Question 129933
Start with the given system

{{{2x+y=28}}}
{{{y=5x}}}




{{{2x+5x=28}}}  Plug in {{{y=5x}}} into the first equation. In other words, replace each {{{y}}} with {{{5x}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{2x+5x=28}}} Distribute



{{{7x=28}}} Combine like terms on the left side



{{{x=(28)/(7)}}} Divide both sides by 7 to isolate x




{{{x=4}}} Divide





Now that we know that {{{x=4}}}, we can plug this into {{{y=5x}}} to find {{{y}}}




{{{y=5(4)}}} Substitute {{{4}}} for each {{{x}}}



{{{y=20}}} Simplify



So our answer is {{{x=4}}} and {{{y=20}}} which also looks like *[Tex \LARGE \left(4,20\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(4,20\right)]. So this verifies our answer.



{{{ drawing( 900, 900, -5, 5, -5, 25,
grid(1),
graph( 900, 900, -5, 5, -5, 25, (28-2x)/1, 5x) 
)}}} Graph of {{{2x+y=28}}} (red) and {{{y=5x}}} (green)