Question

giải pt \(\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\\dfrac{6a+14b}{9}=\dfrac{9}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\6a+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\6\left(\dfrac{5}{6}-b\right)+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\5-6b+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\8b=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{1}{3}\\b=\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

\(\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{14a}{9}+\dfrac{14b}{9}=\dfrac{35}{27}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{9}a=\dfrac{8}{27}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2.\dfrac{1}{3}}{3}+\dfrac{14b}{9}=1\)

\(\Leftrightarrow\dfrac{14b}{9}=1-\dfrac{2}{9}=\dfrac{7}{9}\)

\(\Leftrightarrow b=\dfrac{7}{9}:\dfrac{14}{9}=\dfrac{1}{2}\)

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Answer 3

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\6a+14b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6a+6b=5\\6a+14b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{2}\\a=\dfrac{1}{3}\end{matrix}\right.\)