Solving A System of Equations

Valerie Flores & Lucy MirandaDecember 18, 2013

2-3 A

Problem SituationO 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.

Define VariablesO h represents the cost of the

hardback and p represents the cost of the paperback

O h = cost of hardback

O p = cost of paperback

System of EquationsO The system of equations is

O 2h + 6p = 60 O 2h + 3p = 39.75

Solution MethodYou use the process of elimination.

Steps of Equation Solution

O Set up the equations 2h+6p=60 and 2h+3p=39.75 to subtract them.

1. Then, you cross out the 2h in both equations. Subtract 6p and 3p. Subtract 60 and 39.75 to get 20.25

2. Finally divide by 3. That’s how much a paperback costs. p = 6.75

3. To find h, substitute 6.75 in for p. 2h+6(6.75)=60

Solution to the System of Equations

O ( 9.75 , 6.75 )

Check of Solution

O2(9.75) + 6( 6.75) = 60

O2(9.75) + 3( 6.75) = 39.75

Solution in the Problem Situation

$9.75 is the cost of the hardback.$6.75 is the cost of the paperback.