Question

verify :-(7/3+4/5) + (-3/2) = 7/3 + (4/5 + (-3)/2)
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Answer 1

Answer:

Yes, (7/3+4/5) + (-3/2) = 7/3 + (4/5 + (-3)/2).

Answer 2

Answer:

$\mathrm{Prove\:}-\left(\frac{7}{3}+\frac{4}{5}\right)+\left(-\frac{3}{2}\right)=\frac{7}{3}+\left(\frac{4}{5}+\frac{\left(-3\right)}{2}\right):\quad \mathrm{TRUE}$

Step-by-step explanation:

$-\left(\frac{7}{3}+\frac{4}{5}\right)-\frac{3}{2}=\frac{7}{3}+\frac{4}{5}+\frac{-3}{2}$

$\mathrm{The\:two\:sides\:are\:equal}$

$\Rightarrow \mathrm{True}$

LHS=

$\left(7/3+4/5\right)+\left(-3/2\right)=1\frac{19}{30}$

$\left(7/3+4/5\right)+\left(-3/2\right)$

$=\frac{47}{15}+\left(-3/2\right)$

$=\frac{47}{15}+\frac{-3}{2}$

$=\frac{49}{30}$

$=1\frac{19}{30}$

RHS=

$7/3+\left(4/5+\left(-3\right)/2\right)=1\frac{19}{30}$

$7/3+\left(4/5+\left(-3\right)/2\right)$

$=7/3-\frac{7}{10}$

$=\frac{7}{3}-\frac{7}{10}$

$=\frac{49}{30}$

$=1\frac{19}{30}$

$\Rightarrow\mathrm{Hence\ RHS \ =\ LHS }$