lesson 31 it’s rationally simple - the ogburn...

STUDENT MANUAL ALGEBRA II / LESSON 31

Lesson 31 It’s Rationally Simple

Today we’re going to work on simplifying rational expressions. We worked a bit previously with rational expressions in exponents, but now we’re going to look at simplifying rational expressions with monomials and polynomials. Basically, a rational number is any number written as a fraction that has integers in the numerator and a nonzero integer in the denominator of the expression. An integer can be zero or any other positive or negative whole number. For example, 5/13 and -4/5 are both rational numbers because there are integers in the numerator and nonzero integers (positive or negative whole numbers) in the denominator. Rational expressions often have polynomials in their numerator and denominators. The following would be considered rational expressions:

y + 8/ 2 14/5ab x – 3/ x2 c - a/ 80

For rational expressions, the variables do not need to be set to zero. When we work with these rational expressions, we will assume that there is no variable in the denominator that would result in a solution equal to zero.

For example in 13/7xy, we will not assume that x or y will be equal to zero; otherwise the entire value in the denominator would equal zero.

So, let’s say we want to simplify a rational number. We can simplify any fraction by dividing the greatest common factor between the two integers:

xy/ zy

We know that neither z nor y are not equal to zero.

We can see that the GCF between the numerator and denominator is y, so we can easily take out the y on both parts of the fraction, which is the cancelation rule:

x • y / z • y


The y can cancel out easily because any number divided by itself will equal 1, and any product of a number and 1 will equal the original number.

For example: y/y = 1

x • 1 = x

and z • 1= z

So the simplified rational expression becomes: x/z

We can easily do this with numbers in fractional form as well.

36/48

The GCF of these two numbers is 12

If we factor 12 out of each number, we get:

3 • 12/ 4 • 12

Let’s get rid of 12 on each side because they cancel out to equal 1:

3 • 12/ 4 • 12

Finally, we are left with: ¾

Ah, all this stuff should be old news to you by now! Hopefully you’ve had a lot of practice with such processes in the past. Now let’s expand a bit on these basic concepts:


We can use this factoring method with rational expressions that have monomials and polynomials with a mixture of coefficients, variables, and exponent as well. We’ve already divided polynomials by monomials and by using the long division method, but now we’re going to divide two polynomials by finding the GCF of each term: For example:

2x – 6/ x2 - x - 6

First we must factor each expression:

2(x – 3)/ (x + 2) (x – 3)

Now, we can see that a common factor from both the numerator and denominator is (x – 3), so we can divide both the top and the bottom by (x – 3) and use the cancellation rule to get rid of them:

2 (x – 3) / (x + 2) (x – 3)

Finally, we are left with the simplified expression:

2/(x +2)

Now, it’s very tempting when you initially see that rational expression to just cancel out the 6’s or even to cancel out the 2’s once you have simplified the expression, but be sure to factor all the numbers out first before you start eliminating terms.

Remember, a fraction is considered simplified when there are not more terms that are common. When we look at 2/(x +2) you might think, well 2 is common, but this is not true. Don’t forget that x + 2 represents its own value and number.

Let’s say for example that you really wanted to just get rid of that 2, then you would be simplifying the expression incorrectly. Well use x = 3 for the example:

If x = 3 for this rational expression: 2/(x +2)


Then, we would have 2/ (3 + 2) = 2/5

If we had take those 2’s out, then we would be left with 2/ x + 2

Which would equal 1/3

2/5 and 1/3 are two very different fractions. Hopefully now, you see the value in keeping such expressions as their very own terms.

Okay, now there is a simple rule that you must recall sometimes when simplifying expressions.

(y – x) and (x – y) are always opposites.

In order to make these terms be equal, according to the opposite rational rule, you have to pull out a negative 1 out of one of the terms.

So: (y – x) = -1(x – y)

Let’s do a check to see how this works. We’ll apply this rule to help us solve this rational expression of

(y – x) / (x – y) + 5

When y = 2 and x = 3

(2 – 3) /(3 – 2) +5


(2 – 3) = -1

(3 – 2) = 1

Now, do you see how these two opposite terms are not equal?

So let’s apply our rule:

(2 – 3) /-1(3 – 2) +5

or

-1/-1(1) + 5

Now we have:

-1/-5 and we can solve = 5

The two terms (2 – 3) and (3 – 2) cancelled out once we took the opposite term and multiplied it by – 1.

Let’s try this rule in action:

16 - y2 / y2 – 3y -4

First we can factor both polynomials:


(4 – y) (4 + y) / (y – 4) (y + 1)

So we see that (4 – y) and (y – 4) are opposite terms, so we must multiply the entire polynomial (both terms that were factored out) by -1:

- 1(4 – y) (4 + y) / (y – 4) (y + 1)

- 1(4 – y) = (y – 4)

So the equation becomes:

- 1(y – 4) (4 + y) / (y – 4) (y + 1)

After eliminating the GFC or common factor, we have:

-1(4 + y) / (y + 1)

The -1 makes the entire rational expression negative, so we get:

- 4 + y/ y + 1

And that is the simplified for of the original rational expression!


