Question 557102
{{{drawing(200,200,-5,5,-5,5,
line(-5,0,5,0),
line(-2,5,2,-5),locate(-1,1,1), locate(0,1,2),locate(0.5,-.2,3)
)}}} Angles 1 and 3 are vertical to each other; angle 2 is supplementary to 1 and to 3.
{{{2x+40=x+2y}}} (angles 1 and 3 have the same measure)
{{{(2x+40)+(2y+40)=180}}} (measures of angles 1 and 2 add to 180°)
{{{(2y+40)+(x+2y)=180}}} (measures of angles 2 and 3 add to 180°)
That gives you 3 equations on 2 variables.
Let's simplify:
{{{2x+40=x+2y}}} --> {{{x-2y=-40}}}
{{{(2x+40)+(2y+40)=180}}} --> {{{2x+2y+80=180}}} --> {{{2x+2y=100}}} --> {{{x+y=50}}}
{{{(2y+40)+(x+2y)=180}}} --> {{{x+4y+40=180}}} --> {{{x+4y=140}}}
With two of the equations we may find a solution, which we would need to substitute in the other equation to see that it verifies.
If we take the system
{{{x+y=50}}}
{{{x+4y=140}}}
subtracting the first equation from the second, we get
{{{3y=90}}} --> {{{y=30}}}
and substituting into the first equation we get
{{{x+30=50}}} --> {{{x=20}}}
If {{{x=20}}} with {{{y=30}}} satisfies {{{x-2y=-40}}}, then (x,y)=(20,30) is the solution. If not, there is no solution.
With {{{x=20}}} and {{{y=30}}}, {{{x-2y=20-2*30=20-60=-40}}}, so (20,30)
(or {{{x=20}}} with {{{y=30}}} ) is the solution.