Question 744166
The problem {{{system(X^2+Y^2=25,3X-4X=0)}}} is easy.
The only solution to {{{3X-4X=0}}} is {{{X=0}}}
{{{3X-4X=0}}} --> {{{-X=0}}} --> {{{highlight(X=0)}}}
Substituting {{{X=0}}} into {{{X^2+Y^2=25}}} you get {{{Y^2=25}}},
which has two solutions: {{{highlight(Y=5)}}} and {{{highlight(Y=-5)}}}
The solutions to the system are the two pairs (0,5) and (0,-5), or
{{{highlight(system(X=0,Y=5))}}} and {{{highlight(system(X=0,Y=-5))}}}
 
NOTE:
If the problem had been {{{system(3X-4Y=0,X^2+Y^2=25)}}}, the system could be solved similarly, with a little more work.
The answer is not pretty, so that was not the intended problem.
(Classroom problems are supposed to have pretty answers).
In that case, solving for X to substitute in the other equation we get
{{{3X-4Y=0}}} --> {{{3X=4Y}}} --> {{{X=4Y/3}}} to substitute.
So you have
{{{system(X=4Y/3,X^2+Y^2=25)}}} --> {{{(4Y/3)^2+Y^2=25}}} --> {{{16Y^2/9+Y^2=25}}}
At this point, I would multiply both sides to the equal sign times 9 to get rid of fractions. I would get
{{{16Y^2+9Y^2=225}}} --> {{{16Y^2+9Y^2-225=0}}}
That equation yields two ugly solutions for Y:
{{{Y = (-9 +- sqrt( 9^2-4*16*(-225) ))/(2*16)=(-9 +- sqrt(9^2+14400))/32=(-9 +- sqrt(14481))/32=(-9 +- sqrt(9*1609))/32=(-9 +- 3sqrt(1609))/32}}} 
Each ugly Y can be substituted into {{{X=4Y/3}}} to get the corresponding ugly X values {{{X=(4/3)((-9 +- 3sqrt(1609))/32)=(-3 +- sqrt(1609))/8}}}
for a pair of ugly solutions.