This week in Pre-calc 11 we kept expanding on rational expressions after adding and subtracting rational expressions we moved into multiplying and dividing rational expressions. And there we’ve sealed the unit by learning how to solve rational equations with all the skills we developed throughout. This post will be about solving rational equations in part one.
Here we have…
In this example, I started with the original equation and realized I needed to have a common denominator. I started by cross multiplying to get the common denominator then I multiplied everything by the denominator value so we get rid of the denominator and we’re just left with a simple equation. Then I moved the terms around to leave the variable on one one side and a value on the other side then I divided to isolate a and because we can’t divide 15 by 6 I just simplified 15 over 6 since both are divisible by 3 and finally we have a solution of a=-5/2. NOTE: that solutions cannot be non-permissible values and therefore we have to state that non-permissible values at the beginning were a≠3/2. Substitute the solution for the variable to check your answer, and if there’s an extraneous solution it could have no solution at all.
Notes from this lesson:
- Factor:
Whenever you see something factorable do that to make the equation easier. Don’t forget to state the restrictions at the beginning as I said before.
- Cross Multiplying and Common Denominators:
Cross-multiplying is a method used when you have exactly two fractions, one on each side of an equation. To cross-multiply, you multiply the numerator of each fraction by the denominator of the other fraction. This process eliminates the denominators, allowing you to solve the resulting equation. When dealing with more than two fractions or adding/subtracting fractions, you must find a common denominator. This means you multiply each fraction’s numerator and denominator by any terms needed to make all the denominators the same.
- Rejecting or Accepting Values:
When solving for 𝑥, compare the found values against the non-permissible (restricted) values. If an 𝑥 value matches any of these restricted values, then it is not a valid solution, indicating no solution exists. To verify a solution, substitute the 𝑥 value back into the original equation. If the resulting statement is true, the 𝑥 value is correct; otherwise, it is not a valid solution.