№ 1203 Алгебра = № 55.11 Математика
Розв’яжіть систему рівнянь:
1) $\begin{cases} 4x + 3y = 0, \\ 5x − 7y = −43; \end{cases}$
2) $\begin{cases} 2x + 9y = −59, \\ 5x − 4y = 38; \end{cases}$
3) $\begin{cases} 3p − 7q = 0, \\ 2p + 9q = 41; \end{cases}$
4) $\begin{cases} 6a − 7b = 51, \\ 2a + 3b = −15. \end{cases}$
Розв'язок:
1) $\begin{cases} 4x + 3y = 0, \\ 5x – 7y = –43; \end{cases}$
$\begin{cases} 4x = –3y, \\ 5x – 7y = –43; \end{cases}$
$\begin{cases} x = –\frac{3}{4}y, \\ 5(–\frac{3}{4}y) – 7y = –43 | · (–4); \end{cases}$
$\begin{cases} x = –\frac{3}{4}y, \\ 15y + 28y = 172; \end{cases}$
$\begin{cases} x = –\frac{3}{4}y, \\ 43y = 172; \end{cases}$ $\begin{cases} x = –\frac{3}{4}y, \\ y = 4; \end{cases}$
$\begin{cases} x = –\frac{3}{4} · 4, \\ y = 4; \end{cases}$ $\begin{cases} x = –3, \\ y = 4. \end{cases}$
2) $\begin{cases} 2x + 9y = –59, \\ 5x – 4y = 38; \end{cases}$ $\begin{cases} 2x = –59 – 9y, \\ 5x – 4y = 38; \end{cases}$
$\begin{cases} x = \frac{–59 – 9y}{2}, \\ 5\frac{–59 – 9y}{2} – 4y = 38 · (–2); \end{cases}$
$\begin{cases} x = \frac{–59 – 9y}{2}, \\ 5(59 + 9y) + 8y = –76, \end{cases}$
$\begin{cases} x = \frac{–59 – 9y}{2}, \\ 295 + 45y + 8y = –76; \end{cases}$
$\begin{cases} x = \frac{–59 – 9y}{2}, \\ 53y = –371; \end{cases}$ $\begin{cases} x = \frac{–59 – 9y}{2}, \\ y = –7; \end{cases}$
$\begin{cases} x = \frac{–59 – 9(–7)}{2}, \\ y = –7; \end{cases}$ $\begin{cases} x = 2, \\ y = –7. \end{cases}$
3) $\begin{cases} 3p – 7q = 0, \\ 2p + 9q = 41; \end{cases}$ $\begin{cases} 3p – 7q = 0, \\ 2p = 41 – 9q; \end{cases}$
$\begin{cases} 3(20{,}5 – 4{,}5q) – 7q = 0, \\ p = 20{,}5 – 4{,}5q; \end{cases}$ $\begin{cases} 61{,}5 – 13{,}5q – 7q = 0, \\ p = 20{,}5 – 4{,}5q; \end{cases}$
$\begin{cases} –20{,}5q = –61{,}5, \\ p = 20{,}5 – 4{,}5q; \end{cases}$ $\begin{cases} q = 3, \\ p = 20{,}5 – 4{,}5 · 3; \end{cases}$ $\begin{cases} q = 3, \\ p =7. \end{cases}$
4) $\begin{cases} 6a – 7b = 51, \\ 2a + 3b = –15; \end{cases}$ $\begin{cases} 6a – 7b = 51, \\ 2a = –15 – 3b; \end{cases}$
$\begin{cases} 6a – 7b = 51, \\ a = \frac{–15 – 3b}{2}; \end{cases}$
$\begin{cases} 6\frac{–15 – 3b}{2} – 7b = 51 | · (–2), \\ a = \frac{–15 – 3b}{2}; \end{cases}$
$\begin{cases} 6(15 + 3b) + 14b = –102, \\ a = \frac{–15 – 3b}{2}; \end{cases}$
$\begin{cases} 90 + 18b + 14b = –102, \\ a = \frac{–15 – 3b}{2}; \end{cases}$
$\begin{cases} 32b = –192, \\ a = \frac{–15 – 3b}{2}; \end{cases}$ $\begin{cases} b = –6, \\ a = \frac{–15 – 3 – 6}{2}; \end{cases}$ $\begin{cases} b = –6, \\ a = 1{,}5. \end{cases}$
Відповідь:
1) (–3 ; 4);
2) (2; –7);
3) ($\frac{41}{11}$; $\frac{123}{77}$);
4) (1,5; –6).