Question

Решить методом подстановки систему уравнений
1) Х+2у=4
3х-4у=2
2) 3х+у=4
5х-2у=14
3) 2х+6у=10
4х-у=7

Answer 1

Объяснение:

1)

$\left \{ \begin{array}{lcl} {{x+2y=4,} \\ {3x-4y=2;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=4-2y,} \\ {3(4-2y)-4y=2;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=4-2y,} \\ {12-6y-4y=2;}} \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ \begin{array}{lcl} {{x=4-2y,} \\ {-10y=-10;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=4-2*1,} \\ {y=1;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=2,} \\ {y=1.}} \end{array} \right.$

Ответ: (2;1 )

2)

$\left \{ \begin{array}{lcl} {{3x+y=4,} \\ {5x-2y=14;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=4-3x,} \\ {5x-2(4-3x)=14;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=4-3x,} \\ {5x-8+6x=14;}} \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ \begin{array}{lcl} {{y=4-3x,} \\ {11x=22;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=4-3*2,} \\ {x=2;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=-2,} \\ {x=2.}} \end{array} \right.$

Ответ: (2;- 2 )

3)

$\left \{ \begin{array}{lcl} {{2x+6y=10,} \\ {4x-y=7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{2x+6(4x-7)=10,} \\ {y=4x-7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{2x+24x-42=10,} \\ {y=4x-7;}} \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ \begin{array}{lcl} {{26x=52,} \\ {y=4x-7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=2,} \\ {y=4*2-7;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{x=2,} \\ {y=1.}} \end{array} \right.$

Ответ: (2; 1 )