Solving Quadratic Equations using Factoring: Solving an equation with fractions

Solve for x:

$-3(\frac{1}{2}x+4)=\frac{1}{2}$

We open the parentheses on the left side by the distributive property and use the formula:

$a(x+b)=ax+ab$

$-\frac{3}{2}x-12=\frac{1}{2}$

We multiply all terms by 2 to get rid of the fractions:

$-3x-12\times2=1$

$-3x-24=1$

We will move the minus 24 to the right section and keep the corresponding sign:

$-3x=24+1$

$-3x=25$

Divide both sections by minus 3:

$\frac{-3x}{-3}=\frac{25}{-3}$

$x=-\frac{25}{3}$

$-\frac{25}{3}$

Solve for X:

$-\frac{1}{2}(x+\frac{1}{4})=\frac{1}{8}$

$-\frac{1}{2}$

Solve for X:

$\frac{1}{2}(x+3)=0$

3-

Solve for X:

$\frac{1}{4}(x-8)=1$

12

Solve for X:

$-\frac{1}{4}(x-2)=3$

-10

Solve X:

$-\frac{1}{6}(x+\frac{1}{3})=\frac{1}{2}(\frac{1}{3}x-\frac{1}{9})$

0

Solve for X:

$-\frac{1}{3}(\frac{1}{4}+\frac{1}{2}x)=\frac{1}{12}$

-1

Solve for X:

$\frac{1}{8}(\frac{1}{2}x-\frac{1}{3})=\frac{1}{4}(\frac{1}{4}x+\frac{1}{6})$

There is no solution

Solve for X:

$-\frac{1}{2}(\frac{1}{4}x+\frac{1}{6})=\frac{1}{4}(\frac{1}{2}x+\frac{1}{3})$

$-\frac{2}{3}$

Solve for X:

$\frac{1}{3}(\frac{1}{3}x+\frac{1}{6})=\frac{1}{18}$

0