# Blog post week 15 solving addition and subtraction rational equations

This week I learned how to add and subtract rational expressions and how to solve for a variable value. The first step is to make the denominators the same and to do that you multiply it by the lowest common multiple, and then you multiply the top with the same denominators. In my example I show how I did that and then after that I made an equation and since it was a linear equation I had to move the variables to one side and then collect the like terms and after solve for my variable. So it was simple algebra, the only thing I found confusing was multiplying the denominators with the numerators because they had to be the same. Then slowly I understood how to do it and now have a good understanding of which denominator goes with each numerator.

# Blog post week 14

This week I learned how to multiply and divide rational expressions by factoring to get the restrictions for only the denominator.

My example is how to divide rational expressions. The first step is to factor, the second step is to state the restrictions, the third step is to flip over the fraction (reciprocal) and multiply from across to bottom. Then scratch out what is common and whatever is left that is your answer. I find this really easy because I love working with fractions and it is not that complicated. I find multiplication easier than dividing because then you can factor the expressions and then state your restrictions and the last step to multiply and scratch out the common expressions. There is no reciprocal required. And in my example there is a difference of squares so the factored form will be conjugates which is really easy. With this unit factoring is the key component because it tells the most important part the restrictions for x or any variable.

My division example that I used and simplified.

# Blog post week 13

This week I learned how to graph and solve reciprocal functions. So to solve the function you must factor it to get the roots, or also known as the x ints. So the graph has 2 solutions. And to know if a reciprocal function has no solutions that means that it doesn’t factor and has extraneous roots. And if you don’t have an equation but a graph instead you would know that it doesn’t have any solutions if the graph doesn’t cross the x ints. My example I solved had 2 solutions because it had 2 roots.

My example that I used and explained

# Blog post week 12

This week, I have learned how to graph the absolute value of a linear function. And since you can’t have negatives the graph goes down and then comes up which makes it a shape of a v. My example that I used was y=3x+3. And then I graphed it and labeled the x int and the y int. Also the domain and range. And its a linear function because it does not have an x^2 so it won’t be a parabola. The domain is always going to be that XER and the range can be that y \ge 0. Or any number greater than zero in this case my range is that y is greater or equal to zero. I can’t wait to learn more of these functions and so far graphing them is easy. And its a lot easier to find the x ints when it is in vertex form because then you can easily plot the vertex on the graph and then convert it to general form to get the y int.