Three more - Combine into single equivalent fraction problems.

[gijas]

Combine into a Single Equivalent Fraction.

1.) (2y - 1)/(y(y + 2) + (3/4)

LCD = y(y - 2)(2^2)

I think you multiply (2y - 1)/(y(y + 2) by (2^2)

and (3/4) by (y(y - 2) ?



2.) 3/(x(x + 1) + (4x + 7)/(x^2 + 3x + 2) + (x + 2)/(x^2 - 1)


LCD = (x - 1)x(x + 1)(x + 2)?

I get (3x^4 +14x^3 + 10x^2 - 12x - 6) / (x - 1)x(x + 1)(x + 2)?

But Im not sure which missing LCD to multiply each fraction with since there are three different fractions?



3.) (y^2 + 8y + 15)/(y + 1) Divided By (y + 3)/(y^2 + 6y + 5)

I thought you have to switch around the numerator and denominator in the second fraction since your dividing:

for example.. (y^2 + 6y + 5)/(y + 3)

LCD = (y + 1)(y + 3)

multiply (y^2 + 8y + 15)/ (y + 1) by (y + 3)
multiply (y^2 + 6y + 5)/(y + 3) by (y + 1)

 

[mmm4444bot]

1.) (2y - 1)/(y(y + 2) + (3/4)

I think you multiply (2y - 1)/(y(y + 2) by (2^2)

and (3/4) by (y(y - 2))

You've got the correct factors 4 and y(y - 2), but you need to multiply each ratio by 1; otherwise, you'll change the value of the ratio, and the result will not be equivalent.

Multiply the first ratio by 4/4.

Multiply the second ratio by (y(y - 2))/(y(y - 2)).

2.) 3/(x(x + 1)) + (4x + 7)/(x^2 + 3x + 2) + (x + 2)/(x^2 - 1)


LCD = (x - 1)x(x + 1)(x + 2)?

But Im not sure which missing [factor] to multiply each fraction with since there are three different fractions?

The number of ratios involved is not an issue. You need to multiply each ratio by 1.

Look at the denominator of the ratio, and determine which factor(s) from the LCD are missing. Use those to form 1.

We'll do the first one for you.

3/(x(x + 1))

We see that the denominator is missing the factors (x + 2) and (x - 1) from the LCD.

Hence, we multiply this ratio by ((x + 2)(x - 1))/((x + 2)(x - 1)), which is 1.

The resulting equivalent ratio will have the LCD.

After you get the common denominator (the LCD) on all three ratios, you can combine the new numerators.

3.) (y^2 + 8y + 15)/(y + 1) Divided By (y + 3)/(y^2 + 6y + 5)

I thought you have to switch around the numerator and denominator in the second fraction since your dividing:

for example.. (y^2 + 6y + 5)/(y + 3)

LCD = (y + 1)(y + 3)

The operation changes from division to multiplication, after changing to the reciprocal (i.e., "switching around").

a/b divided by c/d is the same as a/b times d/c

We do not need an LCD, when multiplying two ratios.

Factor the quadratic polynomials. There should be some cancellation.

 

[gijas]

For 1.) (2y - 1)/(y(y + 2) + (3/4)

I get = 2(7y + 1)/(y(y + 2)4


For 3.) (y^2 + 8y + 15)/(y + 1) Divided By (y + 3)/(y^2 + 6y + 5)

I get = 1/(y + 5)^2 or 1/(y + 5)


For 2.) 3/(x(x + 1) + (4x + 7)/(x^2 + 3x + 2) + (x + 2)/(x^2 - 1)

I get = (15x^2 + 15x + 9)/(x - 1) x(x + 1)(x + 1)(x + 2)

 

[mmm4444bot]

For 1.) (2y - 1)/(y(y + 2)) + (3/4)

I get = 2(7y + 1)/(y(y + 2)4)

This answer is not correct. The numerator cannot be linear.

Also, you can move the factor 4 to the front, in the denominator (4y(y + 2)).



For 3.) (y^2 + 8y + 15)/(y + 1) Divided By (y + 3)/(y^2 + 6y + 5)

I get = 1/(y + 5)^2 or 1/(y + 5)

There is only one answer, and it is neither 1/(y + 5)^2 nor 1/(y + 5).

(y + 5) is the factor that cancels!



For 2.) 3/(x(x + 1)) + (4x + 7)/(x^2 + 3x + 2) + (x + 2)/(x^2 - 1)

I get = (15x^2 + 15x + 9)/((x - 1) x(x + 1)(x + 1)(x + 2))

This answer is not correct. The numerator is wrong, and the denominator should be the correct LCD!

NOTE: It would make my life easier, if you were to be more careful with grouping symbols. I get confused with mismatched parentheses.

Let's make sure that you've set up the work properly, before we look for arithmetic mistakes.

