Вправа 1111 алгебра Бевз гдз 7 клас

№ 1111
a) $\left\{\begin{array}{c}x-4 y=5 \\ x-3 y=12\end{array}\right.$ $$
\begin{aligned}
& -y=-7 \\
& y=7 \\
& x-4 \cdot 7=5 \\
& x=5+28 \\
& x=33 ;
\end{aligned}
$$ $$
\begin{aligned}
& \text { б) }\left\{\left.\begin{array}{c}
3 x-y=1 \\
3 x+8 y=19
\end{array} \right\rvert\, \cdot(-1)\right. \\
& \left\{\begin{array}{c}
-3 x+y=-1 \\
3 x+8 y=19
\end{array}\right.
\end{aligned}
$$ Якщо $y=2$, то $3 x-2=1$ $$
\begin{aligned}
& 3 x=3 \\
& x=1
\end{aligned}
$$ Відповідь: $(1 ; 2)$ $$
\begin{aligned}
& \text { в) }\left\{\begin{array}{c}
15 y-8 z=29 \\
3 y+2 z=13
\end{array}+\right. \\
& 27 y=81 \\
& y=81: 27 \\
& y=3 \\
& 12 \cdot 3+8 z=52 \\
& 8 z=52-36 \\
& 8 z=16 \\
& z=2
\end{aligned}
$$ $$
\begin{aligned}
& \text { г) }\left\{\begin{array}{l}
3 x+8 t=29 \\
6 x+5 y=47 ;
\end{array}+\right. \\
& -11 t=-11 \\
& t=1 \\
& 6 x+5 \cdot 1=47 \\
& 6 x=47-5 \\
& 6 x=42 \\
& x=7
\end{aligned}
$$