Question

The sum of two rational numbers is ⅖. If one of them is -4/7 find the other.​

Answer 1

$\large\sf\underline{Given}$

• Sum of two rational numbers =$\sf\:\frac{2}{5}$

• One of the number = $\sf\:\frac{-4}{7}$

$\large\sf\underline{To\:find}$

• The other number .

$\large\sf\underline{Solution}$

Let's assume the required number be x .

So according to the question ,

$\sf➛\:x+(\frac{-4}{7})=\frac{2}{5}$

$\sf➛\:x-\frac{4}{7}=\frac{2}{5}$

$\sf➛\:\frac{7x-4}{7}=\frac{2}{5}$

$\sf➛\:5(7x-4)=2 \times 7$

$\sf➛\:35x-20=14$

$\sf➛\:35x=14+20$

$\sf➛\:35x=34$

$\small{\underline{\boxed{\mathrm\pink{➻\:x=\frac{34}{35}}}}}$

‎ ‎‎‎‎‎‎⠀⠀━━━━━━━━▲━━━━━━━━

$\large\sf\underline{Verification\:of\:my\:answer\::}$

$\sf➛\:\frac{34}{35}+(\frac{-4}{7})$

$\sf➛\:\frac{34}{35}-\frac{4}{7}$

$\sf➛\:\frac{34-20}{35}$

$\sf➛\:\frac{14}{35}$

$\small{\underline{\boxed{\mathrm\pink{➻\:\frac{2}{5}}}}}$

Therefore verified .

‎ ‎‎‎‎‎‎⠀⠀ ━━━━━━━━▼━━━━━━━━

Happy solving ¡!

Answer 2

$\huge\rm\purple{an{\mathfrak{S}}wεr:}$

${}$

$\bf{{\underline{Given}}:}$

• $\sf{Sum~of~two~rational~numbers={\dfrac{2}{5}}}$
• $\sf{One~of~the~number~is~{\dfrac{-4}{7}}}$

$\bf{{\underline{To~find}}:}$

• $\sf{The~other~number}$

$\bf{{\underline{Solution}}:}$

• $\sf{Let~the~number~be~~‛x’}$

$\sf{~~~~~~~~~~ x+{\dfrac{-4}{7}}={\dfrac{2}{5}}}$

$\sf\implies{x={\dfrac{2}{5}}-{\dfrac{-4}{7}}}$

$\sf\implies{x={\dfrac{(2×7)-(-4×5)}{35}}~~~~~~(LCM=35)}$

$\sf\implies{x={\dfrac{14-(-20)}{35}}}$

$\sf\implies{x={\dfrac{14+20}{35}}}$

$\sf\implies{x=\purple{\underline{\boxed{\bf~{\dfrac{34}{35}}~}}}}$

$\bf\therefore{{\underline{Required~answer}}:}$

• $\sf{Unknown~number,~~x~~is~{\purple{\bf {\dfrac{34}{35}}}}}$

$\bf{{\underline{Verification}}:}$

$\sf{~~~~ {\dfrac{-4}{7}}+{\dfrac{34}{35}}}$

$\sf{={\dfrac{(5×-4)+34}{35}}~~~~~~(LCM=35)}$

$\sf{={\dfrac{-20+34}{35}}}$

$\sf{={\dfrac{14}{35}}}$

$\sf{={\dfrac{\cancel{14}~~^{2}}{\cancel{35}~~^{5}}}~~~~~~~~~~~~Taking~7~as~common~factor}$

$\bf{={\purple{\bf {\dfrac{2}{5}}}}=RHS}$

${}$

• $\bf{Hence~verified~!}$