Week 15 – Rational Equations

This week we’ve learned to solve rational equations, which is essentially solving unknown variables, but rather being more complicated since those equations can appear in different forms, such as adding/subtracting and multiplying/dividing. And the answer might even end up being “ugly”, none, and sometimes even with infinite solutions.

Here will be an example, take \frac{4}{5x-2}=\frac{3}{4x-1}

There are always multiple ways to approach a question and solve it, for this equation I’ll be cross multiplying both fractions (because both sides contain 1 fraction only and are dividing expressions) and it should end up like this: (4)(4x-1)=(3)(5x-2)

Next, proceed to multiply them and combine all like terms:

16x-4=15x-6

 

16x-15x=-6+4

 

x=-2

Usually, most people will find the non-permissible values from the start of the question, but I personally like to do this at the end so that I can prove if any of my answers make sense and if they’re extraneous:

The values are as follows: 4x-1=0, x=\frac{1}{4},

 

5x-2=0, x=\frac{2}{5}

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