数学 – 数学の威力を発揮する - 方程式 – 等式の証明(整数の恒等式)

学習環境

数学読本〈1〉数・方程式/方程式/不等式 (松坂和夫(著)、岩波書店)の第3章(数学の威力を発揮する - 方程式)、3.4(等式の証明)、整数の恒等式の問42.を解いてみる。

問42.

$\begin{array}{l} 2 a + 1 = - 5 \\ a = - 3 \\ b = 2 \\ 2 b + a = c \\ c = 1 \\ a = - 3, b = 2, c = 1 \end{array}$
$\begin{array}{l} a + b + c = 10 \\ a - b = 0 \\ b = a \\ - c = - 2 \\ c = 2 \\ 2 a = 8 \\ a = 4, b = 4, c = 2 \end{array}$
$\begin{array}{l} a + b + c = 0 \\ - 4 a - b + c = 0 \\ 3 a - 6 b - 2 c = 30 \\ c = - a - b \\ - 4 a - b - a - b = 0 \\ b = - \frac{5}{2} a \\ c = \frac{3}{2} a \\ 3 a + 15 a - 3 a = 30 \\ a = 2, b = - 5, c = 3 \end{array}$
$\begin{array}{l} x = 2 \\ 8 - 24 + 26 - 4 = d \\ d = 6 \\ x = 0 \\ - 4 = - 8 a + 4 b - 2 c + 6 \\ 4 a - 2 b + c = 5 \\ x = 1 \\ 1 - 6 + 13 - 4 = - a + b - c + 6 \\ a - b + c = 2 \\ x = 3 \\ 27 - 54 + 39 - 4 = a + b + c + 6 \\ a + b + c = 2 \\ c = - 4 a + 2 b + 5 \\ a - b - 4 a + 2 b + 5 = 2 \\ b = 3 a - 3 \\ c = - 4 a + 6 a - 6 + 5 = 2 a - 1 \\ a + 3 a - 3 + 2 a - 1 = 2 \\ 6 a = 6 \\ a = 1, b = 0, c = 1, d = 6 \end{array}$

Kamimura's blog