Question 132471
Start with the given system

{{{y=5x+5}}}
{{{y=15x-1}}}




{{{15x-1=5x+5}}}  Plug in {{{y=15x-1}}} into the first equation. In other words, replace each {{{y}}} with {{{15x-1}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{15x=5x+5+1}}}Add 1 to both sides



{{{15x-5x=5+1}}} Subtract 5x from both sides



{{{10x=5+1}}} Combine like terms on the left side



{{{10x=6}}} Combine like terms on the right side



{{{x=(6)/(10)}}} Divide both sides by 10 to isolate x




{{{x=3/5}}} Reduce





Now that we know that {{{x=3/5}}}, we can plug this into {{{y=15x-1}}} to find {{{y}}}




{{{y=15(3/5)-1}}} Substitute {{{3/5}}} for each {{{x}}}



{{{y=8}}} Simplify



So our answer is {{{x=3/5}}} and {{{y=8}}} which also looks like *[Tex \LARGE \left(\frac{3}{5},8\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(\frac{3}{5},8\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 10, 5x+5, 15x-1) }}} Graph of {{{y=5x+5}}} (red) and {{{y=15x-1}}} (green)