Question 164983
solve by substitution:
{{{sqrt(x)-y=0}}}
{{{2x+y=3}}}
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{{{sqrt(x)-y=0}}}
{{{sqrt(x) = y}}}
square both sides
x = y^2
Sub into the 2nd eqn
{{{2x+y=3}}}
{{{2y^2+y=3}}}
{{{2y^2+y-3 = 0}}}
*[invoke solve_quadratic_equation 2,1,-3]
These are the values for y - the solver always uses x as the variable.
The solver has a problem sometimes.
It's (y-1)*(2y+3), the answers shown are correct.
So for y = 1, x = 1
For y = -3/2, x = 9/4