Math, asked by ishu9354, 7 months ago

Solve y/5 + 1 = 1/5 and check the answer​

Answers

Answered by ƦαíηвσωStαƦ
17

{\mathbf {\blue{S}{\underline{\underline{olution:-}}}}}

\mathfrak{\underline{Given:-}}

\:\:\:\:\: \sf {\dfrac{y}{5} + 1 = \dfrac{1}{15} }

{\mathbf {\blue{E}{\underline{\underline{xplanation:-}}}}}

\longrightarrow \sf {\dfrac{y}{5} + 1-1 = \dfrac{1}{15} - 1} \\\\

\longrightarrow \sf {\dfrac{y}{5} = \dfrac{-14}{15} } \\\\

\longrightarrow \sf {\dfrac{y}{5} \times 5 = \dfrac{-14}{15} \times 5} \\\\

\longrightarrow \sf {y = \dfrac{-14}{3} } \\\\

\:\:\:\:\dag\bf{\underline{\underline \pink{Check\: For\: y:-}}}

\longrightarrow \sf {y = \dfrac{-14}{3} } \\\\

\longrightarrow \sf {L.H.S. = \dfrac{-14}{15} + 1 = \dfrac{1}{15} } \\\\

\longrightarrow \sf {R.H.S. = \dfrac{1}{15} } \\\\

\:\:\:\:\dag\bf{\underline{\underline \pink{So:-}}}

\sf\underline{\purple{\:\:\:\:\: L.H.S. = R.H.S.\:\:\:\:\:\:}}

\:\:\:\:\;\:\;\footnotesize\bold{\underline{\underline{\sf{\red{Hence:-}}}}}

  •  \sf {y = \dfrac{-14}{3} } is a solution or root of the given equation.

\rule{200}{2}

Answered by Anonymous
2

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