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Mathematics
OpenStudy (anonymous):

A university bookstore recently sold a wirebound graph-paper notebook for $2.55, and a college-ruled notebook for $3.22. At the start of spring semester, a combination of 50 of these notebooks were sold for a total of $157.65. How many of each type were sold?

OpenStudy (anonymous):

Let x be the number of $2.55 notebooks and 50-x be the number of $3.22 notebooks. 2.55x+ 3.22(50-x) = 157.65 Solve the equation for x, and you will have the number of $2.55 notebooks. Subtracting that value from 50 will give you the number of other notebooks. Let me know if you need more help.

OpenStudy (anonymous):

yes i do need more help please

OpenStudy (anonymous):

Ok, are you ok with how I arrive at this equation? 2.55x+ 3.22(50-x) = 157.65

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok, so now we need to algebraically solve for x. 2.55x + 161 -3.22x = 157.65 -0.67x=-3.35 x=5 We remember that we let x equal the number of $2.55 notebooks sold. So 5 of them were sold. Since we sold 50 of both combined, we must have sold 45 $3.22 notebooks.

OpenStudy (anonymous):

Thank you very much abery. and you have a great night

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