Question 119687
{{{2x+y-4=0}}}
{{{3x+(5/2)y-10=0}}}
or changing the equations, you get, 
{{{2x+y=4}}}
{{{3x+(5/2)y=10}}}
or in matrix form
{{{(matrix(2,2,2,1,3,5/2))(matrix(2,1,x,y))=(matrix(2,1,4,10))}}}
According to Cramer's rule, then
{{{x=abs(matrix(2,2,4,1,10,5/2))/abs(matrix(2,2,2,1,3,5/2))}}}
{{{y=abs(matrix(2,2,2,4,3,10))/abs(matrix(2,2,2,1,3,5/2))}}}
{{{abs(matrix(2,2,4,1,10,5/2))=4(5/2)-1(10)=0}}}
{{{abs(matrix(2,2,2,4,3,10))=2(10)-4(3)=8}}}
{{{abs(matrix(2,2,2,1,3,5/2))=2(5/2)-3(1)=2}}}
{{{x=0/2=0}}}
{{{y=8/2=4}}}
Check your answers.
{{{2x+y-4=0}}}
{{{2(0)+4-4=0}}}
{{{0=0}}}
True statement.
{{{3x+(5/2)y-10=0}}}
{{{3(0)+(5/2)(4)-10=0}}}
{{{10-10=0}}}
{{{0=0}}}
True statement.
Good asnwers.