Đáp án:

d. \(\left\{ \begin{array}{l}
y = 144\\
x = 120\\
z = 126
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left\{ \begin{array}{l}
x =  - 42 - y\\
2\left( { - 42 - y} \right) = 5y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x =  - 42 - y\\
 - 84 - 2y = 5y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x =  - 42 - y\\
7y =  - 84
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  - 12\\
x =  - 30
\end{array} \right.\\
b.DK:y \ne 0\\
\left\{ \begin{array}{l}
x = \dfrac{5}{9}y\\
3.\dfrac{5}{9}y - 2y = 12
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - 1}}{3}y = 12\\
x = \dfrac{5}{9}y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  - 36\\
x =  - 20
\end{array} \right.\\
c.\left\{ \begin{array}{l}
x = \dfrac{3}{2}y\\
z = \dfrac{3}{5}y\\
\dfrac{3}{2}y - y + \dfrac{3}{5}y =  - 22
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{3}{2}y\\
z = \dfrac{3}{5}y\\
\dfrac{{11}}{{10}}y =  - 22
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  - 20\\
x =  - 30\\
z =  - 12
\end{array} \right.\\
d.\left\{ \begin{array}{l}
x = \dfrac{5}{6}y\\
z = \dfrac{7}{8}y\\
\dfrac{5}{6}y + y - \dfrac{7}{8}y = 138
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{5}{6}y\\
z = \dfrac{7}{8}y\\
\dfrac{{23}}{{24}}y = 138
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 144\\
x = 120\\
z = 126
\end{array} \right.\\
e.\left\{ \begin{array}{l}
2x - 2 = y - 2\\
3y - 6 = 2z - 6\\
x + y + z = 24
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{y}{2}\\
z = \dfrac{3}{2}y\\
\dfrac{y}{2} + y + \dfrac{3}{2}y = 24
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
3y = 24\\
x = \dfrac{y}{2}\\
z = \dfrac{3}{2}y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 8\\
x = 4\\
z = 12
\end{array} \right.\\
f.\left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
z = \dfrac{4}{3}y\\
\dfrac{4}{9}{y^2} + {y^2} + \dfrac{{16}}{9}{y^2} = 29
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
z = \dfrac{4}{3}y\\
{y^2} = 9
\end{array} \right.\\
 \to \left[ \begin{array}{l}
y = 3\\
y =  - 3
\end{array} \right. \to \left[ \begin{array}{l}
x = 2;z = 4\\
x =  - 2;z =  - 4
\end{array} \right.
\end{array}\)