$a=2$, $b=3$ のとき、以下の式の値をそれぞれ求めます。 (1) $4a+3b$ (2) $5a-2b$ (3) $2a-6b+3a+4b$ (4) $4a+3b-a-5b$ (5) $a + \frac{2}{3}b$ (6) $\frac{3a+2b}{3}$ (7) $2(5a-2b)$ (8) $12(\frac{1}{3}a+\frac{1}{4}b)$

1. 問題の内容

$a=2$, $b=3$ のとき、以下の式の値をそれぞれ求めます。

(1) $4a+3b$

(2) $5a-2b$

(3) $2a-6b+3a+4b$

(4) $4a+3b-a-5b$

(5) $a + \frac{2}{3}b$

(6) $\frac{3a+2b}{3}$

(7) $2(5a-2b)$

(8) $12(\frac{1}{3}a+\frac{1}{4}b)$

2. 解き方の手順

(1) $4a+3b = 4(2)+3(3) = 8+9 = 17$

(2) $5a-2b = 5(2)-2(3) = 10-6 = 4$

(3) $2a-6b+3a+4b = (2a+3a)+(-6b+4b) = 5a-2b = 5(2)-2(3) = 10-6 = 4$

(4) $4a+3b-a-5b = (4a-a)+(3b-5b) = 3a-2b = 3(2)-2(3) = 6-6 = 0$

(5) $a + \frac{2}{3}b = 2 + \frac{2}{3}(3) = 2+2 = 4$

(6) $\frac{3a+2b}{3} = \frac{3(2)+2(3)}{3} = \frac{6+6}{3} = \frac{12}{3} = 4$

(7) $2(5a-2b) = 2(5(2)-2(3)) = 2(10-6) = 2(4) = 8$

(8) $12(\frac{1}{3}a+\frac{1}{4}b) = 12(\frac{1}{3}(2)+\frac{1}{4}(3)) = 12(\frac{2}{3}+\frac{3}{4}) = 12(\frac{8}{12}+\frac{9}{12}) = 12(\frac{17}{12}) = 17$

3. 最終的な答え

(1) 17

(2) 4

(3) 4

(4) 0

(5) 4

(6) 4

(7) 8

(8) 17