Question 1129302


the equality like {{{y=3x-2}}} is linear equality in {{{slope}}}-{{{intercept}}} form {{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept

so, in case above {{{m=3}}} is a slope and {{{b=-2}}} is y-intercept 

if you want to graph that line, you need two points (line is defined by two points)

since you already have that y-intercept (means {{{x=0}}} is {{{-2}}}, one point is

({{{0}}},{{{-2}}})

for second point you can choose to find x-intercept, when  {{{y=0}}}

{{{y=3x-2}}}=>{{{2=3x}}}=>{{{x=2/3}}}


({{{2/3}}},{{{0}}})


now, plot the points and draw a line through


{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(0,-2,.12),circle(2/3,0,.12),
locate(0,-2,p(0,-2)),locate(2/3,0,p(2/3,0)),
 graph( 600, 600, -10, 10, -10, 10, 3x-2)) }}}



Two other examples: 

{{{2x-3y=6}}} and {{{ 3x+4y=12}}} are written in a standard form {{{ax+by=c}}}

to graph them, you need two points
you can choose any value for {{{x}}} and calculate {{{y}}}, or you can
find {{{x}}} and {{{y}}}-intercepts,which is easier

you do same as above; set {{{x=0}}} to find y-intercept, 

{{{2*0-3y=6}}} ->{{{-3y=6}}} ->{{{y=6/-3}}} ->{{{y=-2}}} 


and set {{{y=0}}} to find x-intercept

{{{2x-3*0=6}}}->{{{2x=6}}}->{{{x=6/2}}}->{{{x=3}}}

so, points are:

({{{0}}}{{{-2}}}) -> y-intercept

({{{3}}}{{{0}}}) -> x-intercept


plot:

{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(0,-2,.12),circle(3,0,.12),
locate(0,-2,p(0,-2)),locate(3,0.4,p(3,0)),
 graph( 600, 600, -10, 10, -10, 10, (2/3)x-2)) }}}



and {{{ 3x+4y=12}}}

set {{{y=0}}} to find x-intercept 

{{{3x+4*0=12}}}->{{{3x=12}}}->{{{x=4}}}


set  {{{x=0}}} to find y-intercept
{{{ 3*0+4y=12}}}->{{{4y=12}}}->{{{y=3}}}


so, points are:

({{{4}}}{{{0}}}) -> x-intercept

({{{0}}}{{{3}}}) -> y-intercept

plot:

{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(0,3,.12),circle(4,0,.12),
locate(0,3,p(0,3)),locate(4,0,p(4,0)),
 graph( 600, 600, -10, 10, -10, 10, -3x/4+12/4)) }}}