Review example 2 How does the author determine what the?
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
To determine the values of p and q in Example 2, the author follows a step-by-step process outlined in the passage. Here's a breakdown of the process:
1. Translate the given information into equations:
- The first row of the table and the fourth sentence state that the total number of pupae received is 63, so the equation is p+q=63.
- The cost of each morpho granadensis pupa is $4.15, so the cost for p pupae is 4.15p.
- The cost of each battus polydamus pupa is $1.50, so the cost for q pupae is 1.50q.
- From the third row of the table and the problem statement, the total cost is $147.50, so the equation for the total cost is 4.15p+1.50q=147.50.
2. Clear the decimals in equation (2) by multiplying both sides by 100:
- This yields the modified equation: 415p+150q=14,750.
3. Solve the system of equations using the elimination method:
- Multiply equation (1) by -150 to get: -150p-150q=-9,450.
- Add this new equation to equation (2) to eliminate q:
-150p-150q + 415p+150q= -9,450 + 14,750.
- Simplify the result to: 265p=5,300.
4. Solve for p:
- Divide both sides of the equation by 265: p=20.
5. Substitute the value of p into equation (1) and solve for q:
- Replace p with 20 in equation (1): 20+q=63.
- Simplify the equation: q=43.
6. The values of p and q are 20 and 43, respectively:
- The solution to the system of equations is (20, 43), which means p=20 and q=43.
By following this step-by-step process, the author determines the values of p and q in Example 2.