Lesson 31 It’s Rationally Simple

Name:_____________________________________________Date:________________

Now, on your own, simplify each rational expression. Show all your work.

1. y – 3/ (y + 1) (y – 3)

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2. (a + 5) a/ ( a – 5) a

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3. 32x/ 16 x2

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4. x2 y/ x y2 z

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5. y2 – 4/ y + 2

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6. x2 – x/ 2x

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7. x2 – 2x + 1/ x2 – 1

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8. – (3 – a / a – 3)

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9. 5x – 5/ (x – 1) 2

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10. x2 - 2 + 1/ (x + 1)

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11. 2 y2 + y – 3/ 4y + 6

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12. 5y – 10/8 – 2y - y2

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13. 6x2 – 4x/ 9x2 – 12x + 4

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14. 15p2 – 2p – 8/ 12p – 15p

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15. y – 1/ y3 – 1

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16. 4m2 + 4m + 1/ 2m2 – 5m – 3

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17. xy – 2y + 3x – 6/ xy + 3x + 4y + 12

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18. x4 – 16/ x4 + x2 – 12

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19. m2 – 2m + mn – 2n/ m3 – 8

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20. Challenge: (x – 2) b+3 y3a / y5a + 1(x – 2)b

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Lesson 32 Rational Productions Presents…Multiplying

Rational Expressions!

In the previous lesson we learned about simplifying rational expressions. Today we’re going to review how to multiply rational expressions with monomials and polynomials. Multiplying rational expression is generally pretty simple. All you have to do is find the product of both numerators and the product of both denominators, and simplify if possible.

Here is the general rule when w, x, y, z are all real numbers when y and z are not equal to zero:

x/y • w/z

Simply multiply the numerators x and w:

x • w = xw

and multiple y and z:

y • z = yz

There you have it:

x/y • w/z = xw/yz

Pretty simple, right?

Well, we follow the same rule for multiplying monomials and/or polynomials in rational expressions:

3x/ 4y • 5/2y

3x • 5 = 15x

4y • 2y = 8y2

so

3x/ 4y • 5/2y = 15x/ 8y2

Here’s the catch, sometimes, we can simplify factor the expression before we even multiply; this makes it easier when we go to multiply. Here’s an example that’s already been factored:

x/ (x – 1) • (x – 1)/ (x + 1)

Now, let’ simply multiply the numerator and denominators:


x • (x – 1) = x(x – 1) Wait! It may be very tempting here to distribute the x, but look at the denominator, it has the same term, so we’ll keep it in this simplified form for now.

(x – 1)• (x + 1) = (x – 1)(x + 1)

Now, we have:

x(x – 1)/ (x – 1)(x + 1)

and we can easily take out the common factor of (x – 1) from the numerator and denominator.

x (x – 1)/ (x – 1) (x + 1)

Finally we have

x/ (x – 1) • (x – 1)/ (x + 1) = x / x + 1

Now, that was simple enough as well, but what about when the monomials and polynomials in our rational expressions haven’t yet been factored? Well, we’re just going to have to do it ourselves!

For example, if we want to multiply:

4y3/ y2 – 5y • y – 5/12

We then have:

4y3• (y – 5) / (y2 – 5y) • 12

Let’s factor the y2 – 5y in the denominator so that we can pull out a common factor of y – 5 that is also present in the numerator, and then cancel it out:

4y3• (y – 5) / y (y – 5) • 12

Then we have 4y3 / y • 12

4 and 12 have a common factor of 4, so we can factor out 4 from each cancel each of those out:

4 (1) y3 / y • (4 • 3) = y3 / y • 3

and y3 and y have a common factor of y

y3 / y • 3 = y (y2)/ y • 3

So our final answer is: y2 / 3

4y3/ y2 – 5y • y – 5/12 = y2 / 3


Lesson 32 Rational Productions Presents…Multiplying

Rational Expressions! Name:___________________________________________Date:__________________

Now, on your own, multiply and simplify each rational expression. Show all your work.

1. 15/ 30 • 4/ 5 =

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2. x/2 • 1/x = ½

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3. 15y/ 5 • y/ 4 =

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4. y/ 2z • z2/ y =

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5. 3x / 7y • 4 / 5z =

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6. y – 4/ x • x / y + 4 =

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7. 4z2/ 3y • 9zy/ 16 z3=

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8. 3a/ 2x2 • 4x/ 6a3 = 1/ a2x

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9. a + 3/ 3b • 6ab2 / a2 – 9 =

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10. 5x/ 6y2 y • 8y/ 10x3 =

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11. 9xy2 / z • 2z2 / 27x2 y =

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12. (x – 4) / 3y • 9xy/ x2 – 16 =

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13. 5y/ x2 • x2 / 20y2 =

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14. y2 + 2y + 1 / y – 1 • 2y2 – 2y/ y2 – 1 =

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15. y2 – 5y + 6/ y + 5 • 3y + 15/ y2 – 4y + 4 =

