# Week 15 – Precalc 11

One of the things we learned this week in Precalc 11 is how to solve rational equations.

#### Cross-Multiplying

The cross-multiplying method only works if a fraction is equal to another fraction ($\frac{2x-1}{4}=\frac{4x+2}{2}$) and does not work with 3 or more terms ($\frac{2x-1}{4} + \frac{3-2}{x}=\frac{4x+2}{2}$)

#### Examples

Solve: $\frac{x+3}{2x}=\frac{x-5}{3}$
First we need to determine the non-permissible values for x.
$x\neq 0$
Next, cross-multiply. Multiply the numerator for the first fraction to the denominator for the second one, vice versa with the denominator for the first fraction.
$3(x+3)=2x(x-5)$
Then distribute.
$3x+9=2x^2-10x$
Since there is a square, we know that it’s a quadratic and it has 2 possible solutions. Move all the terms on one side.
$0=2x^2-13x-9$
Since we it doesn’t factor with nice numbers we can use the quadratic formula.
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-13)\pm \sqrt{(-13)^2-4(2)(-9)}}{2(2)}$
$x=\frac{13\pm \sqrt{241}}{4}$
$x_1=\frac{13+\sqrt{241}}{4}$
$x_2=\frac{13-\sqrt{241}}{4}$

Solve: $\frac{2x-3}{x+1}={x+6}{x-2}$
$x\neq -1, 2$
$(x-2)(2x-3)=(x+6)(x+1)$
$2x^2-3x-4x+6=x^2+x+6x+6$
$2x^2-7x+6=x^2+7x+6$
$x^2-14x=0$
$x(x-14)=0$
$x_1=0$
$x_2=14$

#### Converting to a common denominator

Works everytime even with 3 or more terms.

##### Example

Solve: $3x- \frac{1}{x}=\frac{1}{2}$
First determine the non-permissible values.
$x\neq 0$
Then determine the Lowest/Least Common Denominator (LCD). LCD=2x
Then multiply the fractions so that they have the same denominators.
$\frac{3x(2x)}{1(2x)} - \frac{1(2)}{x(2)}=\frac{1(x)}{2(x)}$
$\frac{6x^2}{2x} - \frac{2}{2x}=\frac{x}{2x}$
Since the denominators are all the same the numerators also have to be the same, so we could ignore the denominators.
$6x^2-2=x$
Since it’s a quadratic we have to make it equal 0 and then factor.
$6x^2-x-2=0$
$(2x+1)(3x-2)=0$
$x_1=- \frac{1}{2}$
$x_2=\frac{2}{3}$