Đáp án:
6) \(\left[ \begin{array}{l}
y = 15\\
y = - 15
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = - 9
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left\{ \begin{array}{l}
x = \dfrac{9}{{11}}y\\
\dfrac{9}{{11}}y + y = 60
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{9}{{11}}y\\
\dfrac{{20}}{{11}}y = 60
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{9}{{11}}y\\
y = 33
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 27\\
y = 33
\end{array} \right.\\
2)\left\{ \begin{array}{l}
x = \dfrac{4}{7}y\\
y - \dfrac{4}{7}y = 24
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{3}{7}y = 24\\
x = \dfrac{4}{7}y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 56\\
x = 32
\end{array} \right.\\
3)\left\{ \begin{array}{l}
x = \dfrac{8}{{12}}y = \dfrac{2}{3}y\\
z = \dfrac{{15}}{{12}}y = \dfrac{5}{4}y\\
\dfrac{2}{3}y + y - \dfrac{5}{4}y = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
z = \dfrac{5}{4}y\\
\dfrac{5}{{12}}y = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 24\\
x = 16\\
z = 30
\end{array} \right.\\
4)\left\{ \begin{array}{l}
x = \dfrac{8}{5}y\\
z = \dfrac{8}{{20}}y = \dfrac{2}{5}y\\
\dfrac{8}{5}y - y - \dfrac{2}{5}y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{8}{5}y\\
z = \dfrac{2}{5}y\\
\dfrac{1}{5}y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 15\\
x = 24\\
z = 6
\end{array} \right.\\
5)\left\{ \begin{array}{l}
3x - 3 = 2y - 4\\
4y - 8 = 3z - 9\\
x - 2y + 3z = 14
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2y - 1}}{3}\\
z = \dfrac{{4y + 1}}{3}\\
\dfrac{{2y - 1}}{3} - 2y + 3.\dfrac{{4y + 1}}{3} = 14
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2y - 1}}{3}\\
z = \dfrac{{4y + 1}}{3}\\
2y - 1 - 6y + 12y + 3 = 42
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 5\\
x = 3\\
z = 7
\end{array} \right.\\
6)\left\{ \begin{array}{l}
x = \dfrac{3}{5}y\\
\dfrac{3}{5}y.y = 135
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{y^2} = 225\\
x = \dfrac{3}{5}y
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 15\\
y = - 15
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = - 9
\end{array} \right.
\end{array}\)