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16. 3x/ 4y2 • 12y/ 5 x2 • 15x3 / 8y =

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17. x – 1/ x + 1 • 2x + 2/ 3x – 3 • 9/4 =

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18. 2x2 + 7x + 3/ 3 x2 + 7x – 6 • 3x2 + x – 2/ x2 + 2x + 1 =

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19. 2ab2 / a2 + 7a • a + 7/ 8b3 =

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20. y – 7/ 3y2 – 8y + 4 • 6y2 – 24/ 3y – 21 =

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Lesson 33 Factoring and Fractioning Fractions

In the previous lesson we learned about multiplying rational expressions. Today we’re going to review how to multiply and divide rational expressions with monomials and polynomials. Remember from way back when, when you want to divide rational expressions, you must first write out the reciprocal 9switching the numerator and denominator) of the divisor (the second fraction) and then multiply.

For example ¾ ÷ ½ = ¾ • 2/1 ( the reciprocal of 1/2 ).

Multiple 3 • 2 = 6 and 4 • 1 = 4

6/4 = 1 ½

Pretty simple, right?

Here is the general rule for dividing when w, x, y, z are all real numbers when y and z are not equal to zero:

x/y ÷ w/z

The reciprocal of w/z is z/w

x • z = xz

and multiple y and w:

y • w = yw

There you have it:

x/y ÷ w/z = xz/ yw

Now let’s look at an example of diving factored and polynomials and polynomials that need to be factored that are in rational form.

1/ y – 1 ÷ 4 =

1/ y – 1• 1/ 4 =

1/ 4(y – 1)

So 1/ y – 1 ÷ 4 =1/ 4(y – 1)

Ah, that’s not so bad!


There is no need to distribute the number to the rest of the term in the denominator because then you would be undoing any factoring that had already taken place. If that number expression were ever needed again, chances are, you’d want it to be factored as much as possible. Keep your answers in simplified terms.

Now, let’s look at one more example that we’ll need to factor ourselves:

x + 5/ x2 + 5x ÷ 2x2 – x – 1/ 5x =

x + 5/ x2 + 5x • 5x/ 2x2 – x – 1 =

x + 5/ x(x+ 5) • 5x/ (2x + 1) (x – 1)

We can use the cancellation rule, and cancel out the common factors of (x + 5) and x:

x + 5/ x (x + 5) • 5 x / (2x + 1) (x – 1) =

We are finally left with: 5/ (2x + 1) (x – 1)

So x + 5/ x2 + 5x ÷ 2x2 – x – 1/ 5x = 5/ (2x + 1) (x – 1)


Lesson 33 Factoring and Fractioning Fractions

Name:____________________________________________Date:_________________

Now, on your own, simplify, multiply and divide each rational expression below. Show all your work.

1. ¾ ÷ 4/3 =

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2. 2 – y / y + 3 ÷ 15y3 / y + 3 =

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3. 6x/ y3 ÷ 3x2 / 5y =

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4. 2x/y ÷6y2 =

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5. 16x2 / 45y ÷ 48x/ 75y2 =

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6. 3x3 / y ÷ 3x2 / 2y2 =

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7. x – y/ x + y ÷ 6 =

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8. 4m / 4m2 -2m – 8 ÷ 12m3 /9m2 – 16 =

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9. x2 – 3x – 18/ x2 + 6x + 9 ÷ 2x2 -11x – 6/ 2x3 + 6x2 =

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10. y2 – 5y/ 5 + y ÷ y2 – 25/ y + 1=

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11. x + 2/ x2 –4x + 4 ÷ 2x + 4/ x2 –2x =

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12. 10y3 + 13y2 – 3y / 3y - 15y2 ÷ (4y2 + 12y + 9) =

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13. (a/b • c/d) ÷ e/f =

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14. a/b • (c/d ÷ e/f) =

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15. (a/b ÷ c/d) ÷ e/f =

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16. (x2 + 1/ x2 – 1 • x + 1/ 5x) ÷ x – 1/ 10x2 =

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17. 7x2 / 5 – 3x ÷ ( 21/ 2x + 4 • 2x2 /3x – 5) =

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18. x2 – 1/ 3x • x2 /x + 1) ÷ x2 –2x + 1/ 6x =

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19. (y2 – 1/ y2 + 3y ÷ y + 1/ 5y2 ) • y2 + 6y + 9/ y -1 =

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20. (y2 + y – 2/ 6y2 – 11y + 3 ÷ y2 – 1/ 6y – 2) • 2y – 3/ 2y + 4 =

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Lesson 34 Adder Up Adder Up!

In the previous lesson we learned about multiplying and dividing rational expressions. Today we’re going to review how to add polynomials in rational form. Now, when we multiplied the rational expressions, we simply multiplied the numerators and denominators, but when adding rational expressions, things get a bit more complicated… Now if the denominators of the two fractions are already the same, then it’s simple. For example:

3/2y + 7/2y

All we have to do is add the two numerators, and the denominator remains the same:

3 + 7 = 10

3/2y + 7/2y = 10/2y and we can simplify this be pulling out the 2 from both the numerator and denominator and we get: 5/y

Simple enough, right?

