Question

Find 2/7×-4/7-5/9-4/7×2/5​

Answer 1

Answer:

$\dfrac{2}{7}\left(-\dfrac{4}{7}\right)-\dfrac{5}{9}\left(-\dfrac{4}{7}\right)\dfrac{2}{5}=-\dfrac{16}{441}\quad \left(\mathrm{Decimal:\quad }\:-0.03628\dots \right)$

Step-By-Step-Explanation:

$\dfrac{2}{7}\left(-\dfrac{4}{7}\right)-\dfrac{5}{9}\left(-\dfrac{4}{7}\right)\dfrac{2}{5}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a$

$=-\dfrac{2}{7}\times \dfrac{4}{7}+\dfrac{5}{9}\times \dfrac{4}{7}\times \dfrac{2}{5}$

$\dfrac{2}{7}\times \dfrac{4}{7}$

$\mathrm{Multiply\:fractions}:\quad \dfrac{a}{b}\times \dfrac{c}{d}=\dfrac{a\:\times \:c}{b\:\times \:d}$

$=\dfrac{2\times \:4}{7\times \:7}$

$\mathrm{Multiply\:the\:numbers:}\:2\times \:4=8$

$=\dfrac{8}{7\times \:7}$

$\mathrm{Multiply\:the\:numbers:}\:7\times \:7=49$

$=\dfrac{8}{49}$

$\dfrac{5}{9}\times \dfrac{4}{7}\times \dfrac{2}{5}$

$\mathrm{Multiply\:fractions}:\quad \dfrac{a}{b}\times \dfrac{c}{d}=\dfrac{a\:\times \:c}{b\:\times \:d}$

$=\dfrac{5\times \:4\times \:2}{9\times \:7\times \:5}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=\dfrac{4\times \:2}{9\times \:7}$

$\mathrm{Multiply\:the\:numbers:}\:4\times \:2=8$

$=\dfrac{8}{9\times \:7}$

$\mathrm{Multiply\:the\:numbers:}\:9\times \:7=63$

$=\dfrac{8}{63}$

$=-\dfrac{8}{49}+\dfrac{8}{63}$

$\mathrm{Least\:Common\:Multiplier\:of\:}49,\:63$

$49,\:63$

$Least\:Common\:Multiplier\:\left(LCM\right)$

$\mathrm{The\:LCM\:of\:}a,\:b\mathrm{\:is\:the\:smallest\:positive\:number\:that\:is\:divisible\:by\:both\:}a\mathrm{\:and\:}b$

$\mathrm{Prime\:factorization\:of\:}49:\quad 7\times \:7$

$\mathrm{Prime\:factorization\:of\:}63:\quad 3\times \:3\times \:7$

$\mathrm{Multiply\:each\:factor\:the\:greatest\:number\:of\:times\:it\:occurs\:in\:either\:}49\mathrm{\:or\:}63$

$=7\times \:7\times \:3\times \:3$

$\mathrm{Multiply\:the\:numbers:}\:7\times \:7\times \:3\times \:3=441$

$=441$

$Adjust\:Fractions\:based\:on\:the\:LCM$

$=-\dfrac{72}{441}+\dfrac{56}{441}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{-72+56}{441}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-72+56=-16$

$=\dfrac{-16}{441}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{-a}{b}=-\dfrac{a}{b}$

$=-\dfrac{16}{441}$