Question 144264

Start with the given system

{{{-16x+2y=-2}}}
{{{y=8x-1}}}




{{{-16x+2(8x-1)=-2}}}  Plug in {{{y=8x-1}}} into the first equation. In other words, replace each {{{y}}} with {{{8x-1}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{-16x+16x-2=-2}}} Distribute



{{{-2=-2}}} Combine like terms on the left side



{{{0=-2+2}}}Add 2 to both sides



{{{0=0}}} Combine like terms on the right side


Since this equation is always true for any x value, this means x can equal any number. So there are an infinite number of solutions.