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katie__michelle
Group Title
By using substitution:
15x+3y=5
5x+y=6
Show me how to do this.
 one year ago
 one year ago
katie__michelle Group Title
By using substitution: 15x+3y=5 5x+y=6 Show me how to do this.
 one year ago
 one year ago

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zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Let's manipulate the second equation, we'll get it in terms of y=, and then we'll PLUG it into the first equation. \[\large 5x+y=6\]Subtracting 5x from both sides gives us,\[\large y=\color{cadetblue}{65x}\]We want to plug this in for the y in the other equation.\[\large 15x+3\color{cadetblue}{y}=5\] Which gives us,\[\huge 15x+3\color{cadetblue}{(65x)}=5\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
We SUBSTITUTED one equation into the other one. From here it's not too bad. We just have to multiply the 3 by each term in the brackets, and then solve for x. If you're still confused, let me know.
 one year ago

katie__michelle Group TitleBest ResponseYou've already chosen the best response.0
Okayy I will,
 one year ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
See this http://openstudy.com/study#/updates/50d0ff5ce4b0091849d7a6f6
 one year ago

katie__michelle Group TitleBest ResponseYou've already chosen the best response.0
@nickersia Oh thanks.
 one year ago
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