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Destinyyyy

  • one year ago

If possible, solve the system of equations. Use any method. I will fan and metal

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  1. Destinyyyy
    • one year ago
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    4x+9y=10 (1) 3x+5y=4 (2)

  2. NotTim
    • one year ago
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    Use the substitution method. do you know how to do that?

  3. Destinyyyy
    • one year ago
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    Yes but the only examples I have are x with no number

  4. Destinyyyy
    • one year ago
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    I dont get how to do it with a number in front of x.

  5. whpalmer4
    • one year ago
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    \[4x+9y=10\]\[3x+5y=4\] For this you could do substitution, as the other poster suggested, or you could do elimination. I'll walk you through both :-)

  6. Destinyyyy
    • one year ago
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    Okay thank you

  7. whpalmer4
    • one year ago
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    Let's try substitution first. Pick one of the equations, and solve it for \(y\). Or \(x\), whatever you like :-)

  8. Destinyyyy
    • one year ago
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    3x+5y=4 y=-3/5x+4/5

  9. whpalmer4
    • one year ago
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    Okay. Now, take the other equation, and wherever you see \(y\), replace it with \(-\frac{3}{5}x+\frac{4}{5}\) What do you get?

  10. whpalmer4
    • one year ago
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    besides some messy algebra? :-)

  11. Destinyyyy
    • one year ago
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    x=-2

  12. Destinyyyy
    • one year ago
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    So (-2,2)

  13. whpalmer4
    • one year ago
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    Yes!

  14. whpalmer4
    • one year ago
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    Another way to do this is called elimination. With elimination, we add or subtract the equations together to get an equation with fewer variables. Sometimes we can do this right away, other times we need to multiply one or both of the equations by some factor to get the elimination to take place. \[4x+9y=10\]\[3x+5y=4\]To do elimination, we want the coefficients of the variables in one of the columns to be equal but opposite. Let's multiply the first equation by -3, and the second equation by 4. \[(-3)4x + (-3)9y = (-3)10\]\[(4)3x+(4)5y=(4)4\] or\[-12x-27y=-30\]\[\ \ \ 12x+20y=\ \ 16\] Agreed?

  15. Destinyyyy
    • one year ago
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    K. I know how to do graphing, elimination and substitution. I just didn't have any examples with it in the version thats all.. Thanks for your help.

  16. whpalmer4
    • one year ago
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    Ok. Yeah, you just solve for the substitution variable, doesn't matter if the coefficient is something other than the nice, easy 1 :-)

  17. Destinyyyy
    • one year ago
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    Okay. Thank you

  18. whpalmer4
    • one year ago
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    Always happy to hear someone say "yeah, I know how to do that already!" — usually it is "I don't know how to do that" :-)

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