- vortish

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

- jamiebookeater

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- asnaseer

which aspect is troubling you here?

- vortish

all of it

- asnaseer

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.

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## More answers

- vortish

no

- asnaseer

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 "-"

- asnaseer

e.g.:
(1+4-5) + (6-3) - (2+4-1) = 1+4-5 + 6-3 -2-4+1

- asnaseer

does that make sense so far?

- vortish

yes

- asnaseer

good, so now try to use these rules to remove the braces in your expression

- vortish

\[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}\]

- asnaseer

perfect!
now you just need to combine "like" terms

- asnaseer

do you know what "like" terms are?

- vortish

7x^14y^14

- asnaseer

no - that is not quite right

- vortish

fudge

- asnaseer

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 \(x^2y^3\)

- asnaseer

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?

- vortish

no

- asnaseer

is it one specific part here that is unclear or the whole thing?

- vortish

the last explanation

- asnaseer

ok - so you understand how I picked out the two unique combinations of powers of x and y in my initial expression - correct?

- vortish

that is where i am having the issue

- asnaseer

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.

- vortish

yes

- asnaseer

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?

- vortish

ok i think that make sense

- asnaseer

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?

- vortish

yes

- vortish

so i would come up with -3x^4y^3 +7x^3y^4

- asnaseer

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

- vortish

is t hat right

- asnaseer

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?

- vortish

-3A+10B

- asnaseer

correct - now just replace A and B by what they represent

- vortish

ok that makes sense

- asnaseer

the "trick" is to recognise these "unique" combinations of terms - that is what is meant by "like" terms in an expression.

- vortish

got it right thanks asnaseer

- asnaseer

it is similar to saying:
2*apples + 3*pears - 5*apples = -3*apples + 3*pears
i.e. you cannot combine "apples" and "pears"

- asnaseer

ok - glad you got it now

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