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brookester6

  • 2 years ago

Use substitution to solve the problem... x=9y x+4=5(y+4) I already now that you can plug x in to the second equation and then distribute but what next?

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  1. zepdrix
    • 2 years ago
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    \[\large \color{orangered}{x=9y}\]\[\large \color{orangered}{x}+4=5(y+4) \qquad \rightarrow \qquad \color{orangered}{9y}+4=5(y+4)\]Distributing gives us,\[\large 9y+4=5y+20\]So you've gotten this much done so far? Hmm ok. Subtract 5y from each side,\[9y-5y+4=\cancel{5y-5y}+20\]

  2. zepdrix
    • 2 years ago
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    \[\large 4y+4=20\]Subtract 4 from each side,\[4y+\cancel{4-4}=20-4\]

  3. zepdrix
    • 2 years ago
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    Understand the process of combining like terms? :D That's how the 9y and 5y combined.

  4. mathstudent55
    • 2 years ago
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    You already have the first equation solved for x. So plug in what x is equal to, 9y, into the second equation and solve for y.

  5. brookester6
    • 2 years ago
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    yea, now y = 4 and i plug it in to the top equation :) Thanks so much!!

  6. zepdrix
    • 2 years ago
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    yay good job c:

  7. brookester6
    • 2 years ago
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    :D

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