This week in math we learned how to solve linear and quadratic rational equations, testing out different methods. For my blog post, I chose solving linear rational expressions because I find it satisfying to isolate x at the end. I also like that I can make a connection with other subjects I have learned in math this semester, and in years before. This undersyanding is important because it is a subject that will continue to advance, and having the foundational understandings of how to solve a linear rational expression is necessary in the success of your future math self.
Lets look at some examples…
Example a.
Step 1. My first step is to identify an LCM. Since one term has a denominator of x, and another of 9, the LCM which must include a 9 and an x can only be 9x.
Step 2. Identify non-pernissible values. Since our LCM is 9x, we need to identify a number that would make 9x = 0. Our non permissible value = 0; 9(0) = 0.
Step 3. Make the denominator common – while we know our LCM, we need to multiply each denominator by whatever term sets it to 9x and do the same to the numerator. In our first term, 9 is the denominator, so we multiply both the denominator and numerator by x. In our second term, we multiply the top and bottom by 9. Since everything is over the same denominator, there is no need to include the denominator in your work.
Step 4. Reorder the terms – now that we have distributed and gotten our new values for the numerator, we need to re order the terms so that the variable is on one side, and the constant on the other. I want to keep x positive, so I move x to the left with 4x making 3x.
Step 5. Now we can isolate for x – I divide everything by 3 to get rid of the coefficent infront of x, leaving me with x = 3. It is not our non permissible value so there is no need to reject the equation.
Example b.
Step 1. As always, we must determine our LCM. Looking at this equation I see there is a 4a, a 12, and 3a in our denominator’s place. I know that 12 includes 4 and 3, and since there is an a, this means our LCM is 12a.
Step 2. Now we must identify our non-permissible values, so we dont end up with a solution that is extraneous. a = 0, because if you multiplied each denomintaor by 0, you will end up with zero.
Step 3. We must now multiply each denominator by a value that will let it equal 12a. We must multiply the numerator by the same values as the denominator. I distribute the 3 into (a + 5), the a to the 11, and the 4 to the 2. Since each denominator is the same, we do not need to include it in our work.
Step 4. Reorder terms; Since both a values are on the left handside, i can keep them there and add them together. I move the 15 to the right hand side becoming 15 + 8 which equals -7.
Step 5. Isolate for X; I divide both sides by 14 to isolate x, leaving me with a negative fraction as my value for a. a =
Example C.
Step 1. I notice right away that one of my denominators contains a 
Step 2. Now I can find my non permissible values, look at the value being added to x and determine what would make that value equal to zero. Since I have a 4 and a 3, the non permissible values are -4 and -3.
Step 3. Now we identify an LCM; looking at this I know that our LCM is (x + 3)( x + 4)
Step 4. Make denominator equal to (x + 3)(x + 4) . I multiply my first term top and bottom by (x +4), my second term by (x+3) and my third term already contains both parts of the denominator. Now I can distribute and write out my values.
Step 5. Now I can add my like terms; 2x and 3x = 5x, and 8+9 = 17.
Step 6. Rearrange, I want to keep my x value positive, so I shift the constant to the right.
Step 7. All thats left to do is isolate; I divide everything by 5 to isolate x, giving me the answer x = -2.