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Solving A System of Equations

Solving A System of Equations. Valerie Flores & Lucy Miranda December 18, 2013 2-3 A. Problem Situation.

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Solving A System of Equations

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  1. Solving A System of Equations Valerie Flores & Lucy Miranda December 18, 2013 2-3 A

  2. Problem Situation • Daniel has a 60$ gift certificate from a local bookstore. He was planning to spend the entire amount by purchasing 2 hardback books and 6 paperback books. However, he purchased only 2 hardback books and 3 paperback books, and he still has 20.25$left on the gift certificate. Find the cost of a hardcover book and the cost of a paperback book.

  3. Define Variables • h represents the cost of the hardback and p represents the cost of the paperback • h = cost of hardback • p = cost of paperback

  4. System of Equations • The system of equations is • 2h + 6p = 60 • 2h + 3p = 39.75

  5. Solution Method • You use the process of elimination.

  6. Steps of Equation Solution • Set up the equations 2h+6p=60 and 2h+3p=39.75 to subtract them. • Then, you cross out the 2h in both equations. Subtract 6p and 3p. Subtract 60 and 39.75 to get 20.25 • Finally divide by 3. That’s how much a paperback costs. p = 6.75 • To find h, substitute 6.75 in for p. 2h+6(6.75)=60

  7. Solution to the System of Equations • ( 9.75 , 6.75 )

  8. Check of Solution • 2(9.75) + 6( 6.75) = 60 • 2(9.75) + 3( 6.75) = 39.75

  9. Solution in the Problem Situation $9.75 is the cost of the hardback. $6.75 is the cost of the paperback.

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