Rational Equation Problem

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Jason76

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When cross -multiplying, then this is fairly easy:

\(\displaystyle \dfrac{26}{x} = \dfrac{20}{x -3}\)

\(\displaystyle 20x = 26(x - 3)\)

\(\displaystyle 20x = 26x - 78\)

\(\displaystyle -6x = -78\)

\(\displaystyle x = 13\)

But what if we multiplied each term in the original problem by the \(\displaystyle LCD\) of \(\displaystyle x(x - 3)\)?

How would we go about that? If the \(\displaystyle LCD\) was only say \(\displaystyle (x - 3)\), then that would be easy. Even a situation with a \(\displaystyle LCD\) of something like \(\displaystyle (x + 6)(x - 3)\) would be easy of some canceling out scheme was going on.

Let's start out with the \(\displaystyle LCD\) method:

\(\displaystyle x(x -3)\dfrac{26}{x} = x(x - 3)\dfrac{20}{x -3}\)

What would be the next step?
 
Jason76 said:
But what if we multiplied each term in the original problem by the \(\displaystyle LCD\) of \(\displaystyle x(x - 3)\)?
How would we go about that? If the \(\displaystyle LCD\) was only say \(\displaystyle (x - 3)\), then that would be easy. Even a situation with a \(\displaystyle LCD\) of something like \(\displaystyle (x + 6)(x - 3)\) would be easy of some canceling out scheme was going on.
Let's start out with the \(\displaystyle LCD\) method:
\(\displaystyle x(x -3)\dfrac{26}{x} = x(x - 3)\dfrac{20}{x -3}\)
What would be the next step?
Click to expand...

There is no difference.

\(\displaystyle x(x -3)\dfrac{26}{x} = x(x - 3)\dfrac{20}{x -3}\)

\(\displaystyle (x -3)\dfrac{26}{1} = x\dfrac{20}{1}\)
 
Jason76 said:
When cross -multiplying, then this is fairly easy:

\(\displaystyle \dfrac{26}{x} = \dfrac{20}{x -3}\)

\(\displaystyle 20x = 26(x - 3)\)

\(\displaystyle 20x = 26x - 78\)

\(\displaystyle -6x = -78\)

\(\displaystyle x = 13\)

But what if we multiplied each term in the original problem by the \(\displaystyle LCD\) of \(\displaystyle x(x - 3)\)?

How would we go about that? If the \(\displaystyle LCD\) was only say \(\displaystyle (x - 3)\), then that would be easy. Even a situation with a \(\displaystyle LCD\) of something like \(\displaystyle (x + 6)(x - 3)\) would be easy of some canceling out scheme was going on.

Let's start out with the \(\displaystyle LCD\) method:

\(\displaystyle x(x -3)\dfrac{26}{x} = x(x - 3)\dfrac{20}{x -3}\)

What would be the next step?
Click to expand...
Cross-multiplying is just shorthand for multiplying by the most obvious common denominator. Some tutors here object to the whole idea of cross-multiplying because it seems to violate the fundamental rule that any change to one side of an equation must be matched by an exactly equivalent change to the other side of the equation.

\(\displaystyle \dfrac{a}{b} = \dfrac{c}{d} \implies bd * \dfrac{a}{b} = bd * \dfrac{c}{d} = \dfrac{b(ad)}{b} = \dfrac{d(bc)}{d} \implies ad = bc.\)

Cross-multiplying just skips the intermediate steps: \(\displaystyle \dfrac{a}{b} = \dfrac{c}{d} \implies ad = bc.\)

The above is a valid short-cut for multiplying by a common denominator.

The danger with cross-multiplying is that you may think \(\displaystyle ad = bc \implies \dfrac{a}{b} = \dfrac{c}{d}.\)

The latter implication is not valid.
 
I can see that with the "\(\displaystyle LCD\) method" that the answer is the same. In fact, lots of stuff cancels out, so solving is simple. But if you couldn't cancel things out, then you would have to multiply \(\displaystyle x(x + 3)\) and some number.
 
mmm4444bot said:
Are you able to post an example, to show what you have in mind? :)
Click to expand...

Using the "\(\displaystyle LCD\) method". You see lots of stuff also canceling out as with the "cross-multiply method".

\(\displaystyle \dfrac{26}{x} = \dfrac{20}{x - 3}\)

\(\displaystyle x(x -3)\dfrac{26}{x} = x(x -3)\dfrac{20}{x - 3}\)

\(\displaystyle (x - 3)26 = 20x\)

\(\displaystyle 26x - 78 = 20x\)

\(\displaystyle -78 = -6x\)

\(\displaystyle x = 13\)

I can't think of any time that such a situation as with "no canceling out" might occur. But if it did, then how would you multiply say \(\displaystyle x(x - 3)(4)\) ?
 
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Jason76 said:
I can see that with the "\(\displaystyle LCD\) method" that the answer is the same. In fact, lots of stuff cancels out, so solving is simple. But if you couldn't cancel things out, then you would have to multiply \(\displaystyle x(x + 3)\) and some number.
Click to expand...
The whole point of multiplying by a common denominator is so that things will cancel out. Please look at my previous post on WHY cross-multiplication works. It is simply shorthand for multiplying both sides of the equation by the product of the denominators. When you multiply by the product of the denominators, cancellation of denominators is guaranteed. Do you bother to read posts? Do you bother to think about what they mean?
 
Ok, i got the idea about multiplying the \(\displaystyle 4x\) thing. Also, cross-multiplying could be a fast shortcut for some, rather than using the \(\displaystyle LCD\) method.

Anyhow, both cross-multiplying and the \(\displaystyle LCD\) method cause "canceling out" which makes problem solving easier.
 
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There are situations where LCD saves time - e.g.

\(\displaystyle \dfrac{x^2}{4} \ + \ \dfrac{5x^3}{7} \ + \ \dfrac{x}{3} \ = \ \dfrac{x^4}{11}\)
 
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