## Elementary Linear Algebra 7th Edition

{$(x_1,x_2)|4x_1+x_2=0$}
$\frac{2}{3}x_1+\frac{1}{6}x_2=0$ $4x_1+x_2=0$ Express $x_2$ in terms of $x_1$ $x_2= -4x_1$ Substitute for $x_2$ in the first equation. $\frac{2}{3}x_1+\frac{1}{6}(-4x_1)=0$ $\frac{2}{3}x_1-\frac{4}{6}x_1=0$ $\rightarrow$ Multiply by 6 to remove fractions. $4x_1-4x_1=0$ $\rightarrow$ This expression is always true and therefore any pair of values which satisfy 1st or 2nd equation is a possible solution.