Question 1147938
given:

{{{kx+2y=8}}}  
{{{18x+ky=12}}}

find: what positive value of {{{k}}} would make the lines below parallel?


recall that  parallel line have {{{same}}} {{{slope}}}

so, write both equations in slope intercept form

{{{kx+2y=8}}} ......solve for {{{y}}}

{{{2y=-kx+8}}} ...both sides divide by {{{2}}}

{{{y=-(k/2)x+4}}} => slope is {{{-(k/2)}}}


{{{18x+ky=12}}}.....take same steps as above

{{{ky=-18x+12}}}

{{{y=-(18/k)x+12/k}}} => slope is {{{-(18/k)}}}


if {{{slopes}}} are {{{same}}} , we have

{{{-(k/2)=-(18/k)}}}..........solve for {{{k}}}, first cross multiply

{{{-(k)k=-(18)2}}}

{{{-k^2=-36}}}......both sides multiply by {{{-1}}}

{{{k^2=36}}}

{{{k=sqrt(36)}}}

{{{highlight(k=6)}}}=>positive value of {{{k}}} that we need


and, your lines are

{{{6x+2y=8}}}  
{{{18x+6y=12}}}

or in slope intercept form

{{{y=-(6/2)x+4}}}=>{{{y=-3x+4}}}

{{{y=-(18/6)x+12/6}}} =>{{{y=-3x+2}}} 


proof with a graph:

{{{ graph( 600, 600, -10, 10, -10, 10, -3x+4, -3x+2) }}}


as you can see, the lines are parallel if {{{highlight(k=6)}}}