Question

( Com passo a passo, por favor ) Resolva as seguintes equações, sendo U=Q


a) 15 - x + 3 = 4x + 2 - 2x

b) 4 (x+1) = 12

c) $\frac{x}{6}$ + $\frac{2x-18}{3}$ = $\frac{5x-9}{6}$

d) $\frac{3x-5}{2}$ - $\frac{x-1}{3}$ = $\frac{5x-9}{6}$

Answer 1

Resposta:

A) 16/3

B) x = 2

Explicação passo-a-passo:

A) 18 - x = 2x + 2

-x -2x =2 -18

-3x = -16

Obtemos: 16/3

B) x + 1= 3

x = 3 - 1

Obtemos: x = 2

Desculpa Amigo só Consegui Resolver Essas

Answer 2

$\boxed{\begin{array}{l}\tt a)~\sf 15-x+3=4x+2-2x\\\sf 4x-2x+x=15+3-2\\\sf 3x=16\\\\\sf x=\dfrac{16}{3}\\\tt b)~\sf 4\cdot(x+1)=12\\\sf x+1=\dfrac{12}{4}\\\\\sf x+1=3\\\sf x=3-1\\\sf x=2\end{array}}$

$\boxed{\begin{array}{l}\tt c)~\sf\dfrac{x}{6}+\dfrac{2x-18}{3}=\dfrac{5x-9}{6}\\\underline{\rm multiplicando~por~6~os~dois~membros~temos:}\\\sf\backslash\!\!\!6\cdot\dfrac{x}{\backslash\!\!\!6}+\backslash\!\!\!6^2\cdot\dfrac{2x-18}{\backslash\!\!\!3}=\backslash\!\!\!6\cdot\dfrac{5x-9}{\backslash\!\!\!6}\\\underline{\rm a~equac_{\!\!,}\tilde ao~equivalente~\acute e:}\\\sf x+2\cdot(2x-18)=5x-9\\\sf x+4x-36=5x-9\\\sf x+4x-5x=36-9\\\sf0x=45\\\sf x=\dfrac{45}{0}~(absurdo!) \\\sf~portanto~ S=\bigg\{~\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\tt d)~\sf\dfrac{3x-5}{2}-\dfrac{x-1}{3}=\dfrac{5x-9}{6}\\\underline{\rm multiplicando~por~6~a~igualdade~temos:}\\\sf \backslash\!\!\!6^3\cdot\dfrac{3x-5}{\backslash\!\!\!2}-\backslash\!\!\!6^2\cdot\dfrac{x-1}{\backslash\!\!\!3}=\backslash\!\!\!6\cdot\dfrac{5x-9}{\backslash\!\!\!6}\\\\\sf3\cdot(3x-5)-2\cdot(x-1)=5x-9\\\sf9x-15-2x+2=5x-9\\\sf9x-2x-5x=15-2-9\\\sf2x=4\\\sf x=\dfrac{4}{2}\\\\\sf x=2\end{array}}$