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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Let'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,\[\huge 15x+3\color{cadetblue}{(65x)}=5\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2We 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.

nickersia
 2 years ago
Best ResponseYou've already chosen the best response.0See this http://openstudy.com/study#/updates/50d0ff5ce4b0091849d7a6f6

katie__michelle
 2 years ago
Best ResponseYou've already chosen the best response.0@nickersia Oh thanks.
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