{:[44(j+2k)=12],[22 k=-11 j+16]:} Consider the system of equations. How many solutions (j,k) does this system have? Choose 1 answer: (A) 0 (B) Exactly 1 (C) Exactly 2 (D) Infinitely many

Analyze Equations: Let's analyze the system of equations: {:[44(j+2k)=12],[22k=-11j+16]:} We will first simplify the equations to make them easier to work with.Simplify First Equation: For the first equation, we divide both sides by 44 to isolate the term (j+2k): 44(j+2k) = 12 (j+2k) = 12/44 (j+2k) = 3/11Simplify Second Equation: For the second equation, we divide both sides by 22 to isolate k: 22k = -11j + 16 k = (-11j + 16)/22 k = -j/2 + 8/11Substitute k into First: Now we have the simplified system of equations: {:[j+2k=3/11],[k=-j/2+8/11]:} Next, we will substitute the expression for k from the second equation into the first equation.Constant Equation: Substituting k from the second equation into the first equation: j + 2(-j/2 + 8/11) = 3/11 j - j + 16/11 = 3/11 0j + 16/11 = 3/11No Solutions: We see that the j terms cancel out, leaving us with a constant equation: 16/11 = 3/11 This is a contradiction because 16/11 does not equal 3/11. Therefore, there are no solutions to the system of equations.

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{:[44(j+2k)=12],[22 k=-11 j+16]:}
Consider the system of equations. How many solutions
(j,k) does this system have?
Choose 1 answer:
(A) 0
(B) Exactly 1
(C) Exactly 2
(D) Infinitely many

$\begin{aligned} 44(j+2 k) & =12 \\ 22 k & =-11 j+16 \end{aligned}$ $\newline$ Consider the system of equations. How many solutions $(j, k)$ does this system have? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) Exactly $1$ $\newline$ (C) Exactly $2$ $\newline$ (D) Infinitely many

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\begin{aligned} 44(j+2 k) & =12 \\ 22 k & =-11 j+16 \end{aligned}

\newline

Consider the system of equations. How many solutions

(j, k)

does this system have?

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B) Exactly

1

\newline

2

\newline

(D) Infinitely many

Analyze Equations: Let's analyze the system of equations: $\newline$ {:\begin{align*}44(j+2k)&=12,\22k&=-11j+16\end{align*}:} We will first simplify the equations to make them easier to work with.
Simplify First Equation: For the first equation, we divide both sides by $44$ to isolate the term $(j+2k)$ : $\newline$ $44(j+2k) = 12$ $\newline$ $(j+2k) = \frac{12}{44}$ $\newline$ $(j+2k) = \frac{3}{11}$
Simplify Second Equation: For the second equation, we divide both sides by $22$ to isolate $k$ : $22k = -11j + 16$ $k = \frac{-11j + 16}{22}$ $k = -\frac{j}{2} + \frac{8}{11}$
Substitute $k$ into First: Now we have the simplified system of equations: $\newline$ \begin{cases}j+2k=\frac{3}{11}\k=-\frac{j}{2}+\frac{8}{11}\end{cases} $\newline$ Next, we will substitute the expression for $k$ from the second equation into the first equation.
Constant Equation: Substituting $k$ from the second equation into the first equation: $\newline$ $j + 2(-j/2 + 8/11) = 3/11$ $\newline$ $j - j + 16/11 = 3/11$ $\newline$ $0j + 16/11 = 3/11$
No Solutions: We see that the $j$ terms cancel out, leaving us with a constant equation: $\newline$ $\frac{16}{11} = \frac{3}{11}$ $\newline$ This is a contradiction because $\frac{16}{11}$ does not equal $\frac{3}{11}$ . Therefore, there are no solutions to the system of equations.