SAT Question 6e02cd78: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

The point of intersection of the graphs of the given equations is the solution to the system of the two equations. Since $y=(x-1{)}^2$ and $y=2x-3$, it follows that $(x-1{)}^2=2x-3$, or $(x-1)\times (x-1)=2x-3$. Applying the distributive property to the left-hand side of this equation yields $x^2-2x+1=2x-3$. Subtracting $2x$ from and adding $3$ to both sides of this equation yields $x^2-4x+4=0$. Factoring the left-hand side of this equation yields $(x-2)\times (x-2)=0$. By the zero product property, if $(x-2)\times (x-2)=0$, it follows that $x-2=0$. Adding $2$ to both sides of $x-2=0$ yields $x=2$. Substituting $2$ for $x$ in either of the given equations yields $y=1$. For example, substituting $2$ for $x$ in the second given equation yields $y=2\times 2-3$, or $y=1$. Therefore, the point of intersection of the graphs of the given equations is $(2,1)$. The $y$-coordinate of this point is $1$.