1)

\(\displaystyle \frac{2y - 1}{y(y + 2)} \cdot \frac{4}{4} + \frac{3}{4} \cdot \frac{y(y + 2)}{y(y + 2)}\)




2)

\(\displaystyle \frac{3}{x(x + 1)} \cdot \frac{(x + 2)(x - 1)}{(x + 2)(x - 1)} + \frac{4x + 7}{(x + 1)(x + 2)} \cdot \frac{x(x - 1)}{x(x - 1)} + \frac{x + 2}{(x + 1)(x - 1)} \cdot \frac{x(x + 2)}{x(x + 2)}\)




3)

\(\displaystyle \frac{(y + 5)(y + 3)}{y + 1} \cdot \frac{y + 3}{(y + 1)(y + 5)}\)

 

[gijas]

NOTE: It would make my life easier, if you were to be more careful with grouping symbols. I get confused with mismatched parentheses.

Let's make sure that you've set up the work properly, before we look for arithmetic mistakes.

1)

\(\displaystyle \frac{2y - 1}{y(y + 2)} \cdot \frac{4}{4} + \frac{3}{4} \cdot \frac{y(y + 2)}{y(y + 2)}\)




2)

\(\displaystyle \frac{3}{x(x + 1)} \cdot \frac{(x + 2)(x - 1)}{(x + 2)(x - 1)} + \frac{4x + 7}{(x + 1)(x + 2)} \cdot \frac{x(x - 1)}{x(x - 1)} + \frac{x + 2}{(x + 1)(x - 1)} \cdot \frac{x(x + 2)}{x(x + 2)}\)




3)

\(\displaystyle \frac{(y + 5)(y + 3)}{y + 1} \cdot \frac{y + 3}{(y + 1)(y + 5)}\)

Yes, that is correct.

1. = (3y^2+14y-4)/(4y(y+2))

My mistake was to multiply 3 by (y(y+2) where I needed to multiply (y(y+2) = y^2 + 2y then multiply that by 3 which is 3y^2 + 6y

I didn't know I could move the 4 in front of the (y(y+2) in the denominator

2. = (5x^3+10x^2-6)/(x-1)x(x+1)(x+2)

I had a extra (x+1) for the LCD

3. = well since I can cancel out (y+5) from the fraction I can only assume I would be left with 1/(y+1)^2? Dont know..

 

[mmm4444bot]

1. (3y^2 + 14y - 4)/(4y(y+2)) ◄ This is correct.



My mistake was to multiply 3 by (y(y+2))

That should have worked, too.



I needed to multiply (y(y+2)) = y^2 + 2y then multiply that by 3 which is 3y^2 + 6y

The expression 3(y(y+2)) is three factors multiplied together: (3)(y)(y + 2).

You may multiply these factors in any order you like (look up "Commutative Property of Multiplication").

For examples:

Multiply 3 times y first, and you will have 3y times (y + 2)

or

Multiply 3 times (y + 2) first, and you will have y times (3y + 6)

or

Multiply y times (y + 2) first, and you will have 3 times (y^2 + 2y)


The final result will always be 3y^2 + 6y.



I didn't know I could move the 4 in front of the (y(y+2)) in the denominator

Again, it's the Commutative Property of Multiplication that allows us to rearrange the order of factors in a product.

Writing y(y+2)4 is okay, but writing 4y(y+2) is the standard way of expressing the same thing (putting the constant in front).

2. = (5x^3+10x^2-6)/((x-1)x(x+1)(x+2)) ◄ This is correct.


I had a extra (x+1) for the LCD

These things happen; glad you found the error.

Note the grouping symbols in red above. These are very important, when texting algebraic fractions. They indicate that the denominator is everything between them.

Without those grouping symbols around the denominator, your typing says that the denominator is only x-1.

3. well since I can cancel out (y+5) from the fraction I can only assume I would be left with 1/(y+1)^2

I'm not sure to what your phrase "the fraction" refers. See below.

We have a product of two fractions.

\(\displaystyle \frac{(y + 5)(y + 3)}{y + 1} \cdot \frac{y + 3}{(y + 1)(y + 5)}\)



The product of two fractions is: numerator times numerator OVER denominator times denominator, yes?



In symbols:

A/B * C/D = (A*C)/(B*D)



Hence, the product in problem 3 is:

\(\displaystyle \frac{(y + 5)(y + 3)(y + 3)}{(y + 1)(y + 1)(y + 5)}\)



We see the common factor y+5 appearing in both the numerator and denominator, so we may cancel it.

What's left?

 

[gijas]

We have a product of two fractions.

\(\displaystyle \frac{(y + 5)(y + 3)}{y + 1} \cdot \frac{y + 3}{(y + 1)(y + 5)}\)



The product of two fractions is: numerator times numerator OVER denominator times denominator, yes?



In symbols:

A/B * C/D = (A*C)/(B*D)



Hence, the product in problem 3 is:

\(\displaystyle \frac{(y + 5)(y + 3)(y + 3)}{(y + 1)(y + 1)(y + 5)}\)



We see the common factor y+5 appearing in both the numerator and denominator, so we may cancel it.

What's left?

(y + 3)^2
(y + 1)^2

 

[mmm4444bot]

(y + 3)^2
(y + 1)^2

▲ This is correct.

Instead of trying to "draw" fractions, we can type the final expression like this:

(y + 3)^2/(y + 1)^2