Question 201154
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We'll do #1 and you can take care of the rest. It's all the same sequence anyway.


1)Given{{{system(y = 2x -4(red(EQN1)),y = -3x+1(red(EQN2)))}}}


{{{red(EQN1)}}}, Let Fy=0
{{{0=2x-4}}}
{{{4=2x}}} ---> {{{cross(4)2/cross(2)=cross(2)x/cross(2)}}}}
{{{red(x=2)}}}, X-Intercept
Let Fx=0:
{{{y=2(0)-4}}}
{{{red(y=-4)}}}, Y-Intercept
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,2x-4),blue(circle(2,0,.12)),blue(circle(0,-4,.12)))}}}


{{{red(EQN2)}}}, Let Fy=0
{{{0=-3x+1}}}
{{{3x=1}}} ---> {{{cross(3)x/cross(3)=1/3}}}
{{{red(x=1/3)}}}, X-Intercept
Let Fx=0:
{{{y=-3(0)+1}}}
{{{red(y=1)}}}, Y-Intercept
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,-3x+1),blue(circle(1/3,0,.12)),blue(circle(0,1,.12)))}}}


Seeing the two lines, {{{red(EQN1)}}} & {{{red(EQN2)}}}:
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,2x-4,-3x+1),blue(circle(1/3,0,.12)),blue(circle(0,1,.12)),blue(circle(2,0,.12)),blue(circle(0,-4,.12)),green(circle(1,-2,.08)),green(circle(1,-2,.10)),green(circle(1,-2,.15)))}}} ----> Point of interection (1,-2)

*Note: we can calculate the point of Intersection by equating the two eqns.
{{{red(EQN1)=red(EQN2)}}}
{{{2x-4=-3x=1}}}
{{{2x+3x=4+1}}}
{{{5x=5}}} ---->{{{cross(5)x/cross(5)=cross(5)1/cross(5)1}}}
{{{red(x=1)}}}, substitute in either eqn. (we use {{{red(EQN1)}}})
{{{y=2(highlight(1))-4=2-4}}}
{{{red(y=-2)}}}
*Point of Intersection, (1,-2)


Thank you,
Jojo

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