Now, let’s say things get a bit trickier. Let’s try to add the following rational expressions:

x/ x2 – 9 + 3/ x2 – 9

Now, at first glance this is great because both the denominators are the same, but before we begin to add the numerators, we might want to factor out those polynomials in the denominator, just to make sure that we’ll be able to add and simplify correctly:

x/ (x + 3) (x – 3) + 3/ (x + 3) (x – 3)

Now, let’s add the numerators: x + 3

And our new expression becomes:

(x + 3) / (x + 3) (x – 3)

And applying the cancellation rule, we can cancel out x + 3

(x + 3) / (x + 3) (x – 3)

Finally, we are left with 1/ x – 3

So x/ x2 – 9 + 3/ x2 – 9 = 1/ x – 3

Okay, that wasn’t too bad either, right?


Well, what if the denominators are not the same? Then, we must find something called the LCD. Do you remember hearing about LCD? It’s the lowest common denominator of each fraction. Basically, you must look at each of the denominators in an expression, and determine the lowest factor that both terms have in common.

Let’s look at 3/5 + 2/ 6

If we want to add these two fractions together, then we must first find the lowest common denominator between the two denominators 5 and 6.

The LCD is 20, so we can change the denominators to 30, but whatever we have to multiply the denominators by to get the LCD, we must also multiply each numerator by the same number; this process is called finding the equivalent.

For example: 5 • x = 30

Well we know that 5 • 6 = 30

Since we had to multiply 5 by 6 to get the LCD of 30, we must also multiple the numerator 3 by 6.

3 • 6 = 18

Now, that rational expression becomes:

18/30

Before we can add the two fractions, we must do the same thing with the second fraction:

6 • x = 30

Well we know that 6 • 5 = 30

Since we had to multiply 6 by 5 to get the LCD of 30, we must also multiple the numerator 2 by 6.

2 • 6 = 12

So that fraction becomes: 12/30

Now we can add the two fractions:

18/30 + 12/30

Remember, when adding fractions, we don’t add the denominator; they remain the same. We only add the numerators:

18/30 + 12/30 = 30/30, simplified 30/30 = 1

So surprisingly enough, 3/5 + 2/ 6 = 1


Now we can use this same method of finding the LCD and the equivalent fractions to add polynomials in rational form:

Let’s say we want to add x/ 3y2 + 4/ 15x2 y

First we must find the LCD of the denominators of the two fractions:

Oftentimes, it is easiest to break down each term into individual factors and variables:

Factors of 3y2 = 3 • y • y

Factors of 15x2 y = 3 • 5 • x • x • y

Now we can see that the lowest common factor of the two numbers (a term that each term can be a factor of)

= 3 • 5 • x • x • y• y

or 15 x2 y2

Now we must find the equivalent fractions and add. So what do we need to multiply each denominator by to get 15x2 y2?

For the first fraction it’s (5x2):

x(5x2) / 3y2(5x2)

And for the second fraction it’s simply y

4(y)/ 15x2 y(y)

Now we can find the equivalent fractions:

x(5x2) / 3y2(5x2) = 5x3 / 15x2 y2

4(y)/ 15x2 y(y) = 4y/ 15x2y2

And finally we can add the two fractions together.

5x3 / 15x2 y2+ 4y/ 15x2y2 = 5x3 + 4y/ 15x2y2

So, x/ 3y2 + 4/ 15x2 y = 5x3 + 4y/ 15x2y2

These steps may seem a bit tedious, but it’s so important to go through each step carefully in order to find the correct answer.


Lesson 34 Adder Up Adder Up!

Name:_____________________________________________Date:________________

Now, on your own add each rational expression below. Remember, you may need to find the LCD and creating equivalent fractions. Simplify your answers. Show all your work.

1. 3/8 + ½ =

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2. 5/3y + 2/3y =

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3. m/ 3 + m/2 =

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4. 13/3x2 + 2/3x2 =

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5. 2/y + 7/5y =

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6. (1/ x – 1) + (1/ x + 1)=

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7. x/ x2 – 4 + 2/ x2 – 4 =

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8. 1/2x + 3/5x =

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9. ½ + x/x + 2 =

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10. (m/ m – 3) + (m + 5/ m + 2) =

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11. 3/ x2y + 5/ 2x2 y3 =

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12. 3n + n2 / 7n + 1 =

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13. 5y/ 2 + 5y/ 4y – 2=

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14. a/ a – 2 + 4a/ a2 – 4a + 4 =

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15. 2x / 2x – 1 + 5x/ 2x2 – 7x + 3 =

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16. (x2 + 3x/ x2 + 3x + 2) + (x/ x + 2) =

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For the following problems (#’s 17 and 18), evaluate the following expression for each value of y given (12/ y + 5) + (3y – 2/ x + 5).

