Question

Розв'яжiть систему рівнянь способом підстановки:

1) [4(x + y) - 3y = 2,
[9(x-2y) - 6x=-11;

2) [8(x+y) - 12y = 6,
[6(3x-y) +18x=13.​

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{4(x+y)-3y=2} \atop {9(x-2y)-6x=-11}} \right. \\\\\\\left \{ {{4x+4y-3y=2} \atop {9x-18y-6x=-11}} \right. \\\\\\\left \{ {{4x+y=2} \atop {3x-18y=-11}} \right.\\\\\\\left \{ {{y=2-4x} \atop {3x-18\cdot(2-4x)=-11}} \right.\\\\\\\left \{ {{y=2-4x} \atop {3x-36+72x=-11}} \right. \\\\\\\left \{ {{y=2-4x} \atop {75x=-11+36}} \right. \\\\\\\left \{ {{y=2-4x} \atop {75x=25}} \right.$

$\displaystyle\bf\\\left \{ {{y=2-4\cdot \dfrac{1}{3} } \atop {x=\dfrac{1}{3} }} \right. \\\\\\\left \{ {{y= \dfrac{2}{3} } \atop {x=\dfrac{1}{3} }} \right. \\\\\\Otvet: \ \Big(\frac{1}{3} \ ; \ \frac{2}{3} \Big)\\\\\\2)\\\\\left \{ {{8(x+y)-12y=6} \atop {6(3x-y)+18x=13}} \right.\\\\\\\left \{ {{8x+8y-12y=6} \atop {18x-6y+18x=13}} \right. \\\\\\\left \{ {{8x-4y=6} \atop {36x-6y=13}} \right. \\\\\\\left \{ {{2x-y=1,5} \atop {36x-6y=13}} \right.$

$\displaystyle\bf\\\left \{ {{y=2x-1,5} \atop {36x-6\cdot (2x-1,5)=13}} \right.\\\\\\\left \{ {{y=2x-1,5} \atop {36x-12x+9=13}} \right.\\\\\\\left \{ {{y=2x-1,5} \atop {24x=4}} \right.\\\\\\\left \{ {{y=2\cdot \dfrac{1}{6} -1,5} \atop {x=\dfrac{1}{6} }} \right.\\\\\\\left \{ {{y=2\cdot \dfrac{1}{6} -1,5} \atop {x=\dfrac{1}{6} }} \right.\\\\\\\left \{ {{y=-\dfrac{7}{6} } \atop {x=\dfrac{1}{6} }} \right.\\\\\\Otvet: \ \Big(\frac{1}{6} \ ; \ -\frac{7}{6} \Big)$

Answer 2

Решение.

Выражаем из одного уравнения одну из переменных через другую , а потом подставляем во второе уравнение .

$1.\ \ \left\{\begin{array}{l}4(x+y)-3y=2\\9(x-2y)-6x=-11\end{array}\right\ \ \left\{\begin{array}{l}4x+y=2\\3x-18y=-11\end{array}\right\ \ \left\{\begin{array}{l}y=2-4x\\3x-18(2-4x)=-11\end{array}\right\\\\\\\left\{\begin{array}{l}y=2-4x\\75x-36=-11\end{array}\right\ \ \left\{\begin{array}{l}y=2-4x\\75x=25\end{array}\right\ \ \left\{\begin{array}{l}y=\dfrac{2}{3}\\x=\dfrac{1}{3}\end{array}\right$

Ответ:   $\Big(\ \dfrac{1}{3}\ ;\ \dfrac{2}{3}\ \Big)\ .$  

$2.\ \ \left\{\begin{array}{l}8(x+y)-12y=6\\6(3x-y)+18x=13\end{array}\right\ \ \left\{\begin{array}{l}8x-4y=6\\36x-6y=13\end{array}\right\ \ \left\{\begin{array}{l}2y=4x-3\\36x-3(4x-3)=13\end{array}\right\\\\\\\left\{\begin{array}{l}2y=4x-3\\24x=4\end{array}\right\ \ \left\{\begin{array}{l}2y=-\dfrac{7}{3}\\x=\dfrac{1}{6}\end{array}\right\ \ \left\{\begin{array}{l}y=-\dfrac{7}{6}\\x=\dfrac{1}{6}\end{array}\right$

Ответ:  $\Big(\ \dfrac{1}{6}\ ; -1\dfrac{1}{6}\ \Big)$  .