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2 years ago

Euguene sold half his books and then bought 17 more he now has 30

Mathematics

1 answer:

FromTheMoon [43]2 years ago

6 0

Answer:

E started with 26 books, sold 13 books, bought 17 more books, and ended with 30 books

Step-by-step explanation:

E has _ books: x

E Sells 1/2 of that _ books: x/2

E buys 17 books and now had 30: (x/2) + 17 = 30

30 - 17 = 13

13 x 2 = 26

x/2 = 26

13 + 17 = 30 <-- checking work

My guess is that x = 13

You didn't clarify what exactly you are looking for but I hope this helps clarify some steps of the problem anyway!

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Please solve it. It will help me for my exam preparation.

lisov135 [29]

The solution of the given expression is $[p^{2}(p-y)^{y-1} (p-y)^{y-2} -2p(p-y)^{y} (p-y)^{y-2} +7(p-y)^{y} (p-y)^{y-1}]/ (p-y)^{y} (p-y)^{y-1}(p-y)^{y-2}$

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We have to solve the above expression.

We know that expression is a combination of numbers, symbols, variables and coefficients, indeterminants. Expression shows relationship between variables. LCM is the smallest number that is divisible by both the numbers whose LCM has been taken. Coefficients are present in the beginning of variables. Symbols are arithmetic signs like addition, subtraction, multiplication and division, etc.

We have to take LCM of the expression and solve accordingly so that the expression can be solved =

$p^{2} /(p-y)^{y} -2p/(p-y)^{y-1} +7/(p-y)^{y-2}$ = $[p^{2}(p-y)^{y-1} (p-y)^{y-2} -2p(p-y)^{y} (p-y)^{y-2} +7(p-y)^{y} (p-y)^{y-1}]/ (p-y)^{y} (p-y)^{y-1}(p-y)^{y-2}$

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