17. y = 15

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18. y = -4

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For the following problems (#’s 19 and 20), evaluate the following expression for each value of x given (2/ 2 – x) + (4 + x2/ 4 - x2) + (2/ 2 + x).

19. x = 10

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20. x = 8

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Lesson 35 Denominators and Their Differences

In the previous lesson we learned about adding rational expressions. We were able to add the polynomials and other terms in rational form together by first finding the LCD or lowest common denominator, and then by determining the equivalent fractions. Today, we’re not just going to find the sum of the two rational expressions, but we’ll also look at solving for their differences.

Let’s first look at how to subtract rational expressions that already have the same denominator (this will save us a few steps…).

(4x2 / x2 + x – 6) – (2x2 + 4x/ x2 + x – 6)

So now we have:

4x2 – (2x2 + 4x)/ x2 + x – 6

We can combine like terms in the numerator and distribute the negative sign:

2x2 - 4x/ x2 + x – 6

And then we can simplify through factorization:

2x(x – 2) / x2 + x – 6

We’ll need to factor the denominator as well to make sure that the expression is fully simplified:

x2 + x – 6 = (x + 3) (x -2)

Then our problem becomes:

2x(x – 2) /(x + 3) (x -2)

Finally, we can use the cancellation rule and cancel out x – 2

2x(x -2)/(x + 3) (x -2)

Our solution becomes: 2x/ x + 3

So (4x2 / x2 + x – 6) – (2x2 + 4x/ x2 + x – 6) = 2x/ x + 3

Well, despite the denominators being the same, subtracting those rational expressions was still a lot of work. Just be careful when you’re subtracting such expressions, and be sure not to rush through them. Take your time, even when they seem tedious or overwhelming.


Now, let’s look at one final example where we must find the LCD and then equivalent fractions:

(y + 5/ x2 – 9) – (1/ x2 + 3x)

First, let’s factor the denominators to find the LCD:

(x + 5)/ (x + 3)(x – 3) – 1/ x(x + 3)

Now we can determine that the LCD is: x(x + 3)(x – 3)

Let’s find the equivalent fractions:

(x)(x + 5)/ (x)(x + 3)(x – 3) = x(x + 5)/ x(x + 3)(x – 3)

and

(x – 3) 1/ (x – 3) x(x + 3) = (x – 3) / x(x + 3)(x – 3)

And now we can subtract the numerators:

x(x + 5) – (x – 3)

Distribute the x and the negative sign, and find like terms to simplify:

x2 + 5x – x + 3= x2 + 4x + 3

Finally, let’s factor out the numerator to see if there are any similar terms to those in the denominator that we can cancel out:

x2 + 4x + 3 = (x + 3)(x + 1)

Now put it all together again:

(x + 3)(x + 1) / x(x + 3)(x – 3)

Cancel out the like terms, (x + 3)

(x – 3) (x + 1) / x(x + 3) (x – 3)

Finally, you’re left with:

x + 1 / x(x + 3)

It can be simplified no further, so…

(y + 5/ x2 – 9) – (1/ x2 + 3x) = x + 1 / x(x + 3)


Lesson 35 Denominators and Their Differences

Name:____________________________________________Date:_________________

Now, on your own subtract each rational expression below. Remember, you may need to find the LCD and creating equivalent fractions. Simplify your answers. Show all your work.

1. (x – 1/ 3) – (x -2/3) =

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2. (3y/ y + 1) – (2y – 1/ y + 1) =

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3. (x/ x - 3) – 3 =

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4. x/3 – 3/x =

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5. (2x/ x2 –2x – 24) – (14x – 22 / x2 – 2x – 24) =

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6. x/3y2 – y/12x2 =

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7. a/b – c/d =

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8. (2x/ x -1) – (x +3/ x) =

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9. (x/ x2 – x – 2) – (1/ x + 1) = 2/ (x – 2)(x + 1) =

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10. (2y/ y – 1) – (16y/ y2 + 6x – 7) =

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11. (5/ x2 – 4) – (3 – x/ 4 - x2) =

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12. (m/ m2 + m – 6) – (m/ m2 –9) =

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13. 1/a – 2/a2 –3/a2 =

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14. (y/ y – 1) + (1/y) – (1/ y2 – y) =

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15. 97x/ x2 – 25) – (2/ x -5) + (15/ 25 - x2) =

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16. (2y/ y + 1) + (4/ y -1) – (4/ y2 – 1) =

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17. (y/ y + 1) – (4/ y + 4) – (3/y2 +5y + 4) =

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18. (5x/ 2x2 –x -3) + (x/ x + 1) – (x/ 2x -3) =

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19. (2a/ a – b) + (b/ a + b) – (2ab/ a2 – b2) =

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20. (x – 2/ xy + x – 2y – 2) – ( x/ x -2) + y/ y + 1) =

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Lesson 36 Solving and Resolving those

Fractional Equations

In the previous lessons we learned about adding, subtracting, multiplying and dividing rational expressions. We were able to add and subtract the polynomials and other terms in rational form together by first finding the LCD or lowest common denominator, and then by determining the equivalent fractions. Today, we’re going to look at how to solve for a variable in a rational expression. A fractional equation is an equation that has one or more rational expressions. Let’s look at how to solve for y in a fractional equation:

(2/ y +1) – (3/ 2y) = (1/ 10y)

The first thing we need to do is determine the LCD of all three fractions. Here are the three denominators that we must find the LCD for:

(y +1), (2y), (10y)

We can determine that the LCD for all three denominators is: 10y(y + 1)

Remember, we want to always keep everything as simplified as possible, so we’ll keep the LCD in this simplified form.

