anonymous
  • anonymous
Which values for x and y make the statement (x-5)(y + 6)=0 true? A. x=-5, y=6 B. x=5, y=-6 C. x=-5, y=-6 D.x=5, y=6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mayankdevnani
  • mayankdevnani
@chrisbush put all the values in option in given equation.If they satisfy the given equation, then that option is correct
mayankdevnani
  • mayankdevnani
can you get it what i am saying?
anonymous
  • anonymous
No i dont understand because i was never taught this,im now in a corse recovery program for summer and i came across this awesome website (:

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More answers

mayankdevnani
  • mayankdevnani
ok.. i will explain you
mayankdevnani
  • mayankdevnani
first take option A, where x=-5, y=6 then put these values in given equation The given equation is = \(\large (x-5)(y+6)=0\) so our new equation becomes, \[\large (-5-5)(6+6)=0\] now, solve for L.H.S .
mayankdevnani
  • mayankdevnani
@chrisbush
anonymous
  • anonymous
Okay (: now how do i solve the rest of the new equation?
mayankdevnani
  • mayankdevnani
lol.. just like as option A . Only you need to do is that take different options like A,B and C. And solve for L.H.S and if any option's L.H.S is zero then that option is correct answer.
mayankdevnani
  • mayankdevnani
ok.. @chrisbush i hope you understand
zzr0ck3r
  • zzr0ck3r
(-5-5)(6+6) = 0 is not true because (-5-5)(6+6) = -10*12 = -120 != 0 so this is not a solution, so not try the next
anonymous
  • anonymous
Lol im so confused D; i dont know the answerrr
mayankdevnani
  • mayankdevnani
lolzz.... @chrisbush try again and again as @zzr0ck3r said
zzr0ck3r
  • zzr0ck3r
notice if you put in x = 5 then the whole thing is 0 notice if you put in y = -6 then the whole things is 0 so pick the one/ones that have either x=5 or y=-6 or both
zzr0ck3r
  • zzr0ck3r
note: 0*anything = 0
anonymous
  • anonymous
so it would be b?
mayankdevnani
  • mayankdevnani
yes!!!
anonymous
  • anonymous
lol no but thx i understand now
mayankdevnani
  • mayankdevnani
welcome:)
zzr0ck3r
  • zzr0ck3r
I c I was reading it wrong, but you have it.:)
anonymous
  • anonymous
thank you guys, very much appreciated (:
mayankdevnani
  • mayankdevnani
welcome:)

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