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I have a problem in my mathematics (as math usually goes...) and it goes like this:

$Simplify.$

$(x^4-2x^2y-7)+(7x^4+6xy-3)$

So, I reordered the equation for simplicity's sake.

$x^4+7x^4-2x^2y+6xy-7-3$

Then, obvously, I simplified the equation and I concluded that the answer must be:

$8x^8+4x^3y^2-10$

But nope.. the answer is

$8x^8-2x^2y+6xy-10$

Why is this? I thought that you had to add the like terms, which in my mind, $2x^2y$ and $6xy$ where - clearly I was mistaken. Why dont you add the $x$ and $y$ for those two terms, but add the $x$ and $y$ variables in the other terms? This is bugging me as I can't clearly see/explain this.

Would someone mind explaining this to me quickly?
Thanks for your help!

Carlos Carlsen
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2 Answers2

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Hint:

Think at the properties of the operations that you are using.

E.g.:

by definition: $$ x^4+7x^4=1\cdot x^4+7\cdot x^4= $$ by distributivity of the product over the addition: $$ (1+7)\cdot x^4=8x^4 $$

and so one for the other terms... but, for

$$ -2x^2y+6xy $$

can you use some property ?

Emilio Novati
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In general, $x^2 \neq x$. If we multiply both sides by $y$ we get that $x^2 y \neq xy$. In general you can combine terms only when all variables are the same in each term and each variable is raised to the same power as all others of its type - only coefficients can vary. For example, take $5x^3y^2$. Right off the bat we know we can only combine this with another term $C x^3 y^2$, where $C$ is any constant value. When we do this we carry out the distributive property in reverse to get $(5+C)x^3y^2$. Note that the combined terms have the same variables and exponents on said variables as the answer we got... This is why it didn't work when you tried to combine unlike terms.

Brevan Ellefsen
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  • Thanks! Your answer was very clear, but I just need to understand one thing:

    If you have, $x^2y^2 + x^4y^6$, could you combine these two terms because they all have an exponent or must they be the same exponent in order to combine them?

     – Carlos Carlsen Jun 03 '16 at 15:59
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    @CarlosCarlsen sure thing! Hopefully this clarifies it a bit... Let me know if you still don't get it. In the example you give the terms cannot be combined, because the $x$ in the first term is raised to the power of $2$, but has an exponent of $4$ in the second term, and likewise for the $y$ variables. Each variable must have the same exponents as all other variables of its type, I. E. all $x$'s must have the same exponent, all $y$'s must have the same exponent, etc – Brevan Ellefsen Jun 04 '16 at 20:18
  • Thanks for the clarification - I can fully comprehend the addition of exponents now!! :D – Carlos Carlsen Jun 06 '16 at 13:50