解方程 281.45 = x + 53.23 + (x × 0.05)/(1 − 0.05),求 x。
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解出 xxx:281.45=x+53.23+x⋅0.051−0.05281.45 = x + 53.23 + \frac{x\cdot 0.05}{1-0.05}281.45=x+53.23+1−0.05x⋅0.05
先把分母算出来。 1−0.05=0.951-0.05 = 0.951−0.05=0.95,方程于是写成 281.45=x+53.23+0.05x0.95.281.45 = x + 53.23 + \frac{0.05x}{0.95}.281.45=x+53.23+0.950.05x. 它的结构是:一个基础金额 xxx、一笔固定费用 53.2353.2353.23,再加一项按含费总额的 5%5\%5% 算出来的附加费。
把这个分数约成精确值。 0.050.95=595=119=0.052631578…\frac{0.05}{0.95} = \frac{5}{95} = \frac{1}{19} = 0.052631578\ldots0.950.05=955=191=0.052631578… 保留 119\tfrac{1}{19}191 而不是取近似的 0.05260.05260.0526,能让整个计算保持精确。
把常数移到另一边。 281.45−53.23=228.22=x+x19.281.45-53.23 = 228.22 = x+\frac{x}{19}.281.45−53.23=228.22=x+19x.
提出 x。 228.22=x(1+119)=2019 x.228.22 = x\left(1+\frac{1}{19}\right) = \frac{20}{19}\,x.228.22=x(1+191)=1920x. 这一步是关键的化简:系数是干净的分数 2019\tfrac{20}{19}1920,而不是一个别扭的小数。
乘以倒数求解。 x=228.22⋅1920=4336.1820=216.809.x = 228.22\cdot\frac{19}{20} = \frac{4336.18}{20} = 216.809.x=228.22⋅2019=204336.18=216.809.
验证。 216.809+53.23=270.039216.809 + 53.23 = 270.039216.809+53.23=270.039,附加费是 216.80919=11.411\tfrac{216.809}{19} = 11.41119216.809=11.411;合计 270.039+11.411=281.450270.039+11.411 = 281.450270.039+11.411=281.450,与左边完全相符。
x=228.22⋅1920=216.809x = 228.22\cdot\frac{19}{20} = 216.809x=228.22⋅2019=216.809
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