Question

(7/25*-15/28)-(-3/5*4/9)​

Answer 1

Answer:

$\left(\dfrac{7}{25}\cdot \dfrac{-15}{28}\right)-\left(-\dfrac{3}{5}\cdot \dfrac{4}{9}\right)=\dfrac{7}{60}\quad \left(\mathrm{Decimal:\quad }\:0.11667\dots \right)$

Step-by-step explanation:

$\left(\dfrac{7}{25}\cdot \dfrac{-15}{28}\right)-\left(-\dfrac{3}{5}\cdot \dfrac{4}{9}\right)$

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

$=\dfrac{7}{25}\cdot \dfrac{-15}{28}+\dfrac{3}{5}\cdot \dfrac{4}{9}$

$\mathrm{Now,\:let's\:simplify,\:}\dfrac{7}{25}\cdot \dfrac{-15}{28}$

$\dfrac{7}{25}\cdot \dfrac{-15}{28}$

$\mathrm{Cross-cancel\:common\:factor:}\:7$

$=\dfrac{1}{25}\cdot \dfrac{-15}{4}$

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

$=\dfrac{1\cdot \left(-15\right)}{25\cdot \:4}$

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

$=\dfrac{-1\cdot \:15}{25\cdot \:4}$

$\mathrm{Multiply\:the\:numbers:}\:25\cdot \:4=100$

$=\dfrac{-15}{100}$

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

$=-\dfrac{15}{100}$

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

$=-\dfrac{3}{20}$

$\mathrm{So,\:our\:equation\:would\:be;}$

$-\dfrac{3}{20}+\dfrac{3}{5}\cdot \dfrac{4}{9}$

$\mathrm{Now,\:let's\:simplify\:}\dfrac{3}{5}\cdot \dfrac{4}{9}$

$\dfrac{3}{5}\cdot \dfrac{4}{9}$

$\mathrm{Cross-cancel\:common\:factor:}\:3$

$=\dfrac{1}{5}\cdot \dfrac{4}{3}$

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

$=\dfrac{1\cdot \:4}{5\cdot \:3}$

$\mathrm{Multiply\:the\:numbers:}\:1\cdot \:4=4$

$=\dfrac{4}{5\cdot \:3}$

$\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15$

$=\dfrac{4}{15}$

$\mathrm{So,\:now\:our\:equation\:is\:}-\dfrac{3}{20}+\dfrac{4}{15}$

$\textrm{So, to add the fractions we'll need to make the denominator of these fractions equal}$

$\mathrm{Least\:Common\:Multiplier\:of\:}20,\:15$

$\mathrm{Prime\:factorization\:of\:}20:\quad 2\cdot \:2\cdot \:5$

$\mathrm{Prime\:factorization\:of\:}15:\quad 3\cdot \:5$

$\mathrm{Multiply\:each\:factor\:the\:greatest\:number\:of\:times\:it\:occurs\:in\:either\:}20\mathrm{\:or\:}15$

$=2\cdot \:2\cdot \:5\cdot \:3$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2\cdot \:5\cdot \:3=60$

$\mathrm{Now,\:Adjust\:Fractions\:based\:on\:the\:LCM}$

$\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}$ $\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\:60$

$\mathrm{For}\:\dfrac{3}{20}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:3$

$\dfrac{3}{20}=\dfrac{3\cdot \:3}{20\cdot \:3}=\dfrac{9}{60}$

$\mathrm{For}\:\dfrac{4}{15}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:4$

$\dfrac{4}{15}=\dfrac{4\cdot \:4}{15\cdot \:4}=\dfrac{16}{60}$

$=-\dfrac{9}{60}+\dfrac{16}{60}$

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

$=\dfrac{-9+16}{60}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-9+16=7$

$=\dfrac{7}{60}$