Question

resolva as equações modulares.

a) |8x-3| = |2x-4|

b) 20. |8x+3| =7

Answer 1

a) 8x - 3 = 2x - 4 => 8x - 2x = 3 - 4 => 6x = -1 => x = -1/6   ou
    8x - 3 = -2x + 4 => 8x + 2x = 4 + 3 => 10x = 7 => x = 7/10

S = { -1/6, 7,10}

b) |8x + 3| = 7/20

8x + 3 = 7/20 => 160x + 60 = 7 => 160x =7 - 60 => 160x = -53 => x = -53/160
ou
8x + 3 = -7/20 => 160x + 60 = -7 => 160x = -67 => x = -67/160

S = { -53/160, -67/160}

Answer 2

Propriedade modular:

$\mathsf{|a|=|b|~\Longleftrightarrow~a=b~~ou~~a=-b}$

Então: 

$\mathsf{a)~|8x-3|=|2x-4|~\Longleftrightarrow~\begin{Bmatrix}\mathsf{8x-3=2x-4}\\\\\mathsf{ou}\\\\\mathsf{8x-3=-2x+4}\end.}$

Resolvendo as equações:

$\mathsf{8x-3=2x-4\hspace{15}\vee\hspace{15}8x-3=-2x+4}\\\\\mathsf{\hspace{31}6x=-1\hspace{15}\vee\hspace{15}10x=7}\\\\\mathsf{\hspace{35}x=-\frac{1}{6}\hspace{15}\vee\hspace{15}x=\frac{7}{10}}\\\\\\\fbox{ \mathsf{S=\begin{Bmatrix}\mathsf{x\in\mathbb{R}|~~x=-\frac{1}{6}~~ou~~x=\frac{7}{10}}\end{Bmatrix}} }~~~~(\mathsf{resposta})$

Propriedade modular (k > 0):

$\mathsf{|x|=k~\Longleftrightarrow~x=-k~~ou~~x=k}$

Então:

$\mathsf{b)~20\cdot|8x+3|=7~\Leftrightarrow~|8x+3|=\frac{7}{20}~\Longleftrightarrow~\begin{Bmatrix}\mathsf{8x+3=-\frac{7}{20}}\\\\\mathsf{ou}\\\\\mathsf{8x+3=\frac{7}{20}}\end.}$

Resolvendo as equações:

$\mathsf{8x+3=-\frac{7}{20}\hspace{40}\vee\hspace{40}8x+3=\frac{7}{20}}\\\\\mathsf{8x=-\frac{7}{20}-3\hspace{40}\vee\hspace{40}8x=\frac{7}{20}-3}\\\\\mathsf{8x=-\frac{7}{20}-\frac{60}{20}\hspace{35}\vee\hspace{41}8x=\frac{7}{20}-\frac{60}{20}}\\\\\mathsf{8x=-\frac{67}{20}\hspace{57}\vee\hspace{41}8x=-\frac{53}{20}}\\\\\mathsf{x=-\dfrac{\frac{67}{20}}{8}\hspace{60}\vee\hspace{41}x=-\dfrac{\frac{53}{20}}{8}}\\\\\\\mathsf{x=-\frac{67}{160}\hspace{59}\vee\hspace{42}x=-\frac{53}{160}}$

$\fbox{ \mathsf{S=\begin{Bmatrix}\mathsf{x\in\mathbb{R}|~x=-\frac{67}{160}~~ou~~x=-\frac{53}{160}}\end{Bmatrix}} }~~~~\mathsf{(resposta)}$