Question

Giải hệ phương trình:

\(\left\{{}\begin{matrix}3x-2y=10\\x-\dfrac{3}{2}y=3\dfrac{1}{3}\end{matrix}\right.\)

Ps: \(3\dfrac{1}{3}\) là hỗn số nha

Answer 1

Giải:

\(3x-2y=10\)

\(\Leftrightarrow3x=10+2y\)

\(\Leftrightarrow x=\dfrac{10+2y}{3}\)

Lại có:

\(x-\dfrac{3}{2}y=3\dfrac{1}{3}\Leftrightarrow x-\dfrac{3}{2}y=\dfrac{10}{3}\)

\(\Leftrightarrow\dfrac{10+2y}{3}-\dfrac{3}{2}y=\dfrac{10}{3}\)

\(\Leftrightarrow\dfrac{10+2y}{3}-\dfrac{3y}{2}=\dfrac{10}{3}\)

\(\Leftrightarrow\dfrac{20+4y}{6}-\dfrac{9y}{6}=\dfrac{20}{6}\)

\(\Leftrightarrow20+4y-9y=20\)

\(\Leftrightarrow20-5y=20\)

\(\Leftrightarrow y=0\)

\(\Leftrightarrow x=\dfrac{10+2.0}{3}=\dfrac{10}{3}\)

Vậy ...

Answer 2

x=\(\dfrac{10}{3}+\dfrac{3}{2}y\)

thay vào 3x-2y=10 ta đc

\(3\left(\dfrac{10}{3}+\dfrac{3}{2}y\right)-2y=10\)

<=> \(10+\dfrac{9}{2}y-2y=10\)

<=> \(\dfrac{5}{2}y=0\Leftrightarrow y=0\)

=> x=10/3