Friday

October 31, 2014

October 31, 2014

Posted by **micheal** on Tuesday, June 16, 2009 at 12:18am.

Example 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)

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=14,750

263p =5300 adding

p =20. solving for p

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

p+q=63 Equation (1)

20+q=63 Substituting 20 for p

q=43. Solving for q

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

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