Question 689019

1. 

{{{x+y=4}}}, 

{{{2x+y=5}}}......here you have system of two linear functions

*[invoke solve_by_graphing 1, 1, 4, 2, 1, 5]


2. 

{{{x-y=5}}} 

{{{2x+3y=0}}}


*[invoke solve_by_graphing 1, -1, 5, 2, 3, 0]


3. 

{{{x-y=-2}}} 

{{{x+y=4}}}

*[invoke solve_by_graphing 1, -1, -2, 1, 1, 4]


4. 

{{{y=x+3}}} (5,0)...here you have a point (x,y)= (5,0)

{{{0=5+3}}}

{{{0=8}}}.......(5,0) is not solution, point doesn't lie  on given line


{{{drawing(600,600,-10,10,-10,10,grid(0),circle(5,0,0.2),graph(600,600,-10,10,-10,10,x+3))}}}

5. 

{{{y=2x+3}}} (-4,1)

{{{1=2(-4)+3}}}

{{{1=-8+3}}}

{{{1=-5}}}.......point doesn't lie  on given line

{{{drawing(600,600,-10,10,-10,10,grid(0),circle(-4,1,0.2),graph(600,600,-10,10,-10,10,2x+3))}}}