{:[-20 x+12 y=24],[-5x+3y=6]:} Consider the system of equations. How many (x,y) solutions does this system have? Choose 1 answer: (A) No solutions (B) Exactly one solution (C) Infinitely many solutions (D) None of the above

Examine the system of equations: First, let's examine the system of equations: {-20x + 12y = 24, -5x + 3y = 6} We will check if the second equation is a multiple of the first one.Check if second equation is a multiple: Divide the first equation by -4 to see if it matches the second equation: (-20x + 12y) / -4 = 24 / -4 5x - 3y = -6 This is the negative of the second equation, which is -5x + 3y = 6.Divide first equation by -4: Since the second equation is just the negative of the first equation after dividing by -4, this means that the two equations are actually the same line. Therefore, they have infinitely many solutions because every point on the line is a solution to both equations.

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Precalculus

Solve rational equations

Full solution

\begin{aligned} -20 x+12 y & =24 \\ -5 x+3 y & =6 \end{aligned}

\newline

Consider the system of equations. How many

(x, y)

solutions does this system have?

\newline

Choose

1

answer:

\newline

(A) No solutions

\newline

(B) Exactly one solution

\newline

\newline

(D) None of the above

Examine the system of equations: First, let's examine the system of equations: $\newline$ \begin{cases} -20x + 12y = 24,\ -5x + 3y = 6 \end{cases} $\newline$ We will check if the second equation is a multiple of the first one.
Check if second equation is a multiple: Divide the first equation by $-4$ to see if it matches the second equation: $\newline$ $\frac{(-20x + 12y)}{-4} = \frac{24}{-4}$ $\newline$ $5x - 3y = -6$ $\newline$ This is the negative of the second equation, which is $-5x + 3y = 6$ .
Divide first equation by ext{-} $4$ : Since the second equation is just the negative of the first equation after dividing by ext{-} $4$ , this means that the two equations are actually the same line. Therefore, they have infinitely many solutions because every point on the line is a solution to both equations.