$x = 3$、$y = -4$ のとき、次の式の値をそれぞれ求めます。 (1) $2x + 3y$ (2) $\frac{2x + 3y}{3}$ (3) $\frac{4x + 6y}{6}$ (4) $\frac{4x + y}{6}$ (5) $\frac{2}{3}x + y$ (6) $\frac{2}{3}x - y$ (7) $\frac{2x - 3y}{3}$ (8) $\frac{2x - y}{3}$

1. 問題の内容

$x = 3$、$y = -4$ のとき、次の式の値をそれぞれ求めます。

(1) $2x + 3y$

(2) $\frac{2x + 3y}{3}$

(3) $\frac{4x + 6y}{6}$

(4) $\frac{4x + y}{6}$

(5) $\frac{2}{3}x + y$

(6) $\frac{2}{3}x - y$

(7) $\frac{2x - 3y}{3}$

(8) $\frac{2x - y}{3}$

2. 解き方の手順

それぞれの式に、$x = 3$、$y = -4$ を代入して計算します。

(1) $2x + 3y = 2(3) + 3(-4) = 6 - 12 = -6$

(2) $\frac{2x + 3y}{3} = \frac{2(3) + 3(-4)}{3} = \frac{6 - 12}{3} = \frac{-6}{3} = -2$

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

(4) $\frac{4x + y}{6} = \frac{4(3) + (-4)}{6} = \frac{12 - 4}{6} = \frac{8}{6} = \frac{4}{3}$

(5) $\frac{2}{3}x + y = \frac{2}{3}(3) + (-4) = 2 - 4 = -2$

(6) $\frac{2}{3}x - y = \frac{2}{3}(3) - (-4) = 2 + 4 = 6$

(7) $\frac{2x - 3y}{3} = \frac{2(3) - 3(-4)}{3} = \frac{6 + 12}{3} = \frac{18}{3} = 6$

(8) $\frac{2x - y}{3} = \frac{2(3) - (-4)}{3} = \frac{6 + 4}{3} = \frac{10}{3}$

3. 最終的な答え

(1) -6

(2) -2

(3) -2

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

(5) -2

(6) 6

(7) 6

(8) $\frac{10}{3}$