与えられた6つの一次方程式を解く問題です。 (1) $5x + 12 = -8$ (2) $2x - 5 = 9x - 19$ (3) $x + 2(x - 9) = 3(8 - x)$ (4) $\frac{1}{3}x - 2 = -\frac{7}{4}$ (5) $\frac{x - 1}{3} - \frac{x - 4}{5} = 1$ (6) $0.3(x - 7) = 2x + 1.3$

1. 問題の内容

与えられた6つの一次方程式を解く問題です。

(1) $5x + 12 = -8$

(2) $2x - 5 = 9x - 19$

(3) $x + 2(x - 9) = 3(8 - x)$

(4) $\frac{1}{3}x - 2 = -\frac{7}{4}$

(5) $\frac{x - 1}{3} - \frac{x - 4}{5} = 1$

(6) $0.3(x - 7) = 2x + 1.3$

2. 解き方の手順

(1) $5x + 12 = -8$

$5x = -8 - 12$

$5x = -20$

$x = -4$

(2) $2x - 5 = 9x - 19$

$2x - 9x = -19 + 5$

$-7x = -14$

$x = 2$

(3) $x + 2(x - 9) = 3(8 - x)$

$x + 2x - 18 = 24 - 3x$

$3x - 18 = 24 - 3x$

$3x + 3x = 24 + 18$

$6x = 42$

$x = 7$

(4) $\frac{1}{3}x - 2 = -\frac{7}{4}$

$\frac{1}{3}x = -\frac{7}{4} + 2$

$\frac{1}{3}x = -\frac{7}{4} + \frac{8}{4}$

$\frac{1}{3}x = \frac{1}{4}$

$x = \frac{3}{4}$

(5) $\frac{x - 1}{3} - \frac{x - 4}{5} = 1$

両辺に15を掛ける

$5(x - 1) - 3(x - 4) = 15$

$5x - 5 - 3x + 12 = 15$

$2x + 7 = 15$

$2x = 8$

$x = 4$

(6) $0.3(x - 7) = 2x + 1.3$

$0.3x - 2.1 = 2x + 1.3$

$0.3x - 2x = 1.3 + 2.1$

$-1.7x = 3.4$

$x = -2$

3. 最終的な答え

(1) $x = -4$

(2) $x = 2$

(3) $x = 7$

(4) $x = \frac{3}{4}$

(5) $x = 4$

(6) $x = -2$