Question

Simplify the sum. d^2-9 d+20/d^2-3d-10 + d^2-2d-8/d^2+4d-32

Answer 1

Use the solution I showed you for the previous problem I posted. Use it and do the same. They key is to make denominators the same and factoring helps a lot.

Answer 2

So always to try to factor first and don't cancel anything out until after the addition. Look through the steps I typed up for the problem before to see how it works.

Answer 3

Thnx for answering all my questions genius!!! : ) If I could give you more medals I would

Answer 4

I say this because most people on here aren't always willing to put in the effort of typing so much stuff in for an addition or subtraction question like that through the Equation Box as it takes quite a bit of time and effort. So it's best if you learn from the solution before and attempt it yourself. Plus, trying it yourself is the best way to learn rather than having others do it for you.

@OSMiUM_MONkEY

Answer 5

These questions I'm asking are ones that I have already tried multiple times and could not figure out. Don't think I'm a cheap, because I love to earn my grades. Really though thank you for all your help.

Answer 6

\[{{d^2-9 d+2}\over{d^2-3d-10}} + {{d^2-2d-8}\over{d^2+4d-32}}\]factor out all the polynomials that can be factored:\[{{d^2-9 d+2}\over{(d-5)(d+2)}} + {{(d-4)(d+2)}\over{(d+8)(d-4)}}\]as you can see, \((d-4)\) can be cancelled out:\[{{d^2-9 d+2}\over{(d-5)(d+2)}} + {{\cancel{(d-4)}(d+2)}\over{(d+8)\cancel{(d-4)}}}\]\[{{d^2-9 d+2}\over{(d-5)(d+2)}} + {{(d+2)}\over{(d+8)}}\]now find a common denominator:\[{{(d^2-9 d+2)(d+8)}\over{(d-5)(d+2)(d+8)}} + {{(d+2)(d-5)(d+2)}\over{(d+8)(d-5)(d+2)}}\]put it under a single denominator once you've multiplied:\[{(d^3-d^2-70d+16)+(d^3-d^2-16d-20)}\over{d^3+5d^2-34d-80}\]add the numerator:\[{2d^3-2d^2-86d-4}\over{d^3-5d^2-34d-80}\]

Answer 7

@yummydum
How do you cross out something with that line in the equation box the way you did above?

Answer 8

\cancel{5x + 3}
\[\cancel{5x + 3}\]

Answer 9

^yup

Answer 10

It is one of these answers

Answer 11

1 sec

Answer 12

Let's try:\[\cancel{x+7}\]Yay it works :D. Ty.

@yummydym

Answer 13

Well since yummy made a slight error by typing over the problem incorrectly (+2 instead of the +20), here's the correct version:

\[\frac{d^2−9d+20}{d^2−3d−10}+\frac{d^2−2d−8}{d^2+4d−32}\]

factor it as much as possible:

\[\frac{(d-5)(d-4)}{(d-5)(d+2)}+\frac{(d-4)(d+2)}{(d+8)(d-4)}\]

cancelling out gives
\[\frac{\cancel{(d-5)}(d-4)}{\cancel{(d-5)}(d+2)}+\frac{\cancel{(d-4)}(d+2)}{(d+8)\cancel{(d-4)}} = \frac{d-4}{d+2} + \frac{d+2}{d+8}\]

finding a common denominator
\[\frac{(d-4)(d+8)}{(d+2)(d+8)} + \frac{(d+2)(d+2)}{(d+2)(d+8)} \]

now we can add them together, and expand the brackets
\[(d-4)(d+8)+(d+2)(d+2) \over (d+2)(d+8)\]
\[d^2 + 4d - 32 + d^2 + 4d + 4 \over (d+2)(d+8)\]
\[2d^2 + 8d - 28 \over (d+2)(d+8)\]

The doesn't match any of the images you posted, so you'd have to check the question since it should be there as an answer. You did post one of them twice so it might be that you forgot posting one

Answer 14

Thanks so much, I have been working on this problem for n hour, you're the best.

Answer 15

oh wow, i was missing a zero in one of my numerators which is why my answer is so bizarre..
the correct answer would be \(2d^2 + 8d - 28 \over (d+2)(d+8)\)

Answer 16

Typo's happen :p

Answer 17

too often, bro :P