Question

sugar bro can u plz solve this??


linear equations in one variable.plz.​

Answer 1

$\huge{✯}$$\huge{\underline{\underline{\mathcal{\sf{Answer}}}}}$

$\huge{☞}$ First Number = $23\dfrac{1}{7}$

$\huge{☞}$ Second Number = $12\dfrac{6}{7}$

$\huge{✯}$ $\huge{\underline{\underline{\mathcal{\sf{Given}}}}}$

$\dfrac{2}{3}$ of first number is equal to $\dfrac{6}{5}$ of other number.

$\huge{✯}$ $\huge{\underline{\underline{\mathcal{\sf{Steps}}}}}$

Let first number be x.

Other number = 36-x

According to the question,

$\large{⇒}$$\dfrac{2}{3}$ of x = $\dfrac{6}{5}$ of (36-x)

$\large{⇒}$Sending x terms to LHS and constant terms to RHS,

$\large{⇒} \dfrac{2}{3} x = \dfrac{6}{5} (36 - x) \\ \\\large{⇒}\dfrac{2}{3}x = \dfrac{216}{5} - \dfrac{6}{5} x \\ \\ \large{⇒} \dfrac{2}{3} x + \dfrac{6}{5} x = \dfrac{216}{5} \\ \\\large{⇒} \dfrac{10x + 18x}{15} = \dfrac{216}{5} \\ \\ \large{⇒}28x = \dfrac{216 \times 15}{5} \\ \\\large{⇒} x = \dfrac{216\times 15}{5 \times 28} \\ \\ \: \large{⇒} x = \dfrac{162}{7} \\ \\\large{⇒} x = 23 \dfrac{1}{7}$

$\large{⇒}$Second Number = 36 - $\dfrac{162}{7}$

$\large{⇒}$Second Number = $\dfrac{252-162}{7}$

$\large{⇒}$Second Number = $\dfrac{90}{7}$

$\large{⇒}$Second Number = 12$\dfrac{6}{7}$

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