Clearing fractions and setting up for the quadratic formula?

[Amaz1ng]

How do I clear fractions in problems like this and set it up for the quadratic formula:

$$\frac{1}{x-1}+\frac{2}{3}=\frac{2}{x-3}$$

I would like to set this up to be solved with the quadratic formula. Also, is it true that the formula can solve all quadratic equations?

 

[micromass]

Hi Amaz1ng

How do I clear fractions in problems like this and set it up for the quadratic formula:

$$\frac{1}{x-1}+\frac{2}{3}=\frac{2}{x-3}$$

I would like to set this up to be solved with the quadratic formula.

Multiply both sides of the equation with the denumerators of the fractions. This will clear the fractions. Also, before you do anything, you'll have to note that you have to assume that x is not 1 or 3.

Also, is it true that the formula can solve all quadratic equations?

Yes!

 

[Amaz1ng]

whats a denumerator? :shy:

 

[micromass]

The thing below your fraction. Thus in $$\frac{a}{b}$$, a is the numerator and b is the denumerator. So what you need to do is multiply your equation with x-1 and x-3.

 

[Amaz1ng]

btw like this?

$$\frac{1(x-3)}{(x-1)(x-3)}+\frac{2(x-3)}{3(x-3)}=\frac{2*3(x-1)}{3(x-1)(x-3)}$$

 

[micromass]

No, not like that. First multiply both sides of the equation by x-1, what do you get?

 

[micromass]

It's not wrong what you wrote, but it just won't help us forward...

 

[Amaz1ng]

$$\frac{1(x-1)}{(x-1)(x-1)}+\frac{2(x-1)}{3(x-1)}=\frac{2(x-1)}{(x-3)(x-1)}$$

 

[Amaz1ng]

$$\frac{1(x-1)}{(x-1)}+\frac{2(x-1)}{3}=\frac{2(x-1)}{(x-3)}$$

 

[micromass]

But, why do you also divide by x-1? That won't help you forward, you should just multiply.

Maybe you'll grasp it better with some example:

Solve $$\frac{1}{x}+2=9$$
We multiply both sides by x, and we get $$1+2x=9x$$. This gives us x=1/7.

 

[micromass]

$$\frac{1(x-1)}{(x-1)}+\frac{2(x-1)}{3}=\frac{2(x-1)}{(x-3)}$$

YES! very good!
Now, multiply both sides by x-3...

 

[Amaz1ng]

$$\frac{1(x-1)(x-3)}{(x-1)}+\frac{2(x-1)(x-3)}{3}=\frac{2(x-1)(x-3)}{(x-3)}$$


lol :!)

 

[micromass]

Very good! Doesn't that give you your quadratic formula?

 

[Amaz1ng]

Very good! Doesn't that give you your quadratic formula?

I don't see how. lol

 

[Amaz1ng]

I think I see now, but could you show me what exactly you're talking about.

$$x^2-3x-x+3$$

 

[micromass]

Well, if you have both (x-1) in the numerator and the denumerator, you can scratch them. You'll end up with an equation without fractions. Factoring gives you the desired equation!

 

[Averki]

How do I clear fractions in problems like this and set it up for the quadratic formula:

$$\frac{1}{x-1}+\frac{2}{3}=\frac{2}{x-3}$$

I would like to set this up to be solved with the quadratic formula. Also, is it true that the formula can solve all quadratic equations?

This simplest way to clear fractions would be to immediately look at the denominators of all the fractions and find the least common denominator(LCD). In this equation, the LCD is simply all three of the denominators multiplied together, giving you $$3(x-1)(x-3)$$. Now when you multiply all three fractions in the equation by the LCD, you will immediately clear the fractions. From there you just need to do the algebra and use the quadratic formula to find the answers.

 

[Amaz1ng]

Ok but is there anything to cancel out in the middle fraction or..? I'm asking because it doesn't look like it to me. I have a feeling something needs to be done to it but I can't see what. I'm talking about this:

$$3(x-3)+\frac{6(x-1)(x-3)}{3}=6(x-1)$$//edit
or...cancel out the 3 in the bottom and in the 6...yeah I think that's it.

 

[Averki]

Ok but is there anything to cancel out in the middle fraction or..? I'm asking because it doesn't look like it to me. I have a feeling something needs to be done to it but I can't see what. I'm talking about this:

$$3(x-3)+\frac{6(x-1)(x-3)}{3}=6(x-1)$$


//edit
or...cancel out the 3 in the bottom and in the 6...yeah I think that's it.

Exactly. When you multiply the middle fraction of $$\frac{2}{3}$$ by $$3(x-1)(x-3)$$ the three in both expressions cancel each other, making the expression $$2(x-1)(x-3)$$.

 

[SteamKing]

When you learn what's going on, you'll know that a fraction has a numerator above the bar and a denominator below the bar. Denumerator? I think not.

 

[Amaz1ng]

When you learn what's going on, you'll know that a fraction has a numerator above the bar and a denominator below the bar. Denumerator? I think not.

I think that's how they say it in London or something. ;o