Question 167328
(3,1);{x+3y=6
      {4x-5y=7
Let's see thru Slope-Intercept Form: {{{y=mx+b}}}
1st eqn: {{{x+3y=6}}}
{{{3y=6-x}}} ---> {{{cross(3)y/cross(3)=(6-x)/3}}}
{{{y=(6/3)-(1/3)x}}}
{{{y=(-1/3)x+(1/2)}}} ---> {{{slope=-1/3}}}; y-intercept=1/2
f(y)=0:
{{{(1/3)x=1/2}}} ---> {{{cross(1/3)x/cross(1/3)=(1/2)/(1/3)}}}
{{{x=3/2}}}
Let's see the graph:
{{{drawing(300,300,-3,3,-3,3,grid(1),graph(300,300,-3,3,-3,3,(-1/3)x+(1/2)),circle(3/2,0,.2),circle(0,1/2,.20))}}} --> POINTS ARE NOT (3,1), 
Next, 2nd eqn: {{{4x-5y=7}}}
{{{-5y=-4x+7}}} ---> {{{cross(-5)y/cross(-5)=(-4x+7)/-5}}}
{{{y=(-4/-5)x+(7/-5)}}} --> {{{y=(4/5)x-(7/5)}}} --> {{{slope=4/5}}}, y-intercept=-7/5
f(y)=0:
{{{(4/5)x=7/5}}} ---> {{{cross(4/5)x/cross(4/5)=(7/5)/(4/5)}}}
{{{x=(7/5)(5/4)=35/20=7/4}}}
Let's see the graph:
{{{drawing(300,300,-3,3,-3,3,grid(1),graph(300,300,-3,3,-3,3,(4/5)x-(7/5)),circle(7/4,0,.20),circle(0,-7/5,.20))}}} --> POINTS ARE NOT (3,1)
So you see, two line eqn's don't have points(3,1)
The question lies, when substitute to both eqn's points (3,1), does it satisfy the condition. Let's see:
1st eqn: {{{x+3y=6}}}--> subst-->{{{system(x=3,y=1)}}}
{{{3+3(1)=6}}}, good!
2nd eqn: {{{4x-5y=7}}}
{{{4(3)-5(1)=12-5=7}}}, good!
Conclusion, POINTS (3,1) is a solution to both EQNs.
Thank you,
Jojo