anonymous
  • anonymous
a=[bx - c]= d
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
some details please
anonymous
  • anonymous
solve each question for x
anonymous
  • anonymous
i already did it but i'm not sure if it's correct

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freckles
  • freckles
do the brackets mean anything and you do know you have two equal signs right?
anonymous
  • anonymous
is this some kind of algebra or something more exotic,and as freckles said do the brackets have some special you know meaning
anonymous
  • anonymous
i think i wrote it wrong it's supposed to be a [bx -c] = d
freckles
  • freckles
welll okay I would divide both sides by a first isolating the thing in brackets
freckles
  • freckles
and put a little note on the side I suppose saying assuming a is not 0
freckles
  • freckles
bx-c=d/a then just undo the subtract by c by adding c on both sides
freckles
  • freckles
and then divide both sides by b making an assumption that b isn't 0
anonymous
  • anonymous
i did this a [bx - c] =d divide both sides by a then [bx -c] =da bx-c =da then i add c on both sides bx=da+c over b x=da + c over b
freckles
  • freckles
if you divided both sides by a then why does it show you divided a on one side and multiplied a on the opposing side?
freckles
  • freckles
\[a(bx-c)=d \\ \\ \text{ divide } a \text{ on both sides assuming } a \neq 0 \\bx-c=\frac{d}{a} \\ \text{ adding } c \text{ on both sides } bx=c+\frac{d}{a} \\ \text{ now dividing } b \text{ on both sides assuming } b \neq 0 \\ x=\frac{1}{b}(c+\frac{d}{a})\]
anonymous
  • anonymous
if i divide it, i have to put it as a fraction then?
freckles
  • freckles
things you divide by are things that go in the bottom like pretend we have m=3 and we wanted to divide both sides by 2 you can write that as: \[\frac{1}{2}m=\frac{1}{2}(3) \\ \text{ or } \frac{m}{2}=\frac{3}{2}\]
anonymous
  • anonymous
yes i know but i wasnt sure about letters
freckles
  • freckles
the letters represent numbers
anonymous
  • anonymous
words
freckles
  • freckles
they are to be treated just like numbers
anonymous
  • anonymous
ok thanks

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