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Solving Linear Systems

Solving Linear Systems 1.1 - Solution

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a
To tell whether the point is a solution of the system of linear equations, we can substitute it into each of the equations and simplify. If they hold true, it's a solution.
{x+y=4(I)2xy=1(II)\begin{cases}x+y=4 & \, \text {(I)}\\ 2x-y=1 & \text {(II)}\end{cases}
{4+1=?4241=?1\begin{cases}{\color{#0000FF}{4}}+{\color{#009600}{1}}\stackrel{?}{=}4 \\ 2\cdot{\color{#0000FF}{4}}-{\color{#009600}{1}}\stackrel{?}{=}1 \end{cases}
{4+1=?481=?1\begin{cases}4+1\stackrel{?}{=}4 \\ 8-1\stackrel{?}{=}1 \end{cases}
{5471\begin{cases}5\neq4 \\ 7\neq1 \end{cases}
Since the point isn't a solution to either of the equations, it is not a solution to the system.
b
We'll substitute the coordinates into the two equations and see if it's a solution to both of them.
{x4y=0(I)2y=x4+1(II)\begin{cases}x-4y=0 & \, \text {(I)}\\ 2y=\frac{x}{4}+1 & \text {(II)}\end{cases}
{441=?021=?44+1\begin{cases}{\color{#0000FF}{4}}-4\cdot{\color{#009600}{1}}\stackrel{?}{=}0 \\ 2\cdot{\color{#009600}{1}}\stackrel{?}{=}\frac{{\color{#0000FF}{4}}}{4}+1 \end{cases}
{44=?02=?44+1\begin{cases}4-4\stackrel{?}{=}0 \\ 2\stackrel{?}{=}\frac{4}{4}+1 \end{cases}
{44=?02=?1+1\begin{cases}4-4\stackrel{?}{=}0 \\ 2\stackrel{?}{=}1+1 \end{cases}
{0=02=2\begin{cases}0=0 \\ 2=2 \end{cases}
The point is a solution to the linear system.