Алгебра 7 класс контрольные работы Виленкин ответы – страница 55

а) $$ \begin{gathered} +\left\{\begin{array}{c} 3 x-5 y=11 \\ 4 x+5 y=3 \end{array}\right. \\ \hline 7 x=14 \\ x=2 \end{gathered} $$ ⇒ $$ \left\{\begin{array}{c} 3 x-5 y=11 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 3·2 -5 y=11 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 6 -5 y=11 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 5 y=6 - 11 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 5 y= - 5 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} y= - 1 \\ x=2 \end{array}\right. $$
    Ответ: (2; – 1).
б) $$ \left\{\begin{array}{c} 4 x+y=6 \mid \cdot 3 \\ 2 x-3 y=13 \end{array}\right. $$ ⇒ $$ \begin{gathered} +\left\{\begin{array}{c} 12 x+3 y=18 \\ 2 x-3 y=13 \end{array}\right. \\ \hline 14 x=31 \\ x=2 \frac{3}{14} \end{gathered} $$ ⇒ $$ \left\{\begin{array}{c} 4 x+y=6 \\ x=2 \frac{3}{14} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 4 \cdot 2 \frac{3}{14}+y=6 \\ x=2 \frac{3}{14} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} \frac{62}{7}+y=6 \\ x=2 \frac{3}{14} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} y=6-8 \frac{6}{7} \\ x=2 \frac{3}{14} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{l} y=-2 \frac{6}{7} \\ x=2 \frac{3}{14} \end{array}\right. $$
    Ответ: (2 3/14; – 2 6/7).

y = kx + b — общий вид линейного уравнения Подставим в него точки пересечения.
$$ \left\{\begin{array}{c} \mathrm{k} \cdot 2+\mathrm{b}=0 \\ \mathrm{k} \cdot 0+\mathrm{b}=-7 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 2 \mathrm{k}=-\mathrm{b} \\ \mathrm{~b}=-7 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{l} 2 \mathrm{k}=7 \\ \mathrm{~b}=-7 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{l} \mathrm{k}=3,5 \\ \mathrm{~b}=-7 \end{array}\right. $$
у = 3,5х – 7

а) $$ \begin{gathered} +\left\{\begin{array}{c} 4 x+7 y=1 \\ 5 x-7 y=17 \end{array}\right. \\ \hline 9 x=18 \\ x=2 \end{gathered} $$ ⇒ $$ \left\{\begin{array}{c} 4 x+7 y=1 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 4 · 2+7 y=1 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 8+7 y=1 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 7 y=1 - 8 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 7 y= -7 \\ x=2 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} y= - 1 \\ x=2 \end{array}\right. $$
Ответ: (2; – 1).
б) $$ \left\{\begin{array}{c} 5 x-2 y=6 \mid \cdot 2 \\ 3 x+4 y=-1 \end{array}\right. $$ ⇒ $$ \begin{gathered} +\left\{\begin{array}{c} 10 x-4 y=12 \\ 3 x+4 y=-1 \end{array}\right. \\ \hline 13 x=11 \\ x=\frac{11}{13} \end{gathered} $$ ⇒ $$ \left\{\begin{array}{c} 5 x-2 y=6 \\ x=\frac{11}{13} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} 5 \cdot \frac{11}{13}-2 y=6 \\ x=\frac{11}{13} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} \frac{55}{26}-y=3 \\ x=\frac{11}{13} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} y=2 \frac{3}{26}-3 \\ x=\frac{11}{13} \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} y=-\frac{23}{26} \\ x=\frac{11}{13} \end{array}\right. $$
Ответ: (11/13; – 23/26).

y = kx + b — общий вид линейного уравнения Подставим в него точки пересечения.
$$ \left\{\begin{array}{c} \mathrm{k} \cdot(-5)+\mathrm{b}=0 \\ \mathrm{k} \cdot 0+\mathrm{b}=-3 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} -5 \mathrm{k}=-\mathrm{b} \\ \mathrm{~b}=-3 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} -5 \mathrm{k}=3 \\ \mathrm{~b}=-3 \end{array}\right. $$ ⇒ $$ \left\{\begin{array}{c} k=-0,6 \\ b=-3 \end{array}\right. $$
у = – 0,6х – 3