Question Number 226912 by Spillover last updated on 18/Dec/25
Answered by Raphael254 last updated on 19/Dec/25
$$ \\ $$$${b}+{c}\:=\:\mathrm{16} \\ $$$$ \\ $$$${a}+{b}+{c}=\mathrm{24} \\ $$$${a}+\mathrm{16}\:=\:\mathrm{24} \\ $$$${a}\:=\:\mathrm{8} \\ $$$$ \\ $$$$\mathrm{8}\:−\:{b}\:+\:{c}\:=\:{x} \\ $$$$ \\ $$$${there}\:{are}\:{infinite}\:{solutions},\:{when}\:{b}+{c}=\mathrm{16} \\ $$$${and}\:{a}\:=\:\mathrm{8} \\ $$$$ \\ $$$${b}\:=\:\mathrm{3},\:{c}\:=\:\mathrm{13}: \\ $$$$ \\ $$$${a}+{b}+{c}\:=\:\mathrm{8}+\mathrm{3}+\mathrm{13}\:=\:\mathrm{24} \\ $$$${a}+{b}\:=\:\mathrm{16} \\ $$$$\mathrm{8}−\mathrm{3}+\mathrm{13}=\mathrm{18} \\ $$$$ \\ $$$${b},\:=\:\mathrm{7},\:{c}\:=\:\mathrm{9}: \\ $$$$ \\ $$$${a}+{b}+{c}\:=\:\mathrm{8}+\mathrm{7}+\mathrm{9}\:=\:\mathrm{24} \\ $$$${b}+{c}\:=\:\mathrm{16} \\ $$$$\mathrm{8}−\mathrm{7}+\mathrm{9}\:=\:\mathrm{10} \\ $$
Commented by Frix last updated on 20/Dec/25
$$\mathrm{Simply}: \\ $$$$\mathrm{2}\:\mathrm{linear}\:\mathrm{equations},\:\mathrm{3}\:\mathrm{variables} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{unique}\:\mathrm{solution} \\ $$$$ \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:\in\mathbb{R}^{\mathrm{3}} \\ $$$${a}+{b}+{c}=\mathrm{24} \\ $$$${b}+{c}=\mathrm{16} \\ $$$$\Rightarrow\:{a}=\mathrm{8}\wedge{b}={b}=\mathrm{16}−{c} \\ $$$$\Rightarrow\:\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{8}}\\{\mathrm{16}−{c}}\\{{c}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{8}}\\{\mathrm{16}}\\{\mathrm{0}}\end{pmatrix}\:+{c}\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{which}\:\mathrm{represents}\:\mathrm{a}\:\mathrm{line}\:\in\mathbb{R}^{\mathrm{3}} \\ $$$${a}−{b}+{c}=\mathrm{2}{c}−\mathrm{8};\:{c}\in\mathbb{R} \\ $$