Now, we must multiply each numerator by the LCD. Don’t worry about finding equivalent fractions when finding solving.

10y(y + 1) 2/ y +1 – 10y(y + 1) 3/ 2y = 10y(y + 1) 1/ 10y

Now, we can cancel out any portion of the LCD that is provided already in that fraction’s denominator and divide any like terms to simplify and rid ourselves of all the denominators:

10y(y + 1) 2/ (y + 1) – (5y)10y(y + 1) 3/ 2y = 10y(y + 1) 1/ 10y

(10y/2y = 5

Now, we are left with:

10y(2) – 5(y + 1) = (y + 1) (1)

Let’s distribute:

20y – 15y – 15 = y + 1

Now, we combine like terms:


5y – 15 = y + 1

Add 15 to both sides, and subtract y from the right side:

5y – 15 = y + 1

-y +15 -y +15

4y = 16

Divide both sides by 4:

4y/4 = 16/4

y = 4

Now, we have solved for y:

For the equation: (2/ y +1) – (3/ 2y) = (1/ 10y)

y = 4

Sometimes you may find that there are two solutions for a given variable. Remember when we set equations to zero, factored, and had 2 solutions? Well this is sort of like the same thing. We may also find that we need to factor out one or more of the denominators in order to find the LCD. Let’s look at the following equation that we want to solve for x:

(x – 3/ 3x) = (1/ x + 3) + (1/ 3x2 + 9x)

Let’s find the LCD first. We have these three denominators:

(3x), (x + 3), (3x2 + 9x)

We must factor the last denominator in order to find the LCD:

(3x2 + 9x) = 3x(x + 3)

Now we can easily see that the LCD of all three denominators is: 3x(x + 3)

Let’s multiply each fraction be the LCD to simplify and rid the fractions of the denominators:

3x(x + 3) (x – 3/ 3x) = 3x(x + 3)(1/ x + 3) + 3x(x + 3) (1/ 3x2 + 9x)

Now, we can use the cancellation rule to cancel out the denominators:

3x (x + 3) (x – 3/ 3x) = 3x(x + 3)(1/ x + 3) + 3x(x + 3) (1/ 3x(x + 3))

We are now left with:

(x +3) (x – 3) = 3x + 1


Now, we want to use the FOIL method to find the values of the left side of the equation, so that we can easily set it to zero:

x2 - 9 = 3x + 1

Subtract 3x and 1 from the right side, so that the equation is set to zero:

x2 – 3x – 10 = 0

Now, that the equation is set to zero we can factor and solve for the given variables of x:

x2 – 3x – 10 = (x + 2)( x – 5)

(x + 2)(x – 5) = 0

Set each individual term to zero:

x + 2= 0

-2 -2

x = -2

x - 5= 0

+5 +5

x = 5

So we have 2 solutions for x (the opposites of each number):

x = -2, 5

After this, you must perform a check to see whether or not both values could be real solutions. Enter in each value separately into the original equation to see if it is true.


Lesson 36 Solving and Resolving those

Fractional Equations

Name:___________________________________________Date:___________________

Find the LCD that you would use to multiply each fraction by. Do not solve—find only the LCD.

1. (1/ x + 3) = (3) – (2/ x -1) =

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2. (3/y) = (y/6) + 1 – y/ 3y(y+2) =

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3. (1/x2 – 10) = (3/ x + 1) =

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4. (1/m) – (1/ m2 + m) = (1/ m + 1) =

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5. (4/ y2 – y – 6) = (1/y – 3) =

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Now, on your own solve for each variable in the following equations. Be sure to check any or all solutions to see if they are true for the given equation. Remember, just because you find two solutions, that does not mean they will both work. Show all your work.

6. (y + 1/ y – 3) = (3/5)

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7. (x – 1/ 5) + 2 = (2x – 3/ 3)

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8. (1/ y + 1) + (1/ y) = (y + 2/ 2x)

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9. (1/x) + (1/ x2) = (3/4),

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10. (x – 1/ 2x + 1) = (5x + 3/ 6x2 + 3x)

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11. (5 – y/ 2y) + (4/ y – 5) = (2y – 1/ y)

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12. (3/ x + 1) + (15/ x2 –1) = (4/ x – 1)

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13. (5/ y2 – 9) = (3/ y + 3) – (2/ y -3)

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14. (x/ 9) – (4/ 2x + 3) = (1)

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15. (2x/ 4x – 1) + ( x/ 4x + 1) = (1/ 16x2 –1) + (1/2)

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16. (3/ 2m) – (6/ 2m + m2) = (1/ m + 2)

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17. (2/ y + 2) – (6/ y2 + 5y + 6) = 5/ (y + 2) 2

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18. (1/ x-5) – (2x/ x2 –8x + 15) = (x/ x – 3)

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19. (2/ x + 4) = (1/ x – 2) – (2x-1/ x2 + 2x – 8)

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20. (3/ y – 2) – (4/ y – 3) + (6/ y2 –5y + 6)

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Lesson 37 “Look, You’re Very Own Math Book!”

