Question 712086
You should be able to substitute y=kx+3 into xy+20=5y and solve for k.  That should help to find {{{kx^2+3(1-5k)x+5=0}}}.  Knowledge of the general solution to a quadratic equation suggests k must not be zero.


Carry out the steps you find {{{x=(-3(1-5k)-sqrt(225k^2-110k+9))/(2k)}}}
or {{{x=(-3(1-5k)+sqrt(225k^2-110k+9))/(2k)}}}


From that you may want to be sure to know how the discriminant should be greater than or equal to zero.  Solve {{{225k^2-110k+9>=0}}}.  Again use solution to quadratic equation:


{{{k=(11-2*sqrt(10))/45}}} or {{{k=(11+2*sqrt(10))/45}}}  But those values should be first checked as critical points.  The k quadratic should be greater than OR EQUAL TO ZERO.  There are three intervals of k to check the discriminant expression as {{{225k^2-110k+9>=0}}}.


Pushing ahead according to testing k in those intervals, the critical points in decimalized form would be close to k at 0.10389 and at 0.385.  I picked to check values 0, 0.2, and 0.4.  The results of the expression>=0 went like this:


at 0:  {{{9>=0}}}, yes.
at 0.2: {{{-4>=0}}}, no.
at 0.4: {{{1>=0}}}, yes.


Based on that, it seems k should be this:

{{{k<=(11-2*sqrt(10))/45}}}
OR
{{{k>=(11+2*sqrt(10))/45}}}


This was based on expecting complex values of k to cause lack of meaningful solutions to the original set of equations.