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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Medium

In the xy-plane, what is the y-coordinate of the point of intersection of the graphs of $y equals, open parenthesis, x minus 1, close parenthesis, squared$ and $y equals, 2 x minus 3$ ?

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Explanation

The correct answer is 1. The point of intersection of the graphs of the given equations is the solution to the system of the two equations. Since $y equals, open parenthesis, x minus 1, close parenthesis, squared$ and $y equals, 2 x minus 3$ , it follows that $open parenthesis, x minus 1, close parenthesis, squared, equals, 2 x minus 3$ , or $open parenthesis, x minus 1, close parenthesis, times, open parenthesis, x minus 1, close parenthesis, equals, 2 x minus 3$ . Applying the distributive property to the left-hand side of this equation yields $x squared, minus 2 x, plus 1, equals, 2 x minus 3$ . Subtracting $2 x$ from and adding 3 to both sides of this equation yields $x squared, minus 4 x, plus 4, equals 0$ . Factoring the left-hand side of this equation yields $open parenthesis, x minus 2, close parenthesis, times, open parenthesis, x minus 2, close parenthesis, equals 0$ . By the zero product property, if $open parenthesis, x minus 2, close parenthesis, times, open parenthesis, x minus 2, close parenthesis, equals 0$ , it follows that $x minus 2, equals 0$ . Adding 2 to both sides of $x minus 2, equals 0$ yields $x equals 2$ . Substituting 2 for x in either of the given equations yields $y equals 1$ . For example, substituting 2 for x in the second given equation yields $y equals, 2 times 2, minus 3$ , or $y equals 1$ . Therefore, the point of intersection of the graphs of the given equations is $the point with coordinates 2 comma 1$ . The y-coordinate of this point is 1.