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me123
 5 years ago
Could someone please show me how to solve the system by elimination method?
5x+3y=7 7x2y=13
me123
 5 years ago
Could someone please show me how to solve the system by elimination method? 5x+3y=7 7x2y=13

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Certainly. The elimination method allows you to form a new equation with only one of your variables by adding a multiple of one equation to another. First pick a variable you want to get rid of.

me123
 5 years ago
Best ResponseYou've already chosen the best response.0Could you show me how this is done step by step?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, pick a variable you'd like to eliminate (x or y).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so checking the coefficients on x in both equations, we can see that if we multiply the first equation by \(\frac{7}{5}\) We will have a 7 coefficient on our x term. So rewrite the first equation after multiplying it by \(\frac{7}{5}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If anything I say doesn't make sense, stop me and let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what do you get when you multiply the first equation by \(\frac{7}{5}\).

me123
 5 years ago
Best ResponseYou've already chosen the best response.0do you make the 55 a fraction then cross multiply

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, multiply both sides of the whole equation by 7/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{7}{5}(5x+3) = \frac{7}{5}(7)\] And simplify.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0While you're working it out I'm gonna grab some coffee.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You forgot to (a) change the signs (it was a 7) (b) put each of those over 5 (and simplify the 35) (c) keep your x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It should be: \[\frac{7}{5}(5x + 3y) = \frac{7}{5}(7)\] \[\frac{7}{5}(5x) + \frac{7}{5}(3y) = \frac{49}{5}\] \[7x  \frac{21}{5}y = \frac{49}{5}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, now add this equation to the other one. \(7x\  2y = 13\) + \(7x  (21/5)y = 49/5\) ============== For now don't bother to try to combine the fractions and nonfractions, just eliminate the x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you get for the new equation that only has y's?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. Just add the two equations from top to bottom: \(7x \  2y \) = 13 + \(7x  (21/5)y = 49/5\) =============== \(0x  2y  (21/5)y\) = 13 + 49/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now multiply that whole equation by 5 to get rid of the fraction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The equation on the bottom that only has y's that is.

me123
 5 years ago
Best ResponseYou've already chosen the best response.0so if i multiply all by 5 I get 10y 10=65+245

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. 5*(2y) = 10y 5*(21/5)y = 21y 5*13 = 65 (this part you did correctly) 5*49/5 = 49 So the equation would be 10y 21y = 65 + 49

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you combine your terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just leave it as a fraction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. Now plug that in for y into either one of your original equations and solve for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Err wait, no. It's 114/31

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to practice your distribution rules, and working with equations. I suggest http://www.khanacademy.org/video/solvingequationswiththedistributiveproperty?playlist=Algebra%20I%20Worked%20Examples
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