System of Equations Question $y = x^{1/2} \\y = 2x-15$
I'm sorry. I'm confused on how there are two answers for this question ($x=9$ and $x= 25/4$), because when I graph them, the intersection I get is $x = 9$, so how do I show that $x = 9$ when doing the mathematical computation.
 A: When you solve the corresponding equation by squaring, you get also the point of intersection of $y=-\sqrt{x}$ with $y=2x- 15$:

So, by squaring you are solving
$$\left(\pm\sqrt x\right)^2 = (2x-15)^2$$
A: If the system is given in the form $$\begin{aligned} y=&x^{1/2}\\y=&2x-15,\end{aligned}$$ then it has only one solution $(x,y)=(9,3).$ 
Why?
Set $x=a^2$ and solve the system $$\begin{aligned} y=&a\\y=&2a^2-15\end{aligned}$$ 
which gives $\;a=3\;$ or $\;a=-\frac{5}{2},\;$ and then $x=9$ or $x=\frac{25}{4}.$ 
Note that $x=25/4\;$ gives $\;y=-5/2\;$ from the second equation, and consequently $\;x^{1/2}=-5/2\;$ in the first.
However, if $x, y$ are considered to be real, $x^{1/2}\geq 0$ by definition. 
Therefore, $x=25/4$ does not satisfy .

Taking into account the "official" answer $x=9$ or $x=25/4,$ I guess that the first equation of the system in its original form is $$y^2=x.$$ 
A: $$x^{1/2} = y \implies x = y^2$$
Then substitute into the second equation: 
$$y = 2x -15 $$
$$ x = (2x+15)^2$$
Its a little tricky to factorise and you may need to use the quadratic formula to obtain the two roots
A: There are two points of intersection between a parabola and straight line. The two answers are two roots of one quadratic equation in $x$ or $y.$
Square Root is eliminated by squaring and then eliminating $x$
$$ y^2=x;\,y=2x-5\,;\rightarrow 2 y^2-y-15=0$$
$$ (2y+5)(y-3)=0$$
$y$ roots are $(-\dfrac52,3)$ and corresponding  $x$ roots by substitution are their squares $(\dfrac{25}{4},9) $ as shown.

