Question 720757
 {{{x = 2y}}}, {{{y = x + 2}}}, {{{y = 2x}}} and {{{y = x/ 2}}}.....these are all equations of a line because you have {{{x^1}}} and {{{y^1}}}

to plot them  on a graph, all you need is to find two points for each and draw a line through 

easiest way is to find {{{x}}} and {{{y-intercept}}}

1.

{{{x = 2y}}}....to find {{{x-intercept}}} plug in {{{y=0}}}

{{{x = 2*0}}}

{{{x = 0}}}....... {{{x-intercept}}} is at ({{{0}}},{{{0}}})

since {{{y-intercept}}} will be also at {{{0}}},{{{0}}}), we will find a point for {{{x=1}}}

{{{1= 2y}}}

{{{1/2 =y}}}....... the point is at ({{{1}}},{{{1/2}}})

now plot points ({{{0}}},{{{0}}}) and ({{{1}}},{{{1/2}}}) on a Cartesian coordinate system and draw a line through 

{{{ graph( 600, 600, -10, 10, -10, 10, x/2) }}}

2.



{{{y = x + 2}}}....to find {{{x-intercept}}} plug in {{{y=0}}}

{{{0 = x + 2}}}

{{{x= -2}}}....... {{{x-intercept}}} is at ({{{-2}}},{{{0}}})


{{{y = x + 2}}}....to find {{{y-intercept}}} plug in {{{x=0}}}

{{{y = 0 + 2}}}

{{{y= 2}}}....... {{{x-intercept}}} is at ({{{0}}},{{{2}}})


{{{ graph( 600, 600, -10, 10, -10, 10, x+2) }}}

3.

{{{y = 2x}}}....to find {{{x-intercept}}} plug in {{{y=0}}}

{{{0= 2x}}}

{{{x = 0}}}....... {{{x-intercept}}} is at ({{{0}}},{{{0}}})

since {{{y-intercept}}} will be also at {{{0}}},{{{0}}}), we will find a point for {{{x=1}}}

{{{y= 2*1}}}

{{{y =2}}}....... the point is at ({{{1}}},{{{2}}})

now plot points ({{{0}}},{{{0}}}) and ({{{1}}},{{{2}}}) on a Cartesian coordinate system and draw a line through 

{{{ graph( 600, 600, -10, 10, -10, 10, 2x) }}}

4.

{{{y = x/ 2}}}...this is same as 1.  {{{x = 2y}}} when you solve it for {{{y}}}