SOLUTION: I need a real-world example when the solution of a system of linear equations must be in the first quadrant. What do I need to know?

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Question 223231: I need a real-world example when the solution of a system of linear equations must be in the first quadrant. What do I need to know?
Answer by drj(1380) About Me  (Show Source):
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I need a real-world example when the solution of a system of linear equations must be in the first quadrant. What do I need to know?

Here's an example:

Donnie has $5.10 in nickels, dimes, and quarters. He has an equal number of dimes and nickels, with the value of the quarters being $2.40 more than the total value of the dimes and nickels. How many dimes does he have?

Step 1. Let x be the number of nickels

Step 2. Let x be the number of dimes since they are equal in number with the nickels.

Step 3. Let %280.05%2B0.10%29x=0.15x be the dollar values of nickels and dimes.

Step 4. Let 0.25y be the dollar values of quarters.

Step 5. 2.40%2B0.15x=0.25y since the value of the quarters being $2.40 more than the total value of the dimes and nickels. Or rewriting it as

-0.15x%2B0.25y=2.40

Step 6. Also, 0.15x%2B0.25y=5.10 as given by the problem since $5.10 in nickels, dimes, and quarters.

Step 7. Now, we have a linear system of equations in Steps 5 and 6

-0.15x%2B0.25y=2.40 Equation A
0.15x%2B0.25y=5.10 Equation B

Adding Equations A and B

-0.15x%2B0.15x%2B0.25y%2B0.25y=2.40%2B5.10

0.50y=7.50

y=15 quarters

Then solving x from Equation B is 5.10-3.75=0.15x or x=9.

Check Equation A if true -0.15x%2B0.25y=2.40 or -1.35%2B3.75=2.40...which is a true statement.

Step 8. ANSWER: The number of nickels and dimes is 9 each and the number of quarters is 15.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J