Question

please say the correct answer please ​

Answer 1

Answer:

The another rational number is -10/9.

Step-by-step explanation:

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

• The sum of two rational numbers is -4/9.
• One of them is 2/3.

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

• The other rational number.

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

Let: The another rational number be 'x'.

$\:$

According to the question:

We can form an equation to find the value of 'x' i.e, the another rational number.

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\small\underline{\boxed{\bf{\dfrac{2}{3}+x=\dfrac{-4}{9}}}}$

$\:$

Now:

By solving the equation formed,

$\:$

$\:\:\:\:\:\:\::\implies{\sf{ \bigg(\dfrac{2}{3}+x \bigg)=\dfrac{-4}{9}}}$

Transposing the like terms in L.H.S to R.H.S (subtraction)-

$\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{x=\bigg(\dfrac{-4}{9}-\dfrac{2}{3}\bigg)}}$

Simplifying the like terms in R.H.S-

$\:$

• Multiply each side by the opposite side's denominator:

$\\:\:\:\implies{\sf{x=\bigg(\dfrac{3\times -4}{3\times 9} \bigg)- \bigg(\dfrac{9\times 2}{9\times 3}\bigg)}}$

• Simplifying:

$\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{x=\bigg(\dfrac{-12}{27}-\dfrac{18}{27}\bigg)}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\sf{x=\dfrac{\cancel{-30}}{\cancel{27}}}}$

We got,

$\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\boxed{\frak{\red{x=\dfrac{-10}{9}}}}}$

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

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎To verify, substitute the obtained value of 'x' in place of 'x' in the formed equation:

$\:$

$\:\:\:\:\:\:\mapsto{\sf{\dfrac{2}{3}+\purple{x}=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\sf{\dfrac{2}{3}+\purple{\dfrac{-10}{9}}=\dfrac{-4}{9}}}$

$\: \:\:\:\:\ \\ \mapsto{\sf{\bigg(\dfrac{9\times 2}{9\times 3}\bigg)+\bigg(\dfrac{-10\times 3}{9\times 3}\bigg)=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\sf{\bigg(\dfrac{18}{27}+\dfrac{-30}{27}\bigg)=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\sf{\bigg(\dfrac{18-30}{27}\bigg)=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\sf{\dfrac{\cancel{-12}}{\cancel{27}}=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\bf{\dfrac{-4}{9}=\dfrac{-4}{9}}}$

$\:\:\:\:\:\:\mapsto{\sf{L.H.S.=R.H.S.}}$

$\:\:\:\:\:\:\\\mapsto{\sf{Hence,\:the\:value\:of\:x\:is\:correct!}}$

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Answer 2

AnswEr-:

• $\boxed{\sf{\large { The\:number\:is\:= \dfrac{-10}{9}}}}$

EXPLANATION-:

• $\dag{\sf{\large { Given -:\:}}}$

• The sum of two rational numbers is -4/9 .
• If one of them is 2/3 .

• $\dag{\sf{\large { To\:Find -:\:}}}$

• The other rational number .

$\dag{\sf{\large { Solution -:\:}}}$

• $\dag{\sf{\large { Let's \: Assume-:\:}}}$

• The other rational number be x .

$\dag{\sf{\large { Then -:\:}}}$

• $\boxed{\sf{\large { Equation \: Formed\: \: = x + \dfrac{2}{3}=\dfrac{-4}{9}}}}$

$\dag{\sf{\large { Now , \: Soving \:for\: x-:\:}}}$

• $\longrightarrow{\sf{\large { \left(x + \dfrac{2}{3}\right)=\dfrac{-4}{9}}}}$

• $\star{\sf{\large { By \: Transforming \:\dfrac{2}{3} \:to \:RHS}}}$

• $\longrightarrow{\sf{\large { x = \left( \dfrac{-4}{9}- \dfrac{2}{3}\right)}}}$

• $\star{\sf{\large { LCM \:of\:3\:and\:9\:is\:9 .}}}$

• $\longrightarrow{\sf{\large { x =\left( \dfrac{-4- 6}{9} \right)}}}$

• $\longrightarrow{\sf{\large { x = \dfrac{-10}{9}}}}$

$\dag{\sf{\large { Now -:\:}}}$

• $\boxed{\sf{\large { x = \dfrac{-10}{9}}}}$

$\dag{\sf{\large { Then -:\:}}}$

• $\dag{\sf{\large { Putting \:-:\:x = \dfrac{-10}{9}}}}$

• The number is $\implies{\sf{\large { x = \dfrac{-10}{9}}}}$

$\dag{\sf{\large { Hence -:\:}}}$

• $\boxed{\sf{\large { The\:number\:is\:= \dfrac{-10}{9}}}}$

__________________________________

$\dag{\sf{\large { Verification -:\:}}}$

• $\boxed{\sf{\large { Equation \: \: \: = x + \dfrac{2}{3}=\dfrac{-4}{9}}}}$

$\dag{\sf{\large { Here -:\:}}}$

• $\longrightarrow{\sf{\large { x = \dfrac{-10}{9}}}}$

$\dag{\sf{\large { Now -:\:}}}$

• Putting x = -10/9 in Equation-:

• $\longrightarrow{\sf{\large { \left(\dfrac{-10}{9} + \dfrac{2}{3}\right)=\dfrac{-4}{9}}}}$

• $\star{\sf{\large { LCM \:of\:3\:and\:9\:is\:9 .}}}$

• $\longrightarrow{\sf{\large { \left(\dfrac{-10+6}{9} \right)=\dfrac{-4}{9}}}}$

• $\longrightarrow{\sf{\large { \dfrac{-4}{9} =\dfrac{-4}{9}}}}$

$\dag{\sf{\large { Therefore-:\:}}}$

• $\boxed{\sf{\large { LHS = RHS}}}$

• $\boxed{\sf{\large { Hence,\: Verified. }}}$

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