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Barrelracing

  • one year ago

Which of the following expressions is the inverse of the function y= x+5/7 a) y= x+7/5 b) y= x-5/7 c) y=5x-7 d) y=7x-5

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  1. freckles
    • one year ago
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    solve for x

  2. freckles
    • one year ago
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    if the equation is y=x+5/7 to do that you just need to subtract 5/7 on both sides if the equation is y=(x+5)/7 I would first multiply 7 on both sides and if is this equation you have one more step remaining after that

  3. Barrelracing
    • one year ago
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    its the first one but im not sure how to divide the 5/7 to both sides..sorry can u help?

  4. freckles
    • one year ago
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    no subtraction is not division

  5. freckles
    • one year ago
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    Here is an example: \[y=x+a \\ \text{ Subtract } a \text{ on both sides } \\ y-a=x \\ \text{ or other way around } x=y-a \\ \text{ then you can interchange } x \text{ and } y \\ y=x-a \text{ this is the inverse of } y=x+a\]

  6. Barrelracing
    • one year ago
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    sorry read it wrong so it would be 5/7y = x?

  7. freckles
    • one year ago
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    well you didn't exactly subtract 5/7 on the left hand side it appears you multiply 5/7 on left hand side

  8. Barrelracing
    • one year ago
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    im so confused can u just do this one for me so I can use it as an example for my other problems? im not cheating it would be easier for me to understand. im sorry.

  9. freckles
    • one year ago
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    oh so you didn't understand the example above either?

  10. Barrelracing
    • one year ago
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    oh wait got it hold on

  11. Barrelracing
    • one year ago
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    y-5/7=x?

  12. freckles
    • one year ago
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    \[\text{ pretend } a=\frac{2}{3} \\ y=x-\frac{2}{3} \text{ is the inverse of } y=x+\frac{2}{3} \\ \text{ if work needs \to be shown we can go through the process again } \\ \text{ pretend we have } \\ y=x+\frac{2}{3} \\ \text{ subtract } \frac{2}{3} \text{ on both sides } \\ y-\frac{2}{3}=x \\ x=y-\frac{2}{3} \\ \text{ interchange } x \text{ and } y \\ y=x-\frac{2}{3} \\ \text{ So we know } y=x-\frac{2}{3} \text{ is the inverse of } y=x+\frac{2}{3}\]

  13. freckles
    • one year ago
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    and yes your a=5/7

  14. freckles
    • one year ago
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    \[y=x+\frac{5}{7} \\ \text{ subtract} \frac{5}{7} \text{ on both sides } \\ y-\frac{5}{7}=x \\ x=y-\frac{5}{7}\] yep now just interchange x and y

  15. Barrelracing
    • one year ago
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    interchange?

  16. freckles
    • one year ago
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    replace x with y and y with x

  17. freckles
    • one year ago
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    pull the old switch a roo

  18. UsukiDoll
    • one year ago
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    y= x+5/7 switch the x and the y and then solve for y

  19. freckles
    • one year ago
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    \[y=x-\frac{5}{7}\]

  20. freckles
    • one year ago
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    you can solve for x then switch but either way

  21. UsukiDoll
    • one year ago
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    I was taught to switch first then solve for y :)

  22. Barrelracing
    • one year ago
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    y = x-5/7

  23. UsukiDoll
    • one year ago
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    but that works too... even though I got used to one technique @freckles

  24. freckles
    • one year ago
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    here is another example @Barrelracing of finding inverse of a linear function \[y=\frac{x+5}{7} \\ \text{ multiply both sides by } 7 \\ 7y=x+5 \\ \text{ then subtract} 5 \text{ on both sides } 7y-5=x \\ \text{ now interchange } x\text{ and } y \\ 7x-5=y \\ y=7x-5 \text{ is the inverse of } y=\frac{x+5}{7}\]

  25. freckles
    • one year ago
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    there is a reason I like to do the switch step last @UsukiDoll

  26. freckles
    • one year ago
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    but a linear example is not the best example to show why

  27. anonymous
    • one year ago
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    By the way @Barrelracing you look wonderful.

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