Question

system of two linear equations in two variables is given below.
y=1/6x-3
y=-5-1/6x







Which of the following statements is true?
The given system has no solution.
The solution to the given system is (-6, -4).
The solution to the given system is (6, -6).
The given system has infinitely many solutions.

Answer 1

To determine the nature of the given system of equations, let's first analyze the two equations:

1. \( y = \frac{1}{6}x - 3 \)
2. \( y = -5 - \frac{1}{6}x \)

Next, we can start by rewriting the second equation to see if we can express it in a comparable form:

Starting with \( y = -5 - \frac{1}{6}x \):

This is equivalent to:
\[ y = -\frac{1}{6}x - 5 \]

Now we write both equations together:

1. \( y = \frac{1}{6}x - 3 \) (Equation 1)
2. \( y = -\frac{1}{6}x - 5 \) (Equation 2)

Next, we can set the two equations equal to each other to find the possible point of intersection:

\[
\frac{1}{6}x - 3 = -\frac{1}{6}x - 5
\]

To eliminate the fractions, we can multiply through by 6:

\[
x - 18 = -x - 30
\]

Now, add \( x \) to both sides:

\[
2x - 18 = -30
\]

Next, add 18 to both sides:

\[
2x = -12
\]

Now, divide by 2:

\[
x = -6
\]

Now that we found \( x = -6 \), we can substitute this value back into either equation to find the corresponding \( y \):

Using Equation 1:
\[
y = \frac{1}{6}(-6) - 3 = -1 - 3 = -4
\]

So, the solution to the system is:
\[
(-6, -4)
\]

Now let's evaluate the statements provided:

1. **The given system has no solution.** (False)
2. **The solution to the given system is (-6, -4).** (True)
3. **The solution to the given system is (6, -6).** (False)
4. **The given system has infinitely many solutions.** (False)

Thus, the true statement is: **The solution to the given system is (-6, -4).**