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y=3x-15

y=x^(2)-2x-15
If
(x,y) is a solution to the system of equations and
x > 0, what is the value of
x ?
Choose 1 answer:
(A) 1
(B) 5
(C) 6
15

$y=3 x-15$ $\newline$ $y=x^{2}-2 x-15$ $\newline$ If $(x, y)$ is a solution to the system of equations and $x>0$ , what is the value of $x$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $5$ $\newline$ (C) $6$ $\newline$ $15$

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Complex conjugate theorem

Full solution

Q.

y=3 x-15

\newline

y=x^{2}-2 x-15

\newline

If

(x, y)

is a solution to the system of equations and

x>0

, what is the value of

x

?

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

5

\newline

(C)

6

\newline

15

Set Equations Equal: To find the value of $x$ , we need to set the two equations equal to each other because they both equal $y$ . This will allow us to solve for $x$ . So, we set $3x - 15$ equal to $x^2 - 2x - 15$ . $3x - 15 = x^2 - 2x - 15$
Rearrange and Solve: Next, we rearrange the equation to set it to zero and solve for $x$ . $x^2 - 2x - 15 - 3x + 15 = 0$ $x^2 - 5x = 0$
Factor Out $x$ : Now we factor out an $x$ from the equation. $x(x - 5) = 0$
Set Factors Equal: We set each factor equal to zero to find the possible values for $x$ . $x = 0$ or $x - 5 = 0$
Solve for x: Solving for x in the second equation gives us: $\newline$ $x - 5 = 0$ $\newline$ $x = 5$
Discard $x=0$ : Since we are given that $x > 0$ , we discard the solution $x = 0$ and only consider $x = 5$ .

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