Writing Assignment

Scenario: You’re Math teacher has found a contest online for students to enter in their very own instructions for a portion of a new Math Textbook series that is going to be released. Your teacher gives every student in the class the opportunity to submit instructions that could possibly be published in the textbook. The winner of the contest will receive a cash prize, and their submission will be published in the textbook that all the students at your school will be using.

Task: Now it’s your task to create your very own step-by-step instructions for simplifying and solving the following two rational expressions below. As you write your directions, think about how you would like instructions to be written, and think about the ways that would help someone else easily understand your instructions. Be sure to give clear instructions for each step in your simplification and solutions. After each step, explain your process in written form as you justify the logic for each step in your solution.

Here are the two rational expressions that you must give step-by-step instructions for simplifying and solving:

Purpose: The purpose of this assignment is for you to: understand solving rational equations as a process of reasoning and explain the reasoning; explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution; and construct viable instructions to justify steps in a solution method. It’s important for you to understand how clear and concise instructions help to reinforce the message being relayed.

Writer’s Role: A high school math student writing for a contest.

Audience: The Math Textbook Publishing Company and judges of the contest.

Focus Correction Areas:

1. Clear, neat, concise, and logical written instructions.


2. Clear explanation of each step in simplifying and solving simple rational equations as following from the equality of numbers asserted at the previous step.

3. Construct viable instructions to justify steps in a solution method while using correct term usage and verbal justification


Lesson 37 “Look, You’re Very Own Math Book!”

Writing Assignment

Name: Date:

Use the space below to write your description and justification for your work and solve the equation.

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Lesson 38 Cumulative Review for Lessons 31-37

Name:___________________________________________Date:___________________

Please complete the following cumulative review. Ask you instructor if you have any questions. You may want to review some of the instructional sections of previous lessons to help you remember certain processes as you complete the work below. These will be the same concepts covered in the upcoming assessment.

Simplify each rational expression. Show all your work.

1. x2 – 2x + 1/ x2 – 1

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2. – (3 – a / a – 3)

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3. 5x – 5/ (x – 1) 2

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4. x2 - 2 + 1/ (x + 1)

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5. 2 y2 + y – 3/ 4y + 6

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Multiply and simplify each rational expression. Show all your work.

6. 3a/ 2x2 • 4x/ 6a3 = 1/ a2x

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7. a + 3/ 3b • 6ab2 / a2 – 9 =

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8. 5x/ 6y2 y • 8y/ 10x3 =

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9. 9xy2 / z • 2z2 / 27x2 y =

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10. (x – 4) / 3y • 9xy/ x2 – 16 =

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Simplify, multiply and divide each rational expression below. Show all your work.

11. 4m / 4m2 -2m – 8 ÷ 12m3 /9m2 – 16 =

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12. x2 – 3x – 18/ x2 + 6x + 9 ÷ 2x2 -11x – 6/ 2x3 + 6x2 =

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13. y2 – 5y/ 5 + y ÷ y2 – 25/ y + 1=

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14. x + 2/ x2 –4x + 4 ÷ 2x + 4/ x2 –2x =

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15. 10y3 + 13y2 – 3y / 3y - 15y2 ÷ (4y2 + 12y + 9) =

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Add each rational expression below. Simplify your answers. Show all your work.

16. 3/ x2y + 5/ 2x2 y3 =

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17. 3n + n2 / 7n + 1 =

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18. 5y/ 2 + 5y/ 4y – 2=

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For the following problems , evaluate the following expression for each value of y given (12/ y + 5) + (3y – 2/ x + 5).

19. y = 15

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20. y = -4

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Subtract each rational expression below. Simplify your answers. Show all your work.

21. (2x/ x -1) – (x +3/ x) =

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22. (x/ x2 – x – 2) – (1/ x + 1) = 2/ (x – 2)(x + 1) =

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23. (2y/ y – 1) – (16y/ y2 + 6x – 7) =

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24. (5/ x2 – 4) – (3 – x/ 4 - x2) =

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25. (m/ m2 + m – 6) – (m/ m2 –9) =

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Solve for each variable in the following equations. Be sure to check any or all solutions to see if they are true for the given equation. Remember, just because you find two solutions, that does not mean they will both work. Show all your work.

