Вправа 6.64 гдз 7 клас алгебра Біляніна Білянін Андрух

№ 6.64
Розв'язати систему рівнянь: $$
\begin{aligned}
& \text { a) }\left\{\begin{array}{l}
\frac{(x+3)}{2}+\frac{(y-2)}{3}=2, / \cdot 6 \\
\frac{(x-1)}{4}-\frac{(y+1)}{3}=4 ; / \cdot 12
\end{array}\right. \\
& \left\{\begin{array}{l}
3 \cdot(x+3)+2 \cdot(y-2)=12 \\
3 \cdot(x-1)-4 \cdot(y+1)=48
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+9+2 y-4=12 \\
3 x-3-4 y-4=48
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+2 y=12-9+4 \\
3 x-4 y=48+3+4
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+2 y=7 \cdot / \cdot(-1) \\
3 x-4 y=55
\end{array}\right. \\
& \left\{\begin{array}{l}
-3 x-2 y=-7 \\
3 x-4 y=55
\end{array}\right.
\end{aligned}
$$
\begin{aligned}
& -3 x-2 y+3 x-4 y=-7+55, \\
& -6 y=48 \text {, } \\
& y=-8 \text {, } \\
& 3 x+2 \cdot(-8)=7 \text {, } \\
& 3 x-16=7 \text {, } \\
& 3 x=23 \text {, } \\
& x=\frac{23}{3} \text {. } \\
& \left(\frac{23}{3} ;-8\right) \text { - розв’язок системи. } \\
& \text { б) }\left\{\begin{array}{l}
\frac{(x+y)}{2}+\frac{(x-y)}{3}=6, / \cdot 6 \\
\frac{(x+y)}{4}-\frac{(x-y)}{3}=6 . / \cdot 12
\end{array}\right. \\
& \left\{\begin{array}{l}
3 \cdot(x+y)+2 \cdot(x-y)=36, \\
3 \cdot(x+y)-4 \cdot(x-y)=72,
\end{array}\right. \\
& \begin{array}{l}
\left\{\begin{array}{l}
3 x+3 y+2 x-2 y \\
3 x+3 y-4 x+4 y
\end{array}\right. \\
\left\{\begin{array}{l}
5 x+y=36, \\
-x+7 y=72, / \cdot 5
\end{array}\right.
\end{array} \\
& \left\{\begin{array}{l}
5 x+y=36 \\
-5 x+35 y=360,
\end{array}\right. \\
& 5 x+y-5 x+35 y=36+360 \text {, } \\
& 36 y=396 \text {, } \\
& y=11 \text {, } \\
& -x+7 \cdot 11=72 \text {, } \\
& -x=72-77 \text {, } \\
& -x=-5 \text {, } \\
& x=5 \text {. } \\
& (5 ; 11) \text { - розв’язок системи. }
\end{aligned}