Question

what should be added to the sum of minus 5 by 9 and 7 by 18 to get minus one​

Answer 1

Let the 'x' should be added

to the sum of  $\dfrac{-5}{9}$

and $\dfrac{7}{18}$ to get - 1

                                           

According to the question,

we have to solve this equation

$\implies\dfrac{-5}{9}+\dfrac{7}{18}+x=-1$

                                           

Let us solve ,

$\implies\dfrac{-5}{9}+\dfrac{7}{18}+x=-1\\\\\\\implies\dfrac{(-10)+7}{18}+x=-1\\\\\\\implies\dfrac{-3}{18}+x=-1$

$\\\\\\\implies x=-1+\dfrac{3}{18}\\\\\\\implies x=\dfrac{-18+3}{18}\\\\\\\implies x=\dfrac{-15}{18}$

                                           

$\Huge{\boxed{x=\dfrac{-15}{18}}}$

                                           

Therefore required number is

-15/18

                                           

Verification:-

$\implies\dfrac{-5}{9}+\dfrac{7}{18}+\dfrac{-15}{18}=-1\\\\\\\implies\dfrac{(-10)+7+(-15)}{18}=-1\\\\\\\implies\dfrac{(-25)+7}{18}=-1\\\\\\\implies\dfrac{-18}{18}=-1\\\\\\\implies-1=-1$

                                           

L.H.S = R.H.S

hence it is verified

Answer 2

Answer:

$\dfrac{-5}{9}+\dfrac{7}{18}+x=-1\quad :\quad x=-\dfrac{5}{6}\quad \left(\mathrm{Decimal}:\quad x=-0.83333\dots \right)$

Step-by-step explanation:

$\textrm{Let's suppose the value that should be added with the equation be x.}$

$\textrm{Then our equation will look like;}$

$\dfrac{-5}{9}+\dfrac{7}{18}+x=-1$

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

$-\dfrac{5}{9}+\dfrac{7}{18}+x=-1$

$\black{\mathrm{Least\:Common\:Multiplier\:of\:}9,\:18:\quad 18}$

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

$-\dfrac{10}{18}+\dfrac{7}{18}=-1$

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

$\dfrac{-10+7}{18}=-1$

$\gray{\mathrm{Add/Subtract\:the\:numbers:}\:-10+7=-3}$

$\dfrac{-3}{18}=-1$

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

$-\dfrac{3}{18}=-1$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:3}$

$x-\dfrac{1}{6}=-1$

$\gray{\mathrm{Add\:}\dfrac{1}{6}\mathrm{\:to\:both\:sides}}$

$x-\dfrac{1}{6}+\dfrac{1}{6}=-1+\dfrac{1}{6}$

$\gray{\mathrm{Simplify}}$

$x=-\dfrac{5}{6}$