How many solutions are there to the following system of equations: $\newline$ $y=6-x$ $\newline$ $y=x^{2}-6x+6$

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Algebra 2

Composition of linear and quadratic functions: find an equation

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Q. How many solutions are there to the following system of equations:

\newline

y=6-x

\newline

y=x^{2}-6x+6

Set Equations Equal: Set the two equations equal to each other to find the points of intersection. $\newline$ We have the equations: $\newline$ $y = 6 - x$ $\newline$ $y = x^2 - 6x + 6$ $\newline$ Since both expressions are equal to $y$ , we can set them equal to each other to find the $x$ -values where the graphs intersect. $\newline$ $6 - x = x^2 - 6x + 6$
Rearrange and Solve: Rearrange the equation to set it to zero and solve for $x$ . To find the solutions, we need to rearrange the equation into a standard quadratic form. $0 = x^2 - 6x + 6 - (6 - x)$ $0 = x^2 - 6x + x + 6 - 6$ $0 = x^2 - 5x$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ We can factor out an $x$ from the equation. $\newline$ $0 = x(x - 5)$
Apply Zero Product Property: Solve for $x$ using the zero product property. $\newline$ The zero product property states that if a product of two factors is zero, then at least one of the factors must be zero. $\newline$ So we set each factor equal to zero and solve for $x$ : $\newline$ $x = 0$ or $x - 5 = 0$ $\newline$ $x = 0$ or $x = 5$
Verify Solutions: Verify the solutions in the original equations. $\newline$ We need to check if the $x$ -values we found, $x = 0$ and $x = 5$ , satisfy both original equations. $\newline$ For $x = 0$ : $\newline$ $y = 6 - 0 = 6$ $\newline$ $y = 0^2 - 6\times0 + 6 = 6$ $\newline$ Both equations give $y = 6$ , so $x = 0$ is a solution. $\newline$ For $x = 5$ : $\newline$ $y = 6 - 5 = 1$ $\newline$ $x = 0$ $0$ $\newline$ Both equations give $x = 0$ $1$ , so $x = 5$ is a solution.