Question

Rezolvați prin metoda reducerii și metoda substituție:

A) 2x-3y=-1 
x+3y=4

B) -3x+y=5
x-y=7

C) x-2y+4=0
2x+y=-1

D) x+2y=0
2x+y=1

E) x=2
2x-3y=1

Vă rog să mă ajutați!!!!!!! URGENT!!

Answer 1

$a)\left\{\begin{matrix} 2x - 3y = - 1\\ x + 3y = 4\end{matrix}\right.$

$2x + x = - 1 + 4$

$3x = 3$

$x = \frac{3}{3}$

$x = 1$

$x + 3y = 4$

$1 + 3y = 4$

$3y = 4 - 1$

$3y = 3$

$y = \frac{3}{3}$

$y = 1$

$b)\left\{\begin{matrix} - 3x + y = 5\\ x - y = 7\end{matrix}\right.$

$- 3x + x = 5 + 7$

$- 2x = 12 \: | \div ( - 2)$

$x = - 6$

$x - y = 7$

$- 6 - y = 7$

$- y = 7 + 6$

$- y = 13 \: | \times ( - 1)$

$y = - 13$

$c)\left\{\begin{matrix} x - 2y + 4 = 0\\ 2x + y = - 1\end{matrix}\right.$

$\left\{\begin{matrix} x - 2y = - 4\\ 2x + y = - 1 \: | \times 2\end{matrix}\right.$

$\left\{\begin{matrix} x - 2y = - 4\\ 4x + 2y = - 2\end{matrix}\right.$

$x + 4x = - 4 - 2$

$5x = - 6$

$x = - \frac{6}{5}$

$2x + y = - 1$

$2 \times ( - \frac{6}{5} ) + y= - 1$

$- \frac{12}{5} + y = - 1 \: | \times 5$

$- 12 + 5y = - 5$

$5y = - 5 + 12$

$5y = 7$

$y = \frac{7}{5}$

$d)\left\{\begin{matrix} x + 2y = 0 \: | \times ( - 2)\\ 2x + y = 1\end{matrix}\right.$

$\left\{\begin{matrix} - 2x - 4y = 0\\ 2x + y = 1\end{matrix}\right.$

$- 4y + y = 0 + 1$

$- 3y = 1 \: | \times ( - 1)$

$3y = 1$

$y = \frac{1}{3}$

$2x + y = 1$

$2x + \frac{1}{3} = 1 \: | \times 3$

$6x + 1 = 3$

$6x = 3 - 1$

$6x = 2$

$x = \frac{2}{6}$

$x = \frac{1}{3}$

$e)\left\{\begin{matrix} x = 2\\ 2x - 3y = 1\end{matrix}\right.$

$2 \times 2 - 3y = 1$

$4 - 3y = 1$

$- 3y = 1 - 4$

$- 3y = - 3 \: | \div ( - 3)$

$y = 1$

Answer 2

poza
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