$x = \frac{1}{4}$、 $y = \frac{1}{6}$ のとき、以下の式の値を求めなさい。 (1) $x+y$ (2) $2(x+y)$ (3) $2x+2y$ (4) $-3(x+y)$ (5) $-3x-3y$ (6) $-(x+y)$ (7) $-x-y$ (8) $-(x-y)$

1. 問題の内容

$x = \frac{1}{4}$、 $y = \frac{1}{6}$ のとき、以下の式の値を求めなさい。

(1) $x+y$

(2) $2(x+y)$

(3) $2x+2y$

(4) $-3(x+y)$

(5) $-3x-3y$

(6) $-(x+y)$

(7) $-x-y$

(8) $-(x-y)$

2. 解き方の手順

まず、$x$ と $y$ の値をそれぞれ与えられた式の $x$ と $y$ に代入します。その後、四則演算のルールに従って計算を行います。

(1) $x+y = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$

(2) $2(x+y) = 2(\frac{1}{4} + \frac{1}{6}) = 2(\frac{5}{12}) = \frac{10}{12} = \frac{5}{6}$

(3) $2x+2y = 2(\frac{1}{4}) + 2(\frac{1}{6}) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

(4) $-3(x+y) = -3(\frac{1}{4} + \frac{1}{6}) = -3(\frac{5}{12}) = -\frac{15}{12} = -\frac{5}{4}$

(5) $-3x-3y = -3(\frac{1}{4}) - 3(\frac{1}{6}) = -\frac{3}{4} - \frac{1}{2} = -\frac{3}{4} - \frac{2}{4} = -\frac{5}{4}$

(6) $-(x+y) = -(\frac{1}{4} + \frac{1}{6}) = -(\frac{5}{12}) = -\frac{5}{12}$

(7) $-x-y = -\frac{1}{4} - \frac{1}{6} = -\frac{3}{12} - \frac{2}{12} = -\frac{5}{12}$

(8) $-(x-y) = -(\frac{1}{4} - \frac{1}{6}) = -(\frac{3}{12} - \frac{2}{12}) = -(\frac{1}{12}) = -\frac{1}{12}$

3. 最終的な答え

(1) $\frac{5}{12}$

(2) $\frac{5}{6}$

(3) $\frac{5}{6}$

(4) $-\frac{5}{4}$

(5) $-\frac{5}{4}$

(6) $-\frac{5}{12}$

(7) $-\frac{5}{12}$

(8) $-\frac{1}{12}$