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katie__michelle

  • 2 years ago

By using substitution: 15x+3y=5 5x+y=-6 Show me how to do this.

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  1. zepdrix
    • 2 years ago
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    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,\[\large y=\color{cadetblue}{-6-5x}\]We want to plug this in for the y in the other equation.\[\large 15x+3\color{cadetblue}{y}=5\] Which gives us,\[\huge 15x+3\color{cadetblue}{(-6-5x)}=5\]

  2. zepdrix
    • 2 years ago
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    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.

  3. katie__michelle
    • 2 years ago
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    Okayy I will,

  4. nickersia
    • 2 years ago
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    See this http://openstudy.com/study#/updates/50d0ff5ce4b0091849d7a6f6

  5. katie__michelle
    • 2 years ago
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    @nickersia Oh thanks.

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