) by -150

415p + 150q = 14,750

25p =5300 adding

P =20 solving for p

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

The numbers check

#### Solution Summary

Step by step solutions to all the problems are provided.