№55
Вопрос
Номер 1
Решите систему уравнений:
а) $$
\left\{\begin{array}{c}
3 x-5 y=11 \\
4 x+5 y=3
\end{array}\right.
$$
б) $$
\left\{\begin{array}{c}
4 x+y=6 \\
2 x-3 y=13
\end{array}\right.
$$
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 ).
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 ).
3/14
6/7
Вопрос
Номер 2
График линейной функции пересекает оси координат в точках (2; 0) и (0; – 7). Задайте эту функцию формулой.
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
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
Вопрос
Номер 1
Решите систему уравнений:
а) $$
\left\{\begin{array}{c}
4 x+7 y=1 \\
5 x-7 y=17
\end{array}\right.
$$
б) $$
\left\{\begin{array}{c}
5 x-2 y=6 \\
3 x+4 y=-1
\end{array}\right.
$$
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 ).
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 ).
11/13
23/26
Вопрос
Номер 2
График линейной функции пересекает оси координат в точках (– 5; 0) и (0; – 3). Задайте эту функцию формулой.
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
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
