Вправа 6.47 гдз 7 клас алгебра Біляніна Білянін Андрух

№ 6.47
Розв'язати систему рівнянь: $$
\begin{aligned}
& \text { a) }\left\{\begin{array}{l}
\frac{x}{2}-\frac{y}{3}=2, / \cdot 6 \\
5 x-y=34
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x-2 y=12, \\
5 x-y=34,
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x-2 \cdot(5 x-34)=12, \\
y=5 x-34,
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x-10 x+68=12, \\
y=5 x-34,
\end{array}\right. \\
& \left\{\begin{array}{l}
-7 x=-56, \\
y=6 x-34,
\end{array}\right. \\
& \left\{\begin{array}{l}
x=8 \\
y=5 \cdot 8-34,
\end{array}\right. \\
& \left\{\begin{array}{l}
x=8 \\
y=6
\end{array}\right.
\end{aligned}
$$
\begin{aligned}
&\begin{aligned}
& \text { б) }\left\{\begin{array}{l}
6 y-5 x=1, \\
\frac{(x-1)}{2}+\frac{(3 y-x)}{4}=-4 \frac{3}{4} .
\end{array}\right. \\
& \left\{\begin{array}{l}
6 y-5 x=1 \\
\frac{(x-1)}{2}+\frac{(3 y-x)}{4}=-\frac{19}{4}, / \cdot 4
\end{array}\right. \\
& \left\{\begin{array}{l}
6 y-5 x=1 \\
2 x-2+3 y-x=-19,
\end{array}\right. \\
& \left\{\begin{array}{l}
6 y-5 x=1 \\
x+3 y=-17
\end{array}\right. \\
& \left\{\begin{array}{l}
2 \cdot(-x-17)-5 x=1 \\
3 y=-x-17
\end{array}\right. \\
& \left\{\begin{array}{l}
-2 x-34-5 x=1 \\
3 y=-x-17
\end{array}\right. \\
& \left\{\begin{array}{l}
-7 x=-35 \\
3 y=-x-17,
\end{array}\right. \\
& \left\{\begin{array}{l}
x=-5 \\
3 y=5-17,
\end{array}\right. \\
& \left\{\begin{array}{l}
x=-5, \\
3 y=-12,
\end{array}\right. \\
& \left\{\begin{array}{l}
x=-5 \\
y=-4 .
\end{array}\right.
\end{aligned}\\
&\text { Відповідь: (-5; }-4) \text {. }
\end{aligned}