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vortish

subtract the polynomials (2x^4y^3+4x^3y^4)-(5x^4y^3-6x^3y^4)

  • one year ago
  • one year ago

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  1. asnaseer
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    which aspect is troubling you here?

    • one year ago
  2. vortish
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    all of it

    • one year ago
  3. asnaseer
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    ok, first step is to remove the braces. do you know how to remove the braces? especially the braces with the minus sign outside of it.

    • one year ago
  4. vortish
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    no

    • one year ago
  5. asnaseer
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    ok - the rule here is that any expression within braces remains "as-is" when the braces are removed unless the braces have a minus sign outside of them. In that case all the signs inside the braces change - i.e. all "-"'s change to "+" and all "+"'s change to "-"

    • one year ago
  6. asnaseer
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    e.g.: (1+4-5) + (6-3) - (2+4-1) = 1+4-5 + 6-3 -2-4+1

    • one year ago
  7. asnaseer
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    does that make sense so far?

    • one year ago
  8. vortish
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    yes

    • one year ago
  9. asnaseer
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    good, so now try to use these rules to remove the braces in your expression

    • one year ago
  10. vortish
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    \[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}\]

    • one year ago
  11. asnaseer
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    perfect! now you just need to combine "like" terms

    • one year ago
  12. asnaseer
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    do you know what "like" terms are?

    • one year ago
  13. vortish
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    7x^14y^14

    • one year ago
  14. asnaseer
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    no - that is not quite right

    • one year ago
  15. vortish
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    fudge

    • one year ago
  16. asnaseer
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    the way to think about this might be to imagine each unique combination of products of powers of x and y as a separate entity. e.g. in \(3x^3y^2+2x^2y^3-x^3y^2\) we have two unique combinations, namely: \(x^3y^2\) and \(x^2y^3\)

    • one year ago
  17. asnaseer
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    think of each of these combinations as a new "variable". so, in the example I gave, let \(A=x^3y^2\) and \(B=x^2y^3\). we can then write my expression as: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2B-A=2A+2B\) make sense?

    • one year ago
  18. vortish
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    no

    • one year ago
  19. asnaseer
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    is it one specific part here that is unclear or the whole thing?

    • one year ago
  20. vortish
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    the last explanation

    • one year ago
  21. asnaseer
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    ok - so you understand how I picked out the two unique combinations of powers of x and y in my initial expression - correct?

    • one year ago
  22. vortish
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    that is where i am having the issue

    • one year ago
  23. asnaseer
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    ok - lets take a look in more detail. we have the expression: \(3x^3y^2+2x^2y^3-x^3y^2\) do you agree that this contains terms that involve multiples of \(x^3y^2\) and \(x^2y^3\) only? i.e. there are no other "combinations" of powers of x and y within my expression.

    • one year ago
  24. vortish
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    yes

    • one year ago
  25. asnaseer
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    now, the next step I took was to replace all occurrences of \(x^3y^2\) by a new variable that I called \(A\). this leads to: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A\) make sense so far?

    • one year ago
  26. vortish
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    ok i think that make sense

    • one year ago
  27. asnaseer
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    good - the next step was to tackel the remaining unique combination which was \(x^2y^3\) - I replaced all occurrences of this with another new variable that I called \(B\). this leads to: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A=3A+2B-A\) ok so far?

    • one year ago
  28. vortish
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    yes

    • one year ago
  29. vortish
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    so i would come up with -3x^4y^3 +7x^3y^4

    • one year ago
  30. asnaseer
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    good - now we just simplify our new expression by combining all terms involving \(A\) and separately all terms involving \(B\). this leads to: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A=3A+2B-A\) = \(3A-A+2B\) = \(2A+2B\)

    • one year ago
  31. vortish
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    is t hat right

    • one year ago
  32. asnaseer
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    your expression after removing the braces was:\[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}\] so if you first replace \(x^4y^3\) by \(A\) and \(x^3y^4\) by \(B\) then you would get:\[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}=2A+4B-5A+6B\] what do you think that simplifies to in terms of A and B?

    • one year ago
  33. vortish
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    -3A+10B

    • one year ago
  34. asnaseer
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    correct - now just replace A and B by what they represent

    • one year ago
  35. vortish
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    ok that makes sense

    • one year ago
  36. asnaseer
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    the "trick" is to recognise these "unique" combinations of terms - that is what is meant by "like" terms in an expression.

    • one year ago
  37. vortish
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    got it right thanks asnaseer

    • one year ago
  38. asnaseer
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    it is similar to saying: 2*apples + 3*pears - 5*apples = -3*apples + 3*pears i.e. you cannot combine "apples" and "pears"

    • one year ago
  39. asnaseer
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    ok - glad you got it now

    • one year ago
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