Write (type-in) the written description of the step-by-step solution. We are going to call this the PEM. Make sure you type-in the PEM under the corresponding exercise.

Include a photo (or scan) of your solution. If easier, you can hand in the solution on the next class. It is a free choice to include or not your name with your post.

For the Proofreader (CLIP student): Check the text and input corrections of English grammar and composition. Make changes making sure that your new text (if any) is in a different color. Include your name at the end using the same color. Rate the text from one to five stars.

Remember

MAT096 Students

Immersion Program Students

Exercises are assigned to you.
Check list during class and in list page.

Proofread your assigned written
contributions from MAT96 students.

Tuesday, May 28 at noon at least
have written one of your assigned
exercises. Thursday, May 30 before class have
finished the written assigned exercises. Wednesday, June 5 all assignments
should be finished.

Thursday, May 30 by 8pm, have given
feedback to one of your assigned math
explanations, per the instructions above Sunday, June 2 have
given feedback to another math
explanation. Tuesday, June 4 all assignments
should be finished by 10pm.

A file with all the assignments of the project can be downloaded from the link below:

Next, you can find an example of a PEM where proofread corrections are the text colored purple.

Example

Calculate:

Solution (MIlena Cuellar):

This exercise asks us to find the result of operating positive and negative integers and fractions. Since the expression contains parenthesis, I will work on the parenthesis and then the rest of numbers.

Inside the parenthesis we have: five minus one forth. An integer and a fraction.

To be able to subtract these two numbers I convert the five into a fraction by looking at five as five over one. Now, we have two fractions, five over one minus one over four.

To be able to subtract these two fractions, I have to find the Least Common Denominator (LCD) between one and four. That is the smallest number that is divisible by one and four. The LCD is four.

Once we know the LCD, I have to write both fractions with that common denominator four. In other words, I multiply by the result of dividing my LCD by the denominator of each fraction.

For the fraction five over one, we say: (my LCD) four divided by one is four. Now, multiplying four times five (the numerator of the fraction) and four times one (the denominator of the fraction), we get the equivalent fraction 20 over four.

For the fraction one over four, we say: (my LCD) four divided by four is one, multiplying one my numerator and denominator we get the same fraction, one over four.

Now that we have the two fractions with the same denominator, we only have to do: 20 over four minus one over four is the same as 20 minus four over the LCD four. That is 19 over four.

Going back to our original expression. In there, we are left with one minus two times 19 over four. Following the order of operations, we do first the multiplication and then the additions or subtractions.

Two times 19 over four is the same as multiplying two over one by 19 over four. To multiply two fractions, we multiply numerator times numerator over denominator times denominator. In our case, two times 19 is 38 for the numerator over one times four is four for the denominator. Reducing 38 over four by dividing by two numerator and denominator, we get that 38 over four is the same as 19 over two.

After this, I am left with one minus 19 over two. Again, to add the two numbers, I need to convert one into a fraction by writing it over a denominator one. Now we have two fractions, one over one minus 19 over two.

The LCD of one and two is two therefore writing one over one is the same as two over two. Then we have to do the addition two over two minus 19 over two. Since the two fraction have the same denominator, we only operate the numerators and keep the same denominator. That is the same as two minus 19 over two or negative 17 over two.

## PEM is Plain English Math

## Instructions

For the Writer (MAT096 student):For the Proofreader (CLIP student):Check the text and input corrections of English grammar and composition. Make changes making sure that your new text (if any) is in a different color. Include your name at the end using the same color. Rate the text from one to five stars.Remember

MAT096 StudentsImmersion Program StudentsCheck list during class and in list page.

contributions from MAT96 students.

have written one of your assigned

exercises.

Thursday, May 30 before class have

finished the written assigned exercises.

Wednesday, June 5 all assignments

should be finished.

feedback to one of your assigned math

explanations, per the instructions above

Sunday, June 2 have

given feedback to another math

explanation.

Tuesday, June 4 all assignments

should be finished by 10pm.

For MAT096 students, you can find your assignments here:

Next, you can find an example of a PEM where proofread corrections are the text colored purple.

## Example

Calculate:

## Solution (MIlena Cuellar):

This exercise asks us to find the result of operating positive and negative integers and fractions. Since the expression contains parenthesis, I will work on the parenthesis and then the rest of numbers.

Inside the parenthesis we have: five minus one forth. An integer and a fraction.

To be able to subtract these two numbers I convert the five into a fraction by looking at five as five over one. Now, we have two fractions, five over one minus one over four.

To be able to subtract these two fractions, I have to find the Least Common Denominator (LCD) between one and four. That is the smallest number that is divisible by one and four. The LCD is four.

Once we know the LCD, I have to write both fractions with that common denominator four. In other words, I multiply by the result of dividing my LCD by the denominator of each fraction.

For the fraction five over one, we say: (my LCD) four divided by one is four. Now, multiplying four times five (the numerator of the fraction) and four times one (the denominator of the fraction), we get the equivalent fraction 20 over four.

For the fraction one over four, we say: (my LCD) four divided by four is one, multiplying one my numerator and denominator we get the same fraction, one over four.

Now that we have the two fractions with the same denominator, we only have to do: 20 over four minus one over four is the same as 20 minus four over the LCD four. That is 19 over four.

Going back to our original expression. In there, we are left with one minus two times 19 over four. Following the order of operations, we do first the multiplication and then the additions or subtractions.

Two times 19 over four is the same as multiplying two over one by 19 over four. To multiply two fractions, we multiply numerator times numerator over denominator times denominator. In our case, two times 19 is 38 for the numerator over one times four is four for the denominator. Reducing 38 over four by dividing by two numerator and denominator, we get that 38 over four is the same as 19 over two.

After this, I am left with one minus 19 over two. Again, to add the two numbers, I need to convert one into a fraction by writing it over a denominator one. Now we have two fractions, one over one minus 19 over two.

The LCD of one and two is two therefore writing one over one is the same as two over two. Then we have to do the addition two over two minus 19 over two. Since the two fraction have the same denominator, we only operate the numerators and keep the same denominator. That is the same as two minus 19 over two or negative 17 over two.

The final answer is negative 17 over two.

Below you can find the solution of the problem: