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 m

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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 m      Log On


   

Question 134245: 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?
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
Well to do this problem we first need to come up with variables for what we are trying to find. Since its the price of books and magazines why don't we use b for books and m for magazines.

Next we need to convert the information we have into equations using our variables. Lets start with Harriet bought 4 books and 3 magazines for $1.45. Well to get the final price that was given the cashier had to add the number of books times the price per book and the number of magazines times the price per magizine together. Since we have variables for the price of the books and magizines this can now be easily rewritten as 4b + 3m = 1.45. Doing the same thing with June we get 2b+5m = 1.25.

Now that we have two equations with two unknowns we can use the techniques for solving systems of linear equations to solve for b and m. The system we have is:

4b + 3m = 1.45
2b + 5m = 1.25

Now I don't know about you but I HATE decimals so lets just multiply both equations by 100 to get rid of them. This doesn't need to be done and you will get the right answer if you leave them but I find it easier to get rid of them.
Doing so will get us the following system:

400b + 300m = 145
200b + 500m = 125

Now since the b's in both equations look simular lets use the method of elimination(addition) to eliminate b from the equations therefore being able to solve for m. In order to do this we must multiply the second equation by -2.

-2(200b + 500m = 125) => -400b - 1000m = -250

Now we just add our first equation and our new second equation together.

400b + 300m = 145
-400b - 1000m = -250
---------------------
-700m = -105

Now we can go ahead and divide both sides of that new equation by -700(solving it for m).

m = 0.15

so we now know that magazines cost 15 cents per magazine. But how much do books cost? Well we can use that m to solve for b.

from our origonal system of equations we know that:

400b + 300m = 145

now all we need to do is plug in our m value and solve for b.

400b + 300m = 145
400b + 300(.15) = 145
400b + 45 = 145
400b = 100
b = .25

so we now know that books cost 25 cents a piece and that magazines cost 15 cents a piece. We now need to check our work. We do this by plugging in our m and b values into the origonal system and see if we get the same answers.

4b + 3m = 1.45
2b + 5m = 1.25

4(.25)+3(.15) = 1.45
1 + 0.45 = 1.45
1.45 = 1.45

good we got the first equation to work out but this problem isn't done right until both equations check out so lets check the second one.

2(.25) + 5(.15) = 1.25
0.50 + 0.75 = 1.25
1.25 = 1.25

we got both to check out so now we know that for sure that books are 25 cents and magazines are 15 cents.