Week 14 – Rational Expressions (w/ non-permissible values)

This week we’ve learned to solve certain types of rational expressions. A rational expression is when a fraction’s numerator and denominator are both polynomials and doesn’t have any roots or any unknown variables as an exponent, e.g. \frac{x+8}{4-10^x}

Also, a non-permissible value is a number for any unknown variables in the denominator which makes the fraction undefined (denominator becomes 0, which becomes unsolvable). E.g., in \frac{-9-x}{x+6}, X can’t be -6 (x\neq-6) because the denominator will then be 0 and the expression will be unsolvable.

The following will be a quite ugly expression that would be solved. Here we have \frac{6x}{2(x+3)}\div\frac{4x^2}{5(x+3)}-\frac{x}{2}, since we got dividing fractions, we’ll switch the numerator and denominator of the 2nd fraction and then we simplify any common factors between the first and second fraction, to finally multiply them together:

\frac{6x}{2(x+3)}\div\frac{4x^2}{5(x+3)}-\frac{x}{2} \rightarrow \frac{3}{2}\bullet\frac{5}{2x}-\frac{x}{2} \rightarrow \frac{15}{4x}-\frac{x}{2}

The 3rd fraction’s denominator will now be multiplied so that it has a same denominator as \frac{15}{4x} and can be subtracted by it:

\frac{15-2x(x)}{4x}=\frac{15-2x^2}{4x}

From this whole process we can also find out the non-permissible values, and they are x\neq-3,0.

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