Помогите пж срочно даю 100 и тот кто правильно напишет таму ещё 50 дам

Question 1

Question 2

Решите задачу:

$x - 32 = - 7(x - 8) \\ x - 32 = - 7x + 56 \\ x + 7x = 56 + 32 \\ 8x = 88 \\ x = 88 \div 8 \\ x = 11$

$3(4y - 8) = 3y - 6 \\ 12y - 24 = 3y - 6 \\ 12y - 3y = - 6 + 24 \\ 9y = 18 \\ y = 18 \div 9 \\ y = 2$

$- 3.2n + 4.8 = - 2(1.2n + 2.4) \\ - 3.2n + 4.8 = - 2.4n - 4.8 \\ - 3.2n + 2.4n = - 4.8 - 4.8 \\ - 0.8n = - 9.6 \\ n = - 9.6 \div ( - 0.8) \\ n = - 96 \div ( - 8) \\ n = 12$

$x - 5(0.8x - 1.2) = 7.2 \\ x - 4x + 6 = 7.2 \\ - 3x = 7.2 - 6 \\ - 3x = 1.2 \\ x = 1.2 \div ( - 3) \\ x = - 0.4$

$- (5x - 8) = 25 - (11x - 19) \\ - 5x + 8 = 25 - 11x + 19 \\ 11x - 5x = 44 - 8 \\ 6x = 36 \\ x = 36 \div 6 \\ x = 6$

$\frac{3}{4} x - 1.25 = \frac{9}{8} x - \frac{1}{8} \: \: | \times 8 \\ 2 \times 3x - 10 = 9x - 1 \\ 6x - 9x = - 1 + 10 \\ - 3x = 9 \\ x = 9 \div ( - 3) \\ x = - 3$

$\frac{1}{5} ( \frac{5}{9} x + 5) - \frac{1}{7} ( \frac{7}{6} x + 2 \frac{1}{3} ) = 2 \\ \frac{5x}{5 \times 9} + \frac{5}{5} - \frac{7x}{7 \times 6} - \frac{7}{7 \times 3} = 2 \\ \frac{x}{9} + 1 - \frac{x}{6} - \frac{1}{3} = 2 \: \: \: | \times 18\\ 2x + 18 - 3x - 6 = 36 \\ - x = 36 - 12 \\ - x = 24 \\ x = - 24$

Question 3

Решить уравнения:

1)

$x-32=\left(x-8\right)\left(-7\right)\\x-32=-7x+56\\x+7x=56+32\\8x=88\\\frac{8x}{8}=\frac{88}{8}\\x=11$

2)

$3(4y-8)=3y-6\\12y-24=3y-6\\12y-3y=-6+24\\9y=18\\\frac{9y}{9}=\frac{18}{9}\\y=2$

3)

$-3.2n+4.8=-2\cdot (1.2n+2.4)\\-3.2n+4.8=-2.4n-4.8 |\: \cdot(-10)\\32n-48=24n+48\\32n-24n=48+48\\8n=96\\\frac{8n}{8}=\frac{96}{8}\\n=12$

4)

$x-5\cdot (0,8x-1.2)=7.2\\x-(4x-6)=7.2\\x-4x+6=7.2\\-3x=7.2-6\\-3x=1.2\\\frac{-3x}{-3}=\frac{1.2}{-3}\\x=-0.4$

5)

$x-(5x-8)=25-(11x-19)\\-5x+8=25-11x+19\\-5x+11x=44-8\\6x=36\\\frac{6x}{6}=\frac{36}{6}\\x=6$

6)

$\frac{3}{4}x-1.25=\frac{9}{8}x-\frac{1}{8} \\\\\frac{3}{4}x-\frac{9}{8}x=-\frac{1}{8} +\frac{5}{4} \\\\\frac{2\cdot 3-9}{8}x =\frac{5\cdot 2-1}{8} \\\\-\frac{3}{8}x = \frac{9}{8} | \:\cdot (-8)\\\\24x=-72\\\frac{24}{24}x = -\frac{72}{24}\\x=-3$

7)

$\frac{1}{5}(\frac{5}{9}x+5)-\frac{1}{7}(\frac{7}{6}x+2\frac{1}{3} ) = 2\\\\\frac{1}{9}x+1-\frac{1}{6}x-\frac{1}{3}=2\\\\\frac{1}{9}x-\frac{1}{6}x=2-1+\frac{1}{3} \:\big|\cdot 18\\\\2x-3x=18+6\\-x=24 \:\big|\cdot (-1)\\x=-24$