Вправа 1095 алгебра Бевз гдз 7 клас
№ 1095 $$\begin{aligned}
& \text { a) }\left\{\begin{array}{c}
3 x+1=5(3-2 y) \\
4(x-1)=2(8,5-5 y)
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+1=15-10 y \\
4 x-4=17-10 y
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+10 y=14 \\
4 x=21-10 y
\end{array}\right. \\
& \left\{\begin{array}{c}
x=5,25-2,5 y \\
3 \cdot(5,25-2,5 y)+10 y=14
\end{array}\right.
\end{aligned}
$$
$$
\begin{aligned}
& 15,75-7,5 y+10 y=14 \\
& 2,5 y=14-15,75 \\
& y=-1,75: 2,5 \\
& y=-0,7 \\
& x=5,25-2,5 \cdot(-0,7) \\
& x=5,25+1,75 \\
& x=7
\end{aligned}
$$
Відповідь: (7; -0,7) $$
\begin{aligned}
& \text { б) }\left\{\begin{array}{c}
6(x-2 y)=7-9 y \\
8 x+3 y=5(2 x+1)
\end{array}\right. \\
& \left\{\begin{array}{l}
6 x-12 y=7-9 y \\
8 x+3 y=10 x+5
\end{array}\right. \\
& \left\{\begin{array}{l}
6 x-3 y=7 \\
3 y-2 x=5 ;
\end{array}\right. \\
& \left\{\begin{array}{l}
2 x=3 y-5 \\
6 x-3 y=7 ;
\end{array}\right. \\
& \left\{\begin{array}{c}
x=1,5 y-2,5 \\
6 \cdot(1,5 y-2,5)-3 y=7
\end{array}\right.
\end{aligned}
$$
$$
\begin{aligned}
& 9 y-15-3 y=7 \\
& 6 y=22 \\
& y=22: 6 \\
& y=3 \frac{2}{3} \\
& x=\frac{3}{2} \cdot \frac{11}{3}-2,5 \\
& x=5,5-2,5 \\
& x=3 \\
& \text { Відповідь: }\left(3 ; 3 \frac{2}{3}\right)
\end{aligned}
$$
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