How many solutions does the system have? {[6x-y=-1],[6x+y=-1]:} Choose 1 answer: (A) Exactly one solution (B) No solutions (C) Infinitely many solutions

Analyze Equations: To determine the number of solutions for the system of equations, we need to analyze the two given equations: \[ \begin{cases} 6x - y = -1 \\ 6x + y = -1 \end{cases} \] We can start by adding the two equations together to see if we can find a solution for $ x $ and $ y $. \[ (6x - y) + (6x + y) = -1 + (-1) \]Add Equations: When we add the two equations, the $ y $ terms cancel each other out: \[ 6x - y + 6x + y = -2 \] \[ 12x = -2 \] Now we can solve for $ x $ by dividing both sides by 12. \[ x = \frac{-2}{12} \] \[ x = -\frac{1}{6} \]Solve for x: Now that we have the value of $ x $, we can substitute $ x = -\frac{1}{6} $ into one of the original equations to solve for $ y $. Let's use the first equation: \[ 6x - y = -1 \] \[ 6(-\frac{1}{6}) - y = -1 \] \[ -1 - y = -1 \]Substitute x Value: When we simplify the left side, we get: \[ -1 - y = -1 \] Adding $ 1 $ to both sides to solve for $ y $ gives us: \[ -y = 0 \] \[ y = 0 \]Solve for y: We have found a specific solution for $ x $ and $ y $: $ x = -\frac{1}{6} $ and $ y = 0 $. This means that the system of equations has exactly one solution where the two lines intersect at a single point.

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How many solutions does the system have?

{[6x-y=-1],[6x+y=-1]:}
Choose 1 answer:
(A) Exactly one solution
(B) No solutions
(C) Infinitely many solutions

How many solutions does the system have? $\newline$ $\left\{\begin{array}{l} 6 x-y=-1 \\ 6 x+y=-1 \end{array}\right.$ $\newline$ Choose $1$ answer: $\newline$ (A) Exactly one solution $\newline$ (B) No solutions $\newline$ (C) Infinitely many solutions

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Algebra 2

Find the number of solutions to a system of equations

Full solution

Q. How many solutions does the system have?

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\left\{\begin{array}{l} 6 x-y=-1 \\ 6 x+y=-1 \end{array}\right.

\newline

Choose

1

answer:

\newline

(A) Exactly one solution

\newline

(B) No solutions

\newline

Analyze Equations: To determine the number of solutions for the system of equations, we need to analyze the two given equations: $\newline$ $\begin{cases} 6x - y = -1 \\ 6x + y = -1 \end{cases}$ $\newline$ We can start by adding the two equations together to see if we can find a solution for $x$ and $y$ . $\newline$ $(6x - y) + (6x + y) = -1 + (-1)$
Add Equations: When we add the two equations, the $y$ terms cancel each other out: $\newline$ $6x - y + 6x + y = -2$ $\newline$ $12x = -2$ $\newline$ Now we can solve for $x$ by dividing both sides by $12$ . $\newline$ $x = \frac{-2}{12}$ $\newline$ $x = -\frac{1}{6}$
Solve for x: Now that we have the value of $x$ , we can substitute $x = -\frac{1}{6}$ into one of the original equations to solve for $y$ . Let's use the first equation: $\newline$ $6x - y = -1$ $\newline$ $6(-\frac{1}{6}) - y = -1$ $\newline$ $-1 - y = -1$
Substitute x Value: When we simplify the left side, we get: $\newline$ $-1 - y = -1$ $\newline$ Adding $1$ to both sides to solve for $y$ gives us: $\newline$ $-y = 0$ $\newline$ $y = 0$
Solve for y: We have found a specific solution for $x$ and $y$ : $x = -\frac{1}{6}$ and $y = 0$ . This means that the system of equations has exactly one solution where the two lines intersect at a single point.