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Spartan_Of_Ares

  • 3 years ago

can some one help me add these rational expressions?

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  1. math>philosophy
    • 3 years ago
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    Ok

  2. Spartan_Of_Ares
    • 3 years ago
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    |dw:1356272011895:dw|

  3. math>philosophy
    • 3 years ago
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    Find a common denominator

  4. Spartan_Of_Ares
    • 3 years ago
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    x^2+6?

  5. math>philosophy
    • 3 years ago
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    I think you need to either review the distributive property or learn FOIL method

  6. anonymous
    • 3 years ago
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    \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] use \(a=8,b=x+2,c=4,d=x+3\)

  7. anonymous
    • 3 years ago
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    use parentheses and make a direct substitution

  8. Spartan_Of_Ares
    • 3 years ago
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    wait so i cross multiply?

  9. anonymous
    • 3 years ago
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    there is no such thing as "cross multiply" to add fractions, you compute \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] that is how it is always done in this case \(a=8\) so replace \(a\) by \(8\) \(b=x+2\) so replace \(b\) by \(x+2\) \(c=4\) replace \(c\) by \(4\) and \(d=x+3\) replace \(d\) by \(x+3\)

  10. anonymous
    • 3 years ago
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    \[\frac{\overbrace{8}^a\times \overbrace{(x+3)}^d+\overbrace{(x+2)}^b\times \overbrace{4}^c}{\overbrace{(x+2)}^b\times \overbrace{(x+3)}^d}\]

  11. Spartan_Of_Ares
    • 3 years ago
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    do i combine the denominators?

  12. anonymous
    • 3 years ago
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    you have a choice, you can either leave it in factored form as \[(x+2)(x+3)\] or you can multiply out using the distributive law \[(x+2)(x+3)=x^2+2x+3x+6=x^2+5x+6\] your choice

  13. anonymous
    • 3 years ago
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    your real job is to use the distributive property in the numerator, and then combine like terms, because no one want to look at \[8(x+3)+4(x+2)\]

  14. Spartan_Of_Ares
    • 3 years ago
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    well to get a final answer i would need to use \[x ^{2}5x+6\] wouldnt I?

  15. Spartan_Of_Ares
    • 3 years ago
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    so 8x+24 + 4x+8 = 12x+32 ?

  16. anonymous
    • 3 years ago
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    your numerator is correct, i would be \(12x+3 2\)

  17. anonymous
    • 3 years ago
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    and your denominator is \(x^2+5x+6\)

  18. Spartan_Of_Ares
    • 3 years ago
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    would i have any restrictions?

  19. anonymous
    • 3 years ago
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    yes, the same restrictions you had to begin with the denominator cannot be zero, so you know \(x\neq -2\) and also \(x\neq -3\)

  20. Spartan_Of_Ares
    • 3 years ago
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    so it cant be the 2 and 3 from the original denominators?

  21. anonymous
    • 3 years ago
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    it cannot be \(-2\)

  22. anonymous
    • 3 years ago
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    one denominator is \(x+2\) so if \(x=-2\) then you get \(-2+2=0\) and the denominator cannot be 0

  23. Spartan_Of_Ares
    • 3 years ago
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    ok that makes sense thanks :)

  24. anonymous
    • 3 years ago
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    yw

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