Question 794210

{{{-5x+3y=-5}}}......eq.1
{{{y=(5/3)x+1}}}......eq.2
___________________________substitute {{{y}}} from eq.2 in eq.1

{{{-5x+3((5/3)x+1)=-5}}}......eq.1....solve for {{{x}}}

{{{-5x+3(5/3)x+3=-5}}}

{{{-5x+cross(3)(5/cross(3))x+3=-5}}}

{{{-5x+5x+3=-5}}}

{{{-cross(5x)+cross(5x)+3=-5}}}

{{{ 3=-5}}}.......no solution exist, lines have no intersection point

if you write both eq. in slope-intercept form {{{y=mx+b}}} where {{{m=slope}}} and {{{b=y-intercept}}}, you will have

{{{ y=(5/3)x-5/3}}}......eq.1
{{{y=(5/3)x+1}}}......eq.2

as you can see, slopes are same, both lines have {{{m=(5/3)}}} and it means lines are parallel

{{{ graph( 600, 600, -10, 10, -10, 10, (5/3)x-5/3, (5/3)x+1) }}}