Question

\WWCWEQC

Answer 1

the second one cane be answered quickly,
3/2a +5/2a = 8/2a
you can simplify to 4/a

Answer 2

give me a moment to type the method for the first

Answer 3

\[\frac{ x^2+8x+15 }{ x-4 }* \frac{ x^2-16 }{2x+6 }\]
the first step would be to factor the equation to something more manageable
\[\frac{ (x+3)(x+5) }{ (x-4) }*\frac{ (x-4)(x-4) }{ 2(x+3) }\]
\[\frac{ (x+3)(x+5)(x-4)(x-4) }{ 2(x+3)(x-4) }\] (ask if you are confused to what i did there)
then you cancel common terms
\[\frac{ (x+5)(x-4) }{ 2 }\]
you can leave the answer like that or foil it out.

Answer 4

\[\frac{ (x+5)(x-4) }{ 2 }\]

Answer 5

\[\frac{x^2 +8x +15}{x-4} * \frac{x^2-16}{2x + 6}\\ \frac{(x+3)(x +5)}{ ( x-4) }\ *\frac{( x-4) (x+4)}{2(x + 3)} \\\\\frac{( x+5) \cancel{(x + 3)}}{ \cancel{(x-4)} }* \frac{ \cancel{(x-4)}(x+4)}{2\cancel{(x + 3)}}\\\\\frac{(x+5)(x+4)}{2} \]

Answer 6

:)

Answer 7

whoops forgot about the addition. Thanks

Answer 8

once again you need to factor the functions
\[\frac{ x^2-1 }{ x^2-x-2 }-\frac{ x-1 }{ x-2 }\]
\[\frac{ (x-1)(x+1) }{ (x-1)(x-1) }- \frac{ (x-1) }{ (x-2) }\]
\[\frac{ (x+1) }{ (x-1) }-\frac{ (x-1) }{ (x-2) }\]
to finish the problem you need to make the denominators the same and subtract.
Does that help?

Answer 9

\[\frac{ (x+1)(x-2) }{ (x-1)(x-2) }-\frac{ (x-1)(x-1) }{ (x-1)(x-2) }\]
\[\frac{ (x+1)(x-2)-(x-1)(x-1) }{ (x-1)(x-2) }\]