与えられた10個の比例式を解き、$x$, $t$, $b$, $n$ の値を求める問題です。

1. 問題の内容

与えられた10個の比例式を解き、$x$, $t$, $b$, $n$ の値を求める問題です。

2. 解き方の手順

比例式の性質「内項の積は外項の積に等しい」を利用して、各比例式を解きます。

(1) $7:8 = (x-4):6$

$7 \times 6 = 8 \times (x-4)$

$42 = 8x - 32$

$8x = 74$

$x = \frac{74}{8} = \frac{37}{4}$

(2) $2:1 = (8+x):8$

$2 \times 8 = 1 \times (8+x)$

$16 = 8 + x$

$x = 8$

(3) $4:5 = (x-5):8$

$4 \times 8 = 5 \times (x-5)$

$32 = 5x - 25$

$5x = 57$

$x = \frac{57}{5}$

(4) $9:8 = 3:(x-4)$

$9 \times (x-4) = 8 \times 3$

$9x - 36 = 24$

$9x = 60$

$x = \frac{60}{9} = \frac{20}{3}$

(5) $(n-2):4 = 3:2$

$2 \times (n-2) = 4 \times 3$

$2n - 4 = 12$

$2n = 16$

$n = 8$

(6) $6:7 = (t+2):8$

$6 \times 8 = 7 \times (t+2)$

$48 = 7t + 14$

$7t = 34$

$t = \frac{34}{7}$

(7) $2:1 = (7-x):3$

$2 \times 3 = 1 \times (7-x)$

$6 = 7 - x$

$x = 1$

(8) $4:3 = (b-6):2$

$4 \times 2 = 3 \times (b-6)$

$8 = 3b - 18$

$3b = 26$

$b = \frac{26}{3}$

(9) $5:6 = (x-3):5$

$5 \times 5 = 6 \times (x-3)$

$25 = 6x - 18$

$6x = 43$

$x = \frac{43}{6}$

(10) $6:5 = 7:(7-x)$

$6 \times (7-x) = 5 \times 7$

$42 - 6x = 35$

$6x = 7$

$x = \frac{7}{6}$

3. 最終的な答え

(1) $x = \frac{37}{4}$

(2) $x = 8$

(3) $x = \frac{57}{5}$

(4) $x = \frac{20}{3}$

(5) $n = 8$

(6) $t = \frac{34}{7}$

(7) $x = 1$

(8) $b = \frac{26}{3}$

(9) $x = \frac{43}{6}$

(10) $x = \frac{7}{6}$