## Elementary Linear Algebra 7th Edition

There are infinitely many solutions of this system and they are given by $$x=8-3t,\qquad y=t,$$ where $t$ can be any real number.
Follow the steps below: Step 1: Add the 1st equation the second one to eliminate $x$: $$-2x+2x-6y+6y=-16+16.$$ This yields not an equation but a identity $0=0.$ This means that we have to equate one of the variables with parameter $t$, we will set $y=t$. Now we have \begin{align*} 2x+6y=&16\\ y=&t \end{align*} Step 2: Use the back substitution to find $x$: $$2x+6t=16$$ Divide with $2$: $$x+3t=8.$$ This gives $$x=8-3t.$$