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Question

Write an expression to show the price of 2 discounted books when the first one is 20 + n and each one after that is (20 + n)(n - 5).

Solution

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Using the FOIL Method which is given by (a+b)(c+d)=ac+ad+bc+bd,(a+b)(c+d)=ac+ad+bc+bd, the given expression, (20+n)(n5),(20+n)(n-5) , is equivalent to

20(n)+20(5)+n(n)+n(5)=20n100+n25n=n2+15n100.\begin{align*} & 20(n)+20(-5)+n(n)+n(-5) \\&= 20n-100+n^2-5n \\&= n^2+15n-100 .\end{align*}

Hence, the price of the second book is (n2+15n100) dollars.(n^2+15n-100)\text{ dollars}. Adding the price of the first book, (20+n)(20+n) dollars, then the total price of the two books is

(20+n)+(n2+15n100)=n2+(n+15n)+(100+20)=(n2+16n80) dollars.\begin{align*} & (20+n)+(n^2+15n-100) \\&= n^2+(n+15n)+(-100+20) \\&= (n^2+16n-80)\text{ dollars} .\end{align*}

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