# Thread: Simplifying Fractions - Am I right?

1. ## Simplifying Fractions - Am I right?

Instruction: Write a single fraction with one numerator and one denominator.

Problem#1:

1/x + 1/y
I think the answer should be x+y/xy, yes?

Problem #2:

(5a^-1 + 2b^-1) / (3a^-2 - 2b^-2)
I think the answer should be (3a-b) / 5, yes?

Problem #3:
[9-(1/y^2)] / [9+ (6/y) + (1/y^2)]
I think the answer should be 1 / (6/y), yes?

Problem #4 says to factor completely:
18y^3 + 48y^2 + 42y
I think the final answer should be 6y(3y^2 + 8y + 7), yes?

Thank you!

2. Originally Posted by tesla1410
Instruction: Write a single fraction with one numerator and one denominator.

Problem#1:

1/x + 1/y
I think the answer should be 1/x+y, yes?
$a/b + c/d =[ (ad)+(bc)]/ bd$

3. Originally Posted by tesla1410
Instruction: Write a single fraction with one numerator and one denominator.
(5a^-1 + 2b^-1) / (3a^-2 - 2b^-2)
I think the answer should be (3a-b) / 5, yes?
$\frac{5a^{-1}+2b^{-1}}{3a^{-2}-2b^{-2}} = \frac{\frac{5}{a}+\frac{2}{b}}{\frac{3}{a^2}-\frac{2}{b^2}}$

$= \frac{\frac{5}{a}+\frac{2}{b}}{\frac{3}{a^2}-\frac{2}{b^2}} \cdot \frac{a^2b^2}{a^2b^2} = \frac{ \left ( \frac{5}{a}+\frac{2}{b} \right ) a^2b^2}{ \left ( \frac{3}{a^2}-\frac{2}{b^2} \right ) a^2b^2}$

$= \frac{5ab^2 + 2a^2b}{3b^2 - 2a^2}$

$= \frac{ab(5b + 2a)}{3b^2 - 2a^2}$.

I didn't look at 3, since it's the same method as above, but 1 and 4 look good now.

-Dan

4. Originally Posted by tesla1410
Problem #4 says to factor completely:
18y^3 + 48y^2 + 42y
I think the final answer should be 6y(3y^2 + 8y + 7), yes?
this looks good
i dont have time now for the others ...maybe some one else can help
dan

5. Originally Posted by tesla1410
Problem #3:
[9-(1/y^2)] / [9+ (6/y) + (1/y^2)]
I think the answer should be 1 / (6/y), yes?
$
\frac{9-\frac{1}{y^2}}{9+\frac{6}{y}+\frac{1}{y^2}}$

Multiply demoninator and numerator by $y^2$,
$\frac{9y^2-1}{9y^2+6y+1}=\frac{(3y-1)(3y+1)}{(3y+1)(3y+1)}$
Thus,
$\frac{3y-1}{3y+1}$

6. ## Simplifying Fraction - Am I Right? Part 2

You guys are great for helping.

Let me see if I'm getting the hang of it now.

Is this right?

[(x+1)/x] / [(x-3)/4]

To get rid of the fractions, I'll multiply the denominator and numerator by 4x.

The result is 4x + 4 / x^2 - 3x, which can be reduced to

4(x+1)/x(x-3)

Am I right this time?

7. Originally Posted by tesla1410
Am I right this time?
Yes

8. Originally Posted by tesla1410
You guys are great for helping.

Let me see if I'm getting the hang of it now.

Is this right?

[(x+1)/x] / [(x-3)/4]

To get rid of the fractions, I'll multiply the denominator and numerator by 4x.

The result is 4x + 4 / x^2 - 3x, which can be reduced to

4(x+1)/x(x-3)

Am I right this time?

nice work...i think you got it

9. ## Simplify Fractions - Am I Right? Part 3

I'm on a roll. I am fairly certain I have the first 2 problems correct, but the last 2 are trickier.

1. [5-(1/a)] / (a+b)

Final answer is (5a-1) / [a(a+b)]

2. [(2x +1) / (x^2 - 25)] / [(4x^2 -1) / (x-5)]

Final answer is (2x + 1) / [(4x^2 -1)(x+5)]

3. {[1/(x+h)] - (1/x)} / h

For this one, I multiplied the denominator & numerator by x(x+h) and eventually ended with a final answer of [x-(x+h)] / [xh(x + h)], which I don't think can be further simplified.

4. (x-y) / [(the square root of x) + (the square root of y)]

I don't think this can be simplified any further. (?)

Thank you for helping me - I am learning a lot.

10. Originally Posted by tesla1410
4. $\frac{x-y}{\sqrt x+\sqrt y}$

I don't think this can be simplified any further. (?)
There is a rule in math that says "A fraction is not completely simplified until you rationalize the denominator"

11. Originally Posted by tesla1410

3. {[1/(x+h)] - (1/x)} / h

For this one, I multiplied the denominator & numerator by x(x+h) and eventually ended with a final answer of [x-(x+h)] / [xh(x + h)], which I don't think can be further simplified.
The first two did not dim my eye. Perhaps they are correct.
This one did.
It can be simplified.
$[x-(x+h)]=h$

(x-y) / [(the square root of x) + (the square root of y)]
It can.

$x-y=(\sqrt{x})^2-(\sqrt{y})^2$
Difference of two squares.
$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}+\sqrt{y}}=$ $\sqrt{x}-\sqrt{y}$

12. ## Simplifying Fractions - Part 3

Originally Posted by Quick
There is a rule in math that says "A fraction is not completely simplified until you rationalize the denominator"

OK... so let me give this a try.

The original problem is: (x-y) / (square root of x) + (square root of y)

If I rationalize the denominator, I'd be multiplying it and the numerator by the (square root of x) + (square root of y)......right?

So, I end up with:

(x-y)[(square root of x) + (square root of y)] / (x+y)

And this can't be simplified further.

Am I right??

13. Originally Posted by tesla1410
OK... so let me give this a try.

The original problem is: (x-y) / (square root of x) + (square root of y)

If I rationalize the denominator, I'd be multiplying it and the numerator by the (square root of x) + (square root of y)......right?

So, I end up with:

(x-y)[(square root of x) + (square root of y)] / (x+y)

And this can't be simplified further.

Am I right??
Hacker solved this, but I would like to pount out that your calculations are wrong anyway...

14. Originally Posted by tesla1410
OK... so let me give this a try.

The original problem is: (x-y) / (square root of x) + (square root of y)

If I rationalize the denominator, I'd be multiplying it and the numerator by the (square root of x) + (square root of y)......right?

So, I end up with:

(x-y)[(square root of x) + (square root of y)] / (x+y)

And this can't be simplified further.

Am I right??
If you choose to ignore my method and use rationalization instead:
$\frac{x-y}{\sqrt{x}+\sqrt{y}}\cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
The denominator is,
$(\sqrt{x}+\sqrt{y})\cdot (\sqrt{x}-\sqrt{y})=x-y$
Thus,
$\frac{(x-y)(\sqrt{x}+\sqrt{y})}{x-y}$