Month: December 2018

week 15- precalc

rubeng2016 on

This week in math we learned how to simplify an expression when they are adding.

\frac{6}{a-3}+\frac{2}{a+7}

The non-permissible values are A cannot equal 3,-7. The next step is to multiply both fractions by a common denominator

\frac {6 \times (a+7)}{(a-3)\times (a+7)}+\frac{2\times(a-3)}{a+7\times(a-3)}

 

\frac{6a+42+2a-6}{(a-3)(a+7)}

then collect like terms

\frac{8a+36}{(a-3)(a+7)} is what you are left with

week 14 – Precalc

rubeng2016 on

This week in math we learned how to simplify a rational expression and determine the non-permissible values. The expression below is an example of how to do this.

\frac {10y}{(y-3)^3}\div{\frac {y(y+1)}{(y-3)^2}}

The first step is to determine the non-permissible values which for now are only Y cannot equal 3.

\frac {10y}{(y-3)(y-3)(y-3)}\times{\frac {(y-3)(y-3)}{y(y+1)}}

Now that you have flipped the fraction you have to determine the new non-permissible values of the second fraction which are, Y cannot equal 0,-1.

you can now cross out the top of the fraction with the bottom of the fraction. Then you end up with:

\frac {10}{(y-3)(y+1)}