26. (1/ y + 1) + (1/ y) = (y + 2/ 2x)

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27. (1/x) + (1/ x2) = (3/4),

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28. (x – 1/ 2x + 1) = (5x + 3/ 6x2 + 3x)

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29. (5 – y/ 2y) + (4/ y – 5) = (2y – 1/ y)

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30. (3/ x + 1) + (15/ x2 –1) = (4/ x – 1)

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Lesson 39 Learning through Leading:

Presentation Project for Lessons 31-37

Today you will develop another instructional presentation for a lesson that you will present. For your lesson, you may either use some type of presentation software or poster-boards or some other type of visual aid. You must use some form of visual aid with text and examples of expressions to support your instruction.

In your lesson, you must cover the key terms below, and show an example for how each term might relate to simplifying, adding, subtracting, multiplying dividing, and or solving rational expressions that include polynomials. You do not need to show examples of methods for doing each of those examples individually, just show examples as necessary to cover the following terms:

Terms: Integers, rational numbers, cancellation rule, GCF, polynomials, rational expressions, opposite rational rule, LCD, equivalent fraction, reciprocal, divisor, GCF, product, polynomials, fractional equation.

Your work will be graded based on the following rubric:

PRESENTATION RUBRIC

3 2 1

Covers all terms listed in the directions

Covers most terms listed in the directions

Covers few if no terms listed in the directions

3 2 1

Gives excellent examples for all or most terms.

Gives fair examples for most terms.

Gives poor examples for most terms.

3 2 1

Creative, effective, and supportive visual aid

Somewhat creative, effective, and supportive

visual aid

Very little creativity, and non-supportive visual aid

5-6 3-4 1-2

Obvious effort in project Decent effort in project Little effort in project and/or


and/or presentation overall and/or presentation overall presentation overall

Total______/15 Possible Points


Lesson 39 Learning through Leading:

Presentation Project for Lessons 31-37

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Lesson 40 Cumulative Assessment for Lessons 31-37

Name:____________________________________________Date:__________________

Please complete the following cumulative assessment. Ask you instructor if you have any questions during the test. Be sure to show all of your work.

Simplify each rational expression. Show all your work.

1. 6x2 – 4x/ 9x2 – 12x + 4

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2. 15p2 – 2p – 8/ 12p – 15p

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3. y – 1/ y3 – 1

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4. 4m2 + 4m + 1/ 2m2 – 5m – 3

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5. xy – 2y + 3x – 6/ xy + 3x + 4y + 12

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Multiply and simplify each rational expression. Show all your work.

6. 5y/ x2 • x2 / 20y2 =

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7. y2 + 2y + 1 / y – 1 • 2y2 – 2y/ y2 – 1 =

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8. y2 – 5y + 6/ y + 5 • 3y + 15/ y2 – 4y + 4 =

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9. 3x/ 4y2 • 12y/ 5 x2 • 15x3 / 8y =

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10. x – 1/ x + 1 • 2x + 2/ 3x – 3 • 9/4 =

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Simplify, multiply and divide each rational expression below. Show all your work.

11. (a/b ÷ c/d) ÷ e/f =

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12. (x2 + 1/ x2 – 1 • x + 1/ 5x) ÷ x – 1/ 10x2 =

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13. 7x2 / 5 – 3x ÷ ( 21/ 2x + 4 • 2x2 /3x – 5) =

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14. x2 – 1/ 3x • x2 /x + 1) ÷ x2 –2x + 1/ 6x =

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15. (y2 – 1/ y2 + 3y ÷ y + 1/ 5y2 ) • y2 + 6y + 9/ y -1 =

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Add each rational expression below. Simplify your answers. Show all your work.

16. a/ a – 2 + 4a/ a2 – 4a + 4 =

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17. 2x / 2x – 1 + 5x/ 2x2 – 7x + 3 =

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18. (x2 + 3x/ x2 + 3x + 2) + (x/ x + 2) =

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For the following problems, evaluate the following expression for each value of x given (2/ 2 – x) + (4 + x2/ 4 - x2) + (2/ 2 + x).

19. x = 10

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20. x = 8

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Subtract each rational expression below. Simplify your answers. Show all your work.

21. 1/a – 2/a2 –3/a2 =

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22. (y/ y – 1) + (1/y) – (1/ y2 – y) =

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23. 97x/ x2 – 25) – (2/ x -5) + (15/ 25 - x2) =

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24. (2y/ y + 1) + (4/ y -1) – (4/ y2 – 1) =

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25. (y/ y + 1) – (4/ y + 4) – (3/y2 +5y + 4) =

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Solve for each variable in the following equations. Be sure to check any or all solutions to see if they are true for the given equation. Remember, just because you find two solutions, that does not mean they will both work. Show all your work.

26. (5/ y2 – 9) = (3/ y + 3) – (2/ y -3)

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27. (x/ 9) – (4/ 2x + 3) = (1)

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28. (2x/ 4x – 1) + (x/ 4x + 1) = (1/ 16x2 –1) + (1/2)

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29. (3/ 2m) – (6/ 2m + m2) = (1/ m + 2)

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30. (2/ y + 2) – (6/ y2 + 5y + 6) = 5/ (y + 2) 2

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lesson 31 it’s rationally simple - the ogburn...

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