# SOLUTION: At Gwen's garage sale, all books were one price, and all magazines were another price. Harriet bought four books and three magazines for \$1.45, and June bought two books and five

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 Click here to see ALL problems on Linear-systems Question 104979This question is from textbook University of Phoenix Elementary and Intermediate Algebra : At Gwen's garage sale, all books were one price, and all magazines were another price. Harriet bought four books and three magazines for \$1.45, and June bought two books and five magazines for \$1.25. What was the price of a book and what was the price of a magazine? So first of all, I think all of the tutors are awesome!!! Thank you so much for offfering your services. My question on this is I need to write a system of 2 equations in 2 unknowns and then solve it. How would I set this up?This question is from textbook University of Phoenix Elementary and Intermediate Algebra Answer by elima(1433)   (Show Source): You can put this solution on YOUR website!At Gwen's garage sale, all books were one price, and all magazines were another price. Harriet bought four books and three magazines for \$1.45, and June bought two books and five magazines for \$1.25. What was the price of a book and what was the price of a magazine? Harriet bought = 4B, 3M June bought = 2B, 5M Now lets set them as equations, we know the total amount each spent, so we just make what they bought equal to those; Harriet; 4B+3M=1.45 June; 2B+5M=1.25 ============= Now we will go ahead and solve using the elimination method; 4B+3M=1.45 2B+5M=1.25 ============== I will multiply the second equation by -2 so I can eliminate the B; 4B+3M=1.45 -4B-10M=-2.5 -------------- Now add equations together; -7M=-1.05 divide each side by -7; M=.15 Now plug that answer into either equation for M; 4B+3(.15)=1.45 4B+.45=1.45 4B=1.00 B=.25 ============ so our Books = .25 Magazines = .15 :)