求使下列方程组有无穷多解的k值
8x + 5y = 9
kx + 10y = 18
已知:已知方程为 8x + 5y = 9; kx + 10y = 18
求解:求使下列方程组有无穷多解的k值。
解
已知方程组为
8x + 5y = 9
kx + 10y = 18
方程组的形式为 a₁x + b₁y = c₁ 和 a₂x + b₂y = c₂
要使方程组有无穷多解,需要满足条件:
a₁/a₂ = b₁/b₂ = c₁/c₂
这里,a₁ = 8, b₁ = 5, c₁ = -9 和 a₂ = k, b₂ = 10, c₂ = -18
8/k = 5/10 = -9/-18
现在,8/k = 5/10
=> 8 × 10 = 5k
=> k = (8 × 10) / 5
=> k = 16
因此,如果 k = 16,则方程组有无穷多解。
广告