John M. answered • 03/16/18

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- Let b = the number of books he bought for £80
- Let c = the cost of each book
- Then, b*c = £80 {Eqn 1}
- And (b+4) * (c-£1) = £80 {Eqn 2}
- Two equations, two unknowns. So you can find a unique solution. And you can do this several ways. I'll use substitution. For convenience, I'll drop the £ sign.
- Rewrite Eqn 1 as b = 80/c
- Then substitute into Eqn 2: (80/c + 4) * (c- 1) = 80
- Now multiply out using FOIL method: (80/c * c) + (80/c * -1) + (4c) + (4 * -1) = 80
- Rewrite: 80 - 80/c + 4c - 4 = 80
- Rearrange: -80/c + 4c = 4
- Divide both sides of equation by 4: -20/c + c = 1
- Rearrange: -20/c + c - 1 = 0 {Eqn 3}
- Get a common denominator of c: (-20 + c
^{2}- c) / c = 0 {Eqn 4} - Notice in Eqn 4 that the numerator must be equal to zero in order for the equation to hold. So, we can ignore the c in the denominator
- Rewrite: c
^{2}-c-20 = 0 {Eqn 5} - Now factor Eqn 5: (c-5)(c+4) = 0. The solution is c = 5 or c = -4. Since c is a cost, it cannot be negative, so the only solution is 5
- The cost of each book originally is £5
- Substituting back into Eqn 1,
**the number of books bought by the shopkeeper is 16** - Check your work
- Originally, he bought 16 books @ £5/each = £80
- If he bought 4 more, he would have bought 20. Each book would cost £4 (ie, one £ less). Again, he would have paid 20* £4 = £80, which is the same amount.
- One point about the wording of the problem. The problem should have made it clear that all books cost the same amount.