Find the slope of a line parallel to the line whose equation is 10, x, plus, 12, y, equals, minus, 12010x+12y=−120. Fully simplify your answer.

To find the slope of a line parallel to the line given by the equation:

\[
10x + 12y = -120
\]

we first need to express the equation in **slope-intercept form** \((y = mx + b)\), where \(m\) represents the slope.

### Step 1: Rearrange the equation into slope-intercept form
Start by isolating \(y\):

\[
10x + 12y = -120
\]

Subtract \(10x\) from both sides:

\[
12y = -10x - 120
\]

Now, divide everything by 12 to solve for \(y\):

\[
y = \frac{-10x - 120}{12}
\]

Simplify each term:

\[
y = \frac{-10}{12}x - \frac{120}{12}
\]

\[
y = \frac{-5}{6}x - 10
\]

### Step 2: Identify the slope
From the slope-intercept form \(y = \frac{-5}{6}x - 10\), we see that the slope \(m\) is:

\[
m = \frac{-5}{6}
\]

### Step 3: Parallel lines have the same slope
Since parallel lines have the same slope, the slope of the line parallel to this one is also:

\[
\boxed{\frac{-5}{6}}
\]

Would you like more details on how the slope formula works or any other part of the process?

Here are five related questions to explore:

1. What is the slope of a line perpendicular to \(10x + 12y = -120\)?
2. How do you convert a line from standard form to slope-intercept form in general?
3. What does the slope of a line represent geometrically?
4. How do you find the equation of a line given a point and a slope?
5. What is the y-intercept of the line \(10x + 12y = -120\)?

**Tip:** Parallel lines always have the same slope, while perpendicular lines have slopes that are negative reciprocals of each other.

Find the slope of a line parallel to the line whose equation is 10x + 12y = -120. Fully simplify your answer.

Learn how to find the slope of a line parallel to the given line 10x + 12y = -120. This problem explains the process of converting the line equation to slope-intercept form and identifying the slope. Ideal for students in Grades 8-10, covering key algebraic concepts and the parallel line theorem.

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