Вправа 6.60 гдз 7 клас алгебра Біляніна Білянін Андрух

№ 6.60
Розв'язати систему рівнянь: $$
\begin{aligned}
& \text { a) }\left\{\begin{array}{l}
2 x-3 y=-1, / \cdot 4 \\
3 x+4 y=24 ; / \cdot 3
\end{array}\right. \\
& \left\{\begin{array}{l}
8 x-12 y=-4 \\
9 x+12 y=72,
\end{array}\right. \\
& 8 x-12 y+9 x+12 y=-4+72, \\
& 17 x=68, \\
& x=4, \\
& 2 \cdot 4-3 y=-1, \\
& -3 y=-9 \\
& y=3
\end{aligned}
$$ (4; 3) - розв’язок системи. $$
\begin{aligned}
& \text { б) }\left\{\begin{array}{l}
3 x+2 y=2, \\
x+4 y=-2 ; /-(-3)
\end{array}\right. \\
& \left\{\begin{array}{l}
3 x+2 y=2, \\
-3 x-12 y=6,
\end{array}\right. \\
& 3 x+2 y-3 x-12 y=2+6 \text {, } \\
& -10 y=8 \text {, } \\
& y=-0,8 \text {, } \\
& x+4 \cdot(-0,8)=-2 \text {, } \\
& x-3,2=2 \text {, } \\
& \mathrm{x}=1,2 \text {. }
\end{aligned}
$$ (1,2; -0,8) - розв'язок системи. $$
\begin{aligned}
& \text { в) }\left\{\begin{array}{l}
3 x-4 y=11, / \cdot 4 \\
4 x+6 y=26 ; / \cdot(-3)
\end{array}\right. \\
& \left\{\begin{array}{l}
12 x-16 y=44, \\
-12 x-18 y=-78,
\end{array}\right. \\
& -12 x-18 y=-78, \\
& 12 x-16 y-12 x-18 y=44-78, \\
& -34 y=-34,
\end{aligned}
$$
у = 1,
3х – 4 • 1 = 11,
3х = 15,
х = 5.
(5; 1) – розв’язок системи.
\begin{aligned}
& \text { г) }\left\{\begin{array}{l}
4 x-5 y=-22, / \cdot 2 \\
3 x+2 y=18 . / \cdot 5
\end{array}\right. \\
& \left\{\begin{array}{l}
8 x-10 y=-44, \\
15 x+10 y=90,
\end{array}\right.
\end{aligned}
8х – 10у + 15х + 10у = –44 + 90,
23х = 46,
х = 2,
3 • 2 + 2у = 18,
2у = 18 – 6,
2у = 12,
у = 6.
(2; 6) – розв’язок системи.