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Gina, 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
I have to use substitution or elimnation
i 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
Or at least tell me if i started it off right with those equations ^^^^
Let 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
a basically used a b and c but ok
Rearrange the first equation for x: x = 11 - y - z
After 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?
wow thank you so much
ahah no problem. my back hurts now
awe im sorry :(
aha no worries.
ok and that is what some good ole ibuprofen is for
huh, yeah sure. :P