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Solve the system of equations. $\newline$ $y = -40x + 28$ $\newline$ $y = x^2 - 33x + 34$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Solve a system of linear and quadratic equations

Full solution

Q. Solve the system of equations.

\newline

y = -40x + 28

\newline

y = x^2 - 33x + 34

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = -40x + 28$ $\newline$ $y = x^2 - 33x + 34$ $\newline$ To find the solution, we will set the two equations equal to each other since they both equal $y$ . $\newline$ $-40x + 28 = x^2 - 33x + 34$
Rearrange and Identify Quadratic: Now, we will rearrange the equation to set it to zero and identify the quadratic equation. $\newline$ $x^2 - 33x + 34 + 40x - 28 = 0$ $\newline$ $x^2 + 7x + 6 = 0$
Factor Quadratic Equation: Next, we will factor the quadratic equation. We are looking for two numbers that multiply to $6$ and add up to $7$ . $\newline$ $(x + 6)(x + 1) = 0$
Solve for x: Now, we will solve for $x$ by setting each factor equal to zero. $x + 6 = 0$ or $x + 1 = 0$ $x = -6$ or $x = -1$
Find y Values: We have found the x values, now we need to find the corresponding y values by substituting $x$ back into one of the original equations. Let's use $y = -40x + 28$ . For $x = -6$ : $y = -40(-6) + 28$ $y = 240 + 28$ $y = 268$
Substitute $x$ into Equation: For $x = -1$ : $\newline$ $y = -40(-1) + 28$ $\newline$ $y = 40 + 28$ $\newline$ $y = 68$
Write Coordinates: We have found the $y$ values corresponding to each $x$ value. Now we can write the coordinates in exact form. $\newline$ First Coordinate: $(-6, 268)$ $\newline$ Second Coordinate: $(-1, 68)$

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