Question 165026
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Working eqn: POINT SLOPE FORM: {{{highlight(y=mx+b)}}}
1st eqn:{{{2x+3y=13}}}
simplifying,
{{{3y=13-2x}}}, divide both terms by 3
{{{y=(13-2x)/3}}}
{{{y=(13/3)-(2x/3)}}}
We see f(x)=0;
{{{y=13/2}}}, (0,13/3)
For x:
{{{2x+3y=13}}}
{{{2x=13-3y}}} divide both terms by 2
{{{x=(13-3y)/2}}}
{{{x=(13/2)-(3y/2)}}}
We see f(y)=0;
{{{x=13/2}}}, (13/2,0)
See graph
{{{drawing(300,300,-8,8,-8,8,grid(1),graph(300,300,-8,8,-8,8,13/3-2x/3), circle(0,13/3,0.25),circle(13/2,0,0.25))}}}-- see red line (13/2,13/3)
2nd eqn:
{{{4x-2y=2}}}
simplifying,
{{{2y=4x-2}}} divide both terms by 2
{{{2y/2=4x/2-2/2}}} --> {{{cross(2)y/cross(2)=cross(4)x/cross(2)-(cross(2)/cross(2))}}}
{{{y=2x-1}}}
We see f(x)=0;
{{{y=-1}}}, (0,-1)
We see f(y)=0;
{{{y=2x-1}}} ---> {{{y+1=2x}}} --> {{{y/2+1/2=cross(2)x/cross(2)}}}
{{{x=y/2+1/2=0/2+1/2}}}
{{{x=1/2}}}, (1/2,0)
See graph below,
{{{drawing(300,300,-5,5,-5,5,grid(1),graph(300,300,-5,5,-5,5,2x-1), circle(.5,0,.20),circle(0,-1,.20))}}} see line (1/2,-1)
Thank you,
jojo</pre>