Question

Rezolvarea sistemelor prin metoda subtitutiei


a) {x=2
{3x-5y=1

b) {y=-1
{-2x-4y=0

c) {x+y=5
{x-y=-1

d) {-x-5y=4
{3x-2y=5

e) {3x-5y=-4
{6x-2y=8

f) {2x+5y=1
{3x+10y=-1

Answer 1

$a)\left\{\begin{matrix} x = 2\\ 3x - 5y = 1\end{matrix}\right.$

$3x - 5y = 1$

$3 \times 2 - 5y = 1$

$6 - 5y = 1$

$- 5y = 1 - 6$

$- 5y = - 5 \: | \div ( - 5)$

$y = 1$

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

$- 2x - 4y = 0$

$- 2x - 4 \times ( - 1) = 0$

$- 2x + 4 = 0$

$- 2x = - 4 \: | \div ( - 2)$

$x = 2$

$c)\left\{\begin{matrix} x + y = 5\\ x - y = - 1\end{matrix}\right.$

$x + y = 5$

$x = 5 - y$

$x - y = - 1$

$5 - y - y = - 1$

$- y - y = - 1 - 5$

$- 2y = - 6 \: | \div ( - 2)$

$y = 3$

$x + y = 5$

$x + 3 = 5$

$x = 5 - 3$

$x = 2$

$d)\left\{\begin{matrix} - x - 5y = 4\\ 3x - 2y = 5\end{matrix}\right.$

$- x - 5y = 4$

$- x = 4 + 5y \: | \times ( - 1)$

$x = - 4 - 5y$

$3x - 2y = 5$

$3( - 4 - 5y) - 2y = 5$

$- 12 - 15y - 2y = 5$

$- 15y - 2y = 5 + 12$

$- 17y = 17 \: | \div ( - 17)$

$y = - 1$

$x = - 4 - 5y$

$x = - 4 - 5 \times ( - 1)$

$x = - 4 + 5$

$x = 1$

$e)\left\{\begin{matrix} 3x - 5y = - 4\\ 6x - 2y = 8 \: | \div 2 \end{matrix}\right.$

$\left\{\begin{matrix} 3x - 5y = - 4\\ 3x - y = 4\end{matrix}\right.$

$3x - y = 4$

$- y = 4 - 3x \: | \times ( - 1)$

$y = 3x - 4$

$3x - 5y = - 4$

$3x - 5(3x - 4) = - 4$

$3x - 15x + 20 = - 4$

$3x - 15x = - 4 - 20$

$- 12x = - 24 \: | \div ( - 12)$

$x = 2$

$y = 3x - 4$

$y = 3 \times 2 - 4$

$y = 6 - 4$

$y = 2$

$f)\left\{\begin{matrix} 2x + 5y = 1\\ 3x + 10y = - 1\end{matrix}\right.$

$2x + 5y = 1$

$2x = 1 - 5y$

$x = \frac{1 - 5y}{2}$

$3x + 10y = - 1$

$3 \times \frac{1 - 5y}{2} + 10y = - 1$

$\frac{3(1 - 5y)}{2} + 10y = - 1$

$\frac{3 - 15y}{2} + 10y = - 1 \: | \times 2$

$3 - 15y + 20y= - 2$

$- 15y + 20y = - 2 - 3$

$5y = - 5 \: | \div 5$

$y = - 1$

$x = \frac{1 - 5y}{2}$

$x = \frac{1 - 5 \times ( - 1)}{2}$

$x = \frac{1 + 5}{2}$

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

$x = 3$