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МатематикаГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1109
Задание 1109
\[\boxed{\text{1109.}\text{\ }еуроки - ответы\ на\ пятёрку}\]
\[\textbf{а)}\ \left\{ \begin{matrix} \frac{1}{3}x + \frac{1}{4}y - 2 = 0\ \ | \cdot 12 \\ 5x - y = 11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 4x + 3y - 24 = 0 \\ 5x - y = 11\ \ \ | \cdot 3 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 4x + 3y = 24 \\ 15x - 3y = 33 \\ \end{matrix} \right.\ ( + )\]
\[\left\{ \begin{matrix} 19x = 57 \rightarrow x = 3 \\ y = 5x - 11\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[y = 5 \cdot 3 - 11\]
\[y = 15 - 11 = 4\]
\[Ответ:(3;4).\]
\[\textbf{б)}\ \left\{ \begin{matrix} 0,5x + 0,2y = 7\ \ | \cdot 10 \\ \frac{1}{3}x - \frac{1}{10}y = 0\ \ \ | \cdot 30 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 5x + 2y = 70\ \ | \cdot ( - 2) \\ 10x - 3y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} - 10x - 4y = - 140 \\ 10x - 3y = 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( + )\]
\[\left\{ \begin{matrix} - 7y = - 140 \rightarrow y = 20\ \ \ \ \ \\ 10x = 3 \cdot 20 \rightarrow x = \frac{60}{10} = 6 \\ \end{matrix}\ \right.\ \]
\[Ответ:(6;20).\]
\[\textbf{в)}\ \left\{ \begin{matrix} \frac{1}{5}m - \frac{1}{6}n = 0\ \ | \cdot 30 \\ 5m - 4n = 2\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 6m - 5n = 0\ \ | \cdot ( - 4) \\ 5m - 4n = 2\ \ \ \ \ \ \ \ \ | \cdot 5 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} - 24m + 20n = 0 \\ 25m - 20n = 10 \\ \end{matrix} \right.\ ( + )\]
\[\left\{ \begin{matrix} m = 10\ \ \ \ \ \ \ \ \\ 4n = 5m - 2 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} m = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ n = \frac{5 \cdot 10 - 2}{4} = \frac{48}{4} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} m = 10 \\ n = 12 \\ \end{matrix} \right.\ \]
\[Ответ:(10;12).\]
\[\textbf{г)}\ \left\{ \begin{matrix} \frac{1}{6}u - \frac{1}{3}v = - 3\ \ \ \ \ \ \ \ | \cdot 6 \\ 0,2u + 0,1v = 3,9\ \ | \cdot 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} u - 2v = - 18\ \ \ \ \ \ \\ 2u + v = 39\ \ | \cdot 2 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} u - 2v = - 18 \\ 4u + 2v = 78 \\ \end{matrix} \right.\ ( + )\]
\[\left\{ \begin{matrix} 5u = 60 \rightarrow u = 12 \\ v = 39 - 2u\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[v = 39 - 2 \cdot 12 = 39 - 24\]
\[v = 15\]
\[Ответ:(12;15).\]
