Pre-Calculus 11 Week 15 – Solving Rational Equations

This week, we solved rational equations

Note that this is different from last week, where we only simplified rational expressions.

Simplifying looks like this:

\frac {x^2+5x+6}{x+3}

 

\frac {(x+2)(x+3)}{x+3}

 

x+2

And solving looks like this:

\frac {x^2+5x+6}{x+3}=

 

\frac {(x+2)(x+3)}{x+3}

 

x+2

 

x = -2

 

The major difference here is the equal sign that is present when solving vs no equal sign when simplifying.

A common issue with solving ration expressions is how to add together two or more fractions when they have different denominators. Simply remove the denominator like so

\frac {x-1}{x-3}= \frac{x+1}{x-4}

 

\frac {(x-1)(x-3)(x-4)}{x-3} = \frac {(x+1)(x-3)(x-4)}{x-4}

 

x^2-4x-3 = x^2-2x-3

 

= 0=3x-7

 

3x=7

 

x = \frac {3}{7}

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