Решить систему уравнений методом подстановки 3х+у=-6 и -4х+3у=0;
методом сложения 9х-3у=-6 и -4х+3у=0;
методом сложения 2х+5у=8 и -4х+3у=7;
Методом подстановки 8х-0,2у=11 и х+3х=-1,6.

Ответы

Ответ дал: Wynneve

Ответ:

1) $\left( -\frac{18}{13};\ -\frac{24}{13} \right).$

2) $\left( -\frac{6}{5}; \ -\frac{8}{5} \right).$

3) $\left( -\frac{11}{26} ;\ \frac{23}{13}\right).$

4) $\left( -\frac{2}{5} ;\ -71 \right).$

Объяснение:

1. $\left \{ {{3x+y = -6;} \atop {-4x+3y = 0.}} \right. \Rightarrow \left \{ {{y = -(3x +6);} \atop {-4x+3y = 0.}} \right.\\\\-4x -3(3x+6) = 0;\\-4x - 9x - 18 = 0;\\-13x - 18 = 0;\\13x + 18 = 0;\\13x = -18;\\x = -\frac{18}{13};\\y = -(3\cdot\frac{18}{13} + 6) = -\frac{24}{13}.$

2. $\left \{ {{9x-3y= -6;} \atop {-4x + 3y = 0.}} \right.\\\\ (y = \frac{4x}{3}).\\9x - 4x - 3y + 3y = -6 + 0;\\5x = -6;\\x = -\frac{6}{5};\\y = -\frac{4\cdot\frac{6}{5}}{3} = -\frac{8}{5}.$

3. $\left \{ {{2x+5y=8;} \atop {-4x+3y=7.}} \right. \Rightarrow \left \{ {{4x+10y=16;} \atop {-4x+3y=7.}} \right.\\\\\left(x=\frac{8-5y}{2}\right).\\4x-4x+10y+3y=16+7;\\13y = 23.\\y=\frac{23}{13};\\x = \frac{8 - 5\cdot\frac{23}{13}}{2} = -\frac{11}{26}.$

4. $\left \{ {{8x-0{,}2y=11;} \atop {x+3x=-1{,}6.}} \right. \Rightarrow \left \{ {{8x-0{,}2y=11;} \atop {x = -0{,}4.}} \right. \\\\-8\cdot0{,}4 - 0{,}2y = 11;\\-0{,}2y = 14{,}2;\\y = -71.\\(x = -0{,}4 = -\frac{2}{5}).$