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anonymous
 4 years ago
Ok i understood it finally in regular problems but they had to go and give me a word problem and im confused now i thought i started it right but when i keep working through it i get no answers that match the possible ones
anonymous
 4 years ago
Ok i understood it finally in regular problems but they had to go and give me a word problem and im confused now i thought i started it right but when i keep working through it i get no answers that match the possible ones

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Gina, Sam and Robby all rented movies from the same video store. They each rented some dramas, comedies and documentaries. Gina rented 11 movies total. Sam rented twice as many dramas, three times as many comedies, and twice as many documentaries as Gina. He rented 27 movies total. If Robby rented 19 movies total with the same number of dramas, twice as many comedies, and twice as many documentaries as Gina, how many movies of each type did Gina rent? 3 dramas, 5 comedies and 3 documentaries 2 dramas, 6 comedies and 3 documentaries 1 dramas, 4 comedies and 6 documentaries 4 dramas, 3 comedies and 4 documentaries

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to use substitution or elimnation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i started out with these equations from it : a+b+c=11 2a+3b+2c=27 a+2b+2c=19 but, im not sure if these are right or not

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or at least tell me if i started it off right with those equations ^^^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let x represent the number of dramas Let y represent the number of comedies Let z represent the number of documentaries Gina: x + y + z = 11 Sam: 2x + 3y + 2z = 27 Robby: x + 2y +2z = 19

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a basically used a b and c but ok

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Rearrange the first equation for x: x = 11  y  z

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After rearranging you can put this equation in (2) or (3) and eliminate for y or z. Substitute x = 11  y  z into (2) 2(11  y  z) + 3y + 2z = 27 22  2y  2z + 3y + 2z = 27 y = 27  22 y = 5 YAY that's the value of y :P Substitute y = 5 into (1) x + y + z = 11 x + 5 + z = 11 x + z = 6 x = 6  z Substitute x = 6  z and y = 5 into (3) x + 2y + 2z = 19 6  z + 2y + 2z = 19 2y + z = 13 z = 13  2y z = 13  2(5) z = 3 There's your "z" value Substitute y = 5 and z = 3 into (1) to solve for x x + y + z = 11 x + 5 + 3 = 11 x = 11  8 x = 3 Check if your values satisfy the equation: LS = 2x + 3y + 2z = 2(3) + 3(5) + 2(3) = 6 + 15 + 6 = 27 RS = 27 Since LS = RS, the values of x, y and z satisfy the equation. Therefore, x = 3, y = 5 and z = 3. WELL?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow thank you so much

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahah no problem. my back hurts now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok and that is what some good ole ibuprofen is for
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