Not quite seeing what they're doing here

G

Guest

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Code: 
Here is the problem, and the steps to get to the answer. I am just not quite following it. I would be greatly appreciative to anyone who would like to provide a little narrative as to what I'm not seeing.

ax     cx 
--- + ---- = e
 b      d


goes to...

adx    bcx 
---- + ---- = e       <-- seems they add d/d to the first fraction
 bd     bd                    and b/b to the second, but I am not 
                                 sure what rule gives them that right. 
                                 I understand "why" (to make the 
                                 denominators the same) but not "how".


then on to...


adx + bcx
---------- = e     <--- this makes sense because of the 
    bd                        common denominator


but finally...


        ebd
x = ---------        <--- I am just not following this last 
    b(ed -c)                 step at all. I just don't see what 
                                they did. 


Im teaching myself here, in preparation for the SAT 
so I am without benefit of teacher, so many thanks 
to anyone who is willing to give me a bit of narrative to this one.
 
To go from the first step to the second, they didn't "add" anything. They converted the fractions to a common denominator by multiplying by b/b and d/d, as appropriate. So start your studies by reviewing fractions; they occur often, and, in such symbolic contexts, calculators will be of no help, so you'll need to know the methods.

To go to the last step, they multiplied through by bd, factored the x out to get x(ad + bc), and then divided through by the parenthetical. So review solving linear equations and solving literal equations.

Hope that helps a little.

Eliz.
 
Code: 
Okay, Im starting to get it. I see why you can basically add b/b or d/d to those because they amount to "1", or 1x(everthing else) which is just (everything else), so its a nice trick to get common denominators. 

I also follow the step of breaking out

adx + bcx  to x(ad+bc) using the distributive property of multiplication, then moving bd over to the right side to get "ebd" I understand, because you can multiply each side by bd. Fine.

I am still not quite seeing how they get from

x(ad+bc) = ebd

to

x = ebd
     -------
     b(ed - c)

since what goes through my mind is, okay to get (ad+bc) to the right side, divide both sides by (ad+bc), giving us

x = ebd
     ------
    (ad+bc)

Not sure how to get to the "answer" from here though...
[/code]
 
You're not adding; you're multiplying.

If you took (ax)/b and added d/d, you would have (ax)/b + d/d. You'd have to multiply to get (adx)/(bd).

(Plus, adding 1 doesn't leave the term unchanged, whereas multiplying it by 1 does.)

Eliz.
 
Poor choice of words. I didn't mean mathematically "add" I meant "stick in there", I understand you're multiplying. Either way, I understand all of that, just not the last step. Any direction there?
 
Oh... I hadn't noticed that last bit... :shock:

There's a very good reason for not understanding how they got that: it's garbage! Your answer looks much more sensible.

Sorry about the confusion! :oops:

Eliz.
 
backToCollege said:
since what goes through my mind is, okay to get (ad+bc) to the right side, divide both sides by (ad+bc), giving us
x = ebd
------
(ad+bc)
Not sure how to get to the "answer" from here though...
Click to expand...

You are correct. And it stops there: it IS the answer.
 
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