Solve the system of equations. y = 3x - 5 y = x^2 - 15x + 40 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Substitute y Equation: Substitute y from the first equation into the second equation. Since y = 3x - 5, we can replace y in the second equation with 3x - 5. This gives us 3x - 5 = x^2 - 15x + 40.Rearrange and Solve: Rearrange the equation to set it to zero and solve for x. This means we will subtract 3x and add 5 to both sides, resulting in 0 = x^2 - 15x + 3x + 40 - 5, which simplifies to 0 = x^2 - 12x + 35.Factor Quadratic Equation: Factor the quadratic equation x^2 - 12x + 35. The factors of 35 that add up to -12 are -7 and -5. So, the factored form is (x - 7)(x - 5) = 0.Solve for x: Solve for x by setting each factor equal to zero. This gives us x - 7 = 0 or x - 5 = 0, which means x = 7 or x = 5.Substitute x = 7: Substitute x = 7 into the first equation y = 3x - 5 to find the corresponding value of y. This gives us y = 3(7) - 5, which simplifies to y = 21 - 5, resulting in y = 16.Substitute x = 5: Substitute x = 5 into the first equation y = 3x - 5 to find the corresponding value of y. This gives us y = 3(5) - 5, which simplifies to y = 15 - 5, resulting in y = 10.

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Solve the system of equations. $\newline$ $y = 3x - 5$ $\newline$ $y = x^2 - 15x + 40$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a nonlinear system of equations

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Q. Solve the system of equations.

\newline

y = 3x - 5

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y = x^2 - 15x + 40

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

\newline

(______,______)

\newline

(______,______)

Substitute $y$ Equation: Substitute $y$ from the first equation into the second equation. Since $y = 3x - 5$ , we can replace $y$ in the second equation with $3x - 5$ . This gives us $3x - 5 = x^2 - 15x + 40$ .
Rearrange and Solve: Rearrange the equation to set it to zero and solve for $x$ . This means we will subtract $3x$ and add $5$ to both sides, resulting in $0 = x^2 - 15x + 3x + 40 - 5$ , which simplifies to $0 = x^2 - 12x + 35$ .
Factor Quadratic Equation: Factor the quadratic equation $x^2 - 12x + 35$ . The factors of $35$ that add up to $-12$ are $-7$ and $-5$ . So, the factored form is $(x - 7)(x - 5) = 0$ .
Solve for x: Solve for x by setting each factor equal to zero. This gives us $x - 7 = 0$ or $x - 5 = 0$ , which means $x = 7$ or $x = 5$ .
Substitute $x = 7$ : Substitute $x = 7$ into the first equation $y = 3x - 5$ to find the corresponding value of $y$ . This gives us $y = 3(7) - 5$ , which simplifies to $y = 21 - 5$ , resulting in $y = 16$ .
Substitute $x = 5$ : Substitute $x = 5$ into the first equation $y = 3x - 5$ to find the corresponding value of $y$ . This gives us $y = 3(5) - 5$ , which simplifies to $y = 15 - 5$ , resulting in $y = 10$ .