Question

Resolver a expressão:
0,3+(2/3-1/4)-(-7/12+4/3)=

Resolver as equações em R:
a) 3 (2x - 3) + 2 (x + 1) = 3x +8
b) 2x +3 (x - 4) = 4x + 9

Answer 1

1) $0,3+\left(\dfrac{2}{3}-\dfrac{1}{4}\right)-\left(-\dfrac{7}{12}+\dfrac{4}{3}\right)$

Note que:

$\dfrac{2}{3}-\dfrac{1}{4}=\dfrac{8-3}{12}=\dfrac{5}{12}$

$\dfrac{-7}{12}+\dfrac{4}{3}=\dfrac{-7+16}{12}=\dfrac{9}{12}$

Então:

$0,3+\left(\dfrac{2}{3}-\dfrac{1}{4}\right)-\left(-\dfrac{7}{12}+\dfrac{4}{3}\right)=0,3+\dfrac{5}{12}-\dfrac{9}{12}$

Mas, $0,3=\dfrac{3}{10}=\dfrac{3\cdot1,2}{10\cdot1,2}=\dfrac{3,6}{12}$

Assim:

$0,3+\left(\dfrac{2}{3}-\dfrac{1}{4}\right)-\left(-\dfrac{7}12}+\dfrac{4}{3}\right)=\dfrac{3,6}{12}+\dfrac{5}{12}-\dfrac{9}{12}=\dfrac{3,6+5-9}{12}=\dfrac{-0,4}{12}$

Veja que, $-0,4=\dfrac{-4}{10}=\dfrac{-2}{5}$. Deste modo:

$0,3+\left(\dfrac{2}{3}-\dfrac{1}{4}\right)-\left(-\dfrac{7}{12}+\dfrac{4}{3}\right)=\dfrac{\frac{-2}{5}}{12}=\dfrac{-2}{5}\cdot\dfrac{1}{12}=\dfrac{-2}{60}=\dfrac{-1}{30}$


a) $3(2x-3)+2(x+1)=3x+8$

$6x-9+2x+2=3x+8$

$8x-3x=8+7$

$5x=15$

$x=3$


b) $2x+3(x-4)=4x+9$

$2x+3x-12=4x+9$

$5x-4x=12+9$

$x=21$