Question 971836
Substitution Method should be easiest to use.

Put both into standard form or whichever choice of forms for each you want, first.


<b>1</b>
{{{kx-x-y=5}}}
or
{{{-y=5-(k-1)x}}}
{{{highlight_green(y=-5+(k-1)x)}}}


<b>2</b>
{{{(k+1)x+(1-k)y=3x+1}}}
{{{(k+1)x+(y-ky)=3x+1}}}
{{{(k+1)x-3x+y-ky=1}}}
SUBSTITUTE,
{{{(k+1)x-3x+((k-1)x-5)-k((k-1)x-5)=1}}}
{{{(k+1)x-3x+kx-x-5-(k^2*x-kx-5k)=1}}}
{{{(k+1)x-3x+kx-x-5-k^2*x+kx+5k=1}}}
{{{(k+1)x-3x+kx-x-k^2*x+kx=1-5k+5}}}
{{{x(k+1-3+k-1-k^2+k)=6-5k}}}
{{{x(-k^2+3k-3)=6-5k}}}
{{{x(k^2-3k+3)=5k-6}}}, which was result of multiplication of negative 1 to both sides
{{{highlight(x=(5k-6)/(k^2-3k+3))}}}---------For ONE OF the variables.


Go back to y in terms of x and k, and substitute the formula for x, and simplify for the formula for y.