Упр.17.6 Часть 1 ГДЗ Мордкович Семенов 7 класс (Алгебра)

а) {(3x + 2y)/6 = 3; 2x + 3y = 2};
б) {x/3 — y/4 = 4; (11x + 9y)/6 = 1};
в) {x/2 — y/5 = 3; (6x + 7)/5 = 2x — y};
г) {2x — 3y = -24; (2x + y)/4 = -2};
д) {x/6 + y/5 = 1/2; (7x + 6y)/3 = -1};
е) {x + y = (1 — 3y)/4; x/5 + y/3 = 1/5}.

1. а)

   $$\frac{3x+2y}{6}=3,\quad 2x+3y=2$$
   $$3x+2y=18$$
   $$2y=18-3x,\quad y=9-\frac{3}{2}x$$
   $$2x+3\left(9-\frac{3}{2}x\right)=2$$
   $$2x+27-\frac{9}{2}x=2$$
   $$-\frac{5}{2}x=-25,\quad x=10$$
   $$y=9-\frac{3}{2}\cdot 10=-6$$

   Ответ: $$(10;\,-6)$$

2. б)

   $$\frac{x}{3}-\frac{y}{4}=4,\quad \frac{11x+9y}{6}=1$$
   $$4x-3y=48$$
   $$x=\frac{3}{4}y+12$$
   $$11\left(\frac{3}{4}y+12\right)+9y=6$$
   $$\frac{33}{4}y+132+9y=6$$
   $$\frac{69}{4}y=-126,\quad y=-\frac{168}{23}$$
   $$x=\frac{3}{4}\cdot\left(-\frac{168}{23}\right)+12=\frac{150}{23}$$

   Ответ: $$\left(\frac{150}{23};\,-\frac{168}{23}\right)$$

3. в)

   $$\frac{x}{2}-\frac{y}{5}=3,\quad \frac{6x+7}{5}=2x-y$$
   $$5x-2y=30$$
   $$6x+7=10x-5y$$
   $$-4x+5y=-7$$
   $$y=\frac{5x-30}{2}$$
   $$-4x+5\cdot\frac{5x-30}{2}=-7$$
   $$-8x+25x-150=-14$$
   $$17x=136,\quad x=8$$
   $$y=\frac{5\cdot 8-30}{2}=5$$

   Ответ: $$(8;\,5)$$

4. г)

   $$2x-3y=-24,\quad \frac{2x+y}{4}=-2$$
   $$2x+y=-8$$
   $$y=-2x-8$$
   $$2x-3(-2x-8)=-24$$
   $$8x+24=-24$$
   $$8x=-48,\quad x=-6$$
   $$y=-2\cdot(-6)-8=4$$

   Ответ: $$(-6;\,4)$$

5. д)

   $$\frac{x}{6}+\frac{y}{5}=\frac{1}{2},\quad \frac{7x+6y}{3}=-1$$
   $$5x+6y=15$$
   $$7x+6y=-3$$
   $$2x=-18,\quad x=-9$$
   $$5\cdot(-9)+6y=15$$
   $$6y=60,\quad y=10$$

   Ответ: $$(-9;\,10)$$

6. е)

   $$x+y=\frac{1-3y}{4},\quad \frac{x}{5}+\frac{y}{3}=\frac{1}{5}$$
   $$4x+4y=1-3y$$
   $$4x+7y=1$$
   $$3x+5y=3$$
   $$x=\frac{1-7y}{4}$$
   $$3\cdot\frac{1-7y}{4}+5y=3$$
   $$3-21y+20y=12$$
   $$-y=9,\quad y=-9$$
   $$x=\frac{1-7(-9)}{4}=16$$

   Ответ: $$(16;